dbFin

Important!

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Munkres, Section 12 Topological Spaces

No exercises.

Munkres, Section 13 Basis for a Topology

1 For every $x$ there is an open set $U_x$ such that $x\in U_x\subseteq A$ , therefore, $U=\cup_{x\in A}U_x$ is open and $A=\cup_{x\in A}\{x\}\subseteq U\subseteq A$ , i.e. $U=A$ .

2 Let us enumerate the topologies by columns, i.e. we give numbers 1-3 for the first column from top to bottom, 4-6 for the second column, and 7-9 for the third column. Let “>” be the partial relation between any two topologies that indicates that the topology on the left side is finer (larger) than the topology on the right side. The finer (the larger) topology is the one with the “dust” not the “gravel”. Here we list all maximal ordered subsets of the set of topologies of Figure 12.1 ordered by “>”: 1<2<6<9, 1<2<7<8<9, 1<3<4<6<9, 1<3<7<8<9, 1<5<9.

3 The proof is very similar to Example 3 of §12: the empty set and $X$ are in the collection because their complements are $X$ and the empty set, the complement of any union of open sets is the intersection of the countable complements of the sets, so it is countable as well, finally, the complement of the finite intersection of open sets is the union of the countable complements, so it is countable. Now $\mathbb{T}_\infty$ is the trivial topology if $X$ is finite, but if $X$ is infinite then it is not a topology, as, for example, we can partition $X$ into three disjoint sets $X=X_1\cup X_2\cup \{x\}$ such that the first two are infinite, and then $X_1$ and $X_2$ are open but their union is not.

4 (a) The empty set, $X$ , unions of open sets in the intersection and their finite intersections all lie in every topology in the collection, and therefore, in their intersection. The union does not have to be a topology. For example, the union of two topologies with bases consisting of half-open intervals, one $[)$ (the lower limit topology) and the other $(]$ (the upper limit topology), is not a topology, as, for example, the intersection of $[0,1)$ and $(-1,0]$ is not in the union.

(b) The unique largest contained in all the topologies in the collection is obviously the intersection of the topologies. The smallest topology containing all the topologies in the collection is the one generated by the union as the subbasis.

(c) In Figure 12.1 these are topologies 4 and 5. Their intersection is the largest topology contained in both, it is $\{\emptyset, X, \{a\}\}$ . Their union $\{\emptyset, X, \{a\}, \{a,b\}, \{b,c\}\}$ is a subbasis for the smallest topology containing both, which is $\{\emptyset, X, \{a\}, \{b\}, \{a,b\}, \{b,c\}\}$ .

5 Every topology containing the collection $\mathcal{A}$ must contain all unions of its sets, i.e. the topology generated by it. If the collection is a subbasis, then every topology containing it must contain all finite intersections of its sets, i.e. the basis generated by this subbasis.

6 For $[x,y)$ there is no open set in $\mathbb{R}_K$ containing $x$ that lies in $[x,y)$ . For $0\in(-1,1)-K$ there is no open set in $\mathbb{R}_l$ that lies in $(-1,1)-K$ .

7 Let ">" be the "being finer" relation on the set of topologies (as in the Exercise 2). Then 4>2>1>3, and 1>5, but 3 and 5 are not comparable. The relations are easy to see using lemma 13.3 (we need to check whether it is true that every open set of one topology containing a point contains an open set from the other topology such that it contains the point). To see that 3 and 5 are not comparable consider point $1$ and two open sets $\mathbb{R}-\{0\}\in\mathcal{T}_3$ and $(-\infty, 2)\in\mathcal{T}_5$ . Neither of the open sets contains an open set from the other topology that contains $1$ . (See the diagram on the right.)

8 (a) The standard topology is finer than the topology generated by $\mathcal{B}$ . To see that they are equivalent consider any set $U$ open under the standard topology. Take any point $x\in U$ . Since $U$ is open, and the set of all open intervals is a basis for the standard topology, there is an interval $(a,b)$ that contains $x$ and lies in $U$ . There are two rational points $s$ and $t$ such that $a<s<x<t<b$ . The interval $(s,t)$ contains $x$ , lies in $U$ and is a basis element in $\mathcal{B}$ .

(b) The lower limit topology is finer than the topology generated by $\mathcal{C}$ . Now, for point $\sqrt{2}\in[\sqrt{2}, 2)$ there is basis element in $\mathcal{C}$ that contains $\sqrt{2}$ and is a subset of $[\sqrt{2}, 2)$ .

Munkres, Section 14 The Order Topology

No exercises.

Munkres, Section 15 The Product Topology on XxY

No exercises.

Munkres, Section 16 The Subspace Topology

1 $B$ is open in $A\subseteq Y$ iff $B=A\cap C$ and $C$ is open in $Y\subseteq X$ iff $B=A\cap C$ and $C=Y\cap D$ and $D$ is open in $X$ iff $B=A\cap D$ and $D$ is open in $X$ iff $B$ is open in $A\subseteq X$ .

2 It is finer but not necessarily strictly finer. It is finer, because if we change the topology $\mathcal{T}$ to $\mathcal{T}'$ then all sets that were open are still open, and therefore their intersections with $Y$ are still open in $Y$ . It is not necessarily strictly finer as the new open sets in $\mathcal{T}'-\mathcal{T}$ may not produce new open sets in the subspace topology. For example, a one-point subset will have the same subspace topology regardless of the topology on $X$ .

3 $A$ , $B$ and $E$ are open in $Y$ , and $A$ and $E$ are open in $\mathbb{R}$ .

4 Let $U$ be an open set, and $x\in\pi_1(U)$ . Then there exists $y$ such that $x\times y\in U$ . Since $U$ is open, there is a basis set $A\times B$ in $U$ that contains $x\times y$ . Since it is a basis set, $A$ is open in $X$ . Moreover, $A\subseteq \pi_1(U)$ . Therefore, $\pi_1(U)$ is open. Similarly for $\pi_2(U)$ .

5 (a) Every basis set in $X\times Y$ is a basis set in $X'\times Y'$ .

(b) Yes. If $U$ is open in $X$ , $x\in X$ , $V$ is open in $Y$ , $y\in Y$ then $U\times V$ is open in $X\times Y$ and, therefore, open in $X'\times Y'$ . Therefore, there exists a basis set $A\times B$ in $X'\times Y'$ such that it is a subset of $U\times V$ and it contains $x\times y$ . Therefore, there are open sets $A\in\mathcal{T}'$ and $B\in\mathcal{U}'$ such that $x\in A\subseteq U$ and $y\in B\subseteq V$ . So, $U$ is open in $X'$ and $V$ is open in $Y'$ .

6 A collection of sets $(a,b)$ such that $a,b\in\mathbb{Q}$ is a basis for the standard topology on $\mathbb{R}$ (see Exercise 8(a) of §13).

7 No, for example, $\{0\}$ is a convex subset of the real line.

8 A basis for $\mathbb{R}_l\times\mathbb{R}$ is a collection of sets $[a,b)\times(c,d)$ . A basis for $\mathbb{R}_l\times\mathbb{R}_l$ is a collection of sets $[a,b)\times[ c, d )$ . In both cases the intersection of a basis set and the line is either $[)$ or $()$ or $(]$ or $[]$ , and vice versa. Now not all combinations of these intervals are possible in all cases. If the line is vertical, then the respective topologies are $\mathbb{R}$ and $\mathbb{R}_l$ or $\mathbb{R}_u$ (depending on the direction of the line). If it is horizontal, then in both cases the topology is either $\mathbb{R}_l$ or $\mathbb{R}_u$ . If it has some slope then the first topology may generate either ( $[)$ and $()$ ) or ( $(]$ and $()$ ), depending on the direction of the line. That is, it is either the lower limit or the upper limit topology. For the second topology, if the line is downward sloping, we have $[]$ , $[)$ , $(]$ and $()$ , i.e. the discrete topology. Finally, if the line is upward sloping then the second topology generates a basis with either $[)$ or $(]$ , i.e. either $\mathbb{R}_l$ or $\mathbb{R}_u$ .

Direction	$\mathbb{R}_l\times\mathbb{R}$	$\mathbb{R}_l\times\mathbb{R}_l$
$\uparrow$	$\mathbb{R}$	$\mathbb{R}_l$
$\nearrow$	$\mathbb{R}_l$	$\mathbb{R}_l$
$\rightarrow$	$\mathbb{R}_l$	$\mathbb{R}_l$
$\searrow$	$\mathbb{R}_l$	$\mathbb{R}_d$
$\downarrow$	$\mathbb{R}$	$\mathbb{R}_u$
$\swarrow$	$\mathbb{R}_u$	$\mathbb{R}_u$
$\leftarrow$	$\mathbb{R}_u$	$\mathbb{R}_u$
$\nwarrow$	$\mathbb{R}_u$	$\mathbb{R}_d$

9 Every interval in the dictionary order topology on $\mathbb{R}\times\mathbb{R}$ is a union of open sets in $\mathbb{R}_d\times\mathbb{R}$ . Therefore, the later is finer than the former. A basis for the latter topology is a collection of sets $\{x\}\times(a,b)$ which is an interval in the dictionary order topology. The topology is strictly finer then $\mathbb{R}^2$ — see exercise 5(a): for example, $\{0\}\times\mathbb{R}$ is open in it, but not in $\mathbb{R}^2$ .

10 The first two topologies are not comparable. Indeed, $(0,1)\in[0,1]\times(0.5,1]$ has no open neighborhood in the second topology, and $(0,0.5)\in\{0\}\times(0,1)$ has no open neighborhood in the first topology. The third topology is strictly finer than the first and the second one. Indeed, by 9, the third topology is generated by sets $\{x\}\times((a,b)\cap[0,1])$ : every basis element $((a,b)\cap[0,1])\times((c,d)\cap[0,1])$ of the first topology can be generated as the union of open sets in the third topology, and every basis set $(a,b)<(x,y)<(c,d)$ in the second topology can be generated as the union as well. The fact that the third topology is strictly finer follows from the fact that the first and the second topologies are not comparable.

Munkres, Section 17 Closed Sets and Limit Points

1 The proof is similar to the Theorem 17.1, just the other direction.

2 If $Y$ is closed then we have: $A$ is closed in $Y$ iff $A=Y\cap B$ and $B$ is closed in $X$ iff $A\subseteq Y$ and $A$ is closed in $X$ .

3 $A\times B=X\times Y - ((X-A)\times Y \cup X\times (Y-B))$ is closed.

4 $U-A=U\cap (X-A)$ is open in $X$ . $A-U=A\cap (X-U)$ is closed in $X$ .

5 $[a,b]=X-((-\infty,a)\cup(b,+\infty))$ is closed and contains $(a,b)$ , so it contains the closure of $(a,b)$ . It equals the closure if both endpoints are limiting, i.e. if $(a,b)$ is not empty and for every $x\in(a,b)$ there are $s,t\in(a,b)$ such that $a<s<x<t<b$ . This is equivalent to the requirement that $a$ has no immediate successor, and $b$ has no immediate predecessor (this requirement is sufficient as discussed above, and it is also necessary as otherwise if, for example, $a$ has immediate successor $c$ then $(-\infty,c)$ is an open set containing $a$ that does not intersect $(a,b)$ ).

6 (a) $\={B}$ is closed and contains $A$ , therefore, it contains the closure of $A$ .

(b) and (c) Every closed set containing the union contains each set, and therefore, its closure. The intersection of such closed sets (the left side) contains all closures of all sets (the right side). Now, the other direction.
For any finite number of sets $\{A_i\}$ : $\overline{\cup_i A_i}=$ $\cap_{E\supseteq (\cup_i A_i):X-E\in\mathcal{T}}E=$ $\cap_{C_i\supseteq A_i:X-C_i\in\mathcal{T},i=1,2, . . .}(\cup_i C_i)=$ $\cup_i\cap_{C_i\supseteq A_i:X-C_i\in\mathcal{T}}C_i=$ $\cup_i\overline{A_i}$ (in fact, this is the proof for both directions in the case of a finite number of sets).
For infinite number of sets: some collections of closed sets $C_i$ have non-closed unions so that the intersection over all closed $E$ is the intersection over smaller collection of sets. This suggests that we may have the right side as a strict subset of the left side if there is a collection of closed sets of $A_i$ such that their union is not closed and there is no closed set that is contained in the union and contains all the sets. For example, if we take $A_n=\{\frac{1}{n}\}$ and $A=\cup_n A_n$ then $\={A}=A\cup\{0\}\supset\cup_n\={A}_n=A$ .

7 The problem of the proof is that for different $U$ ’s the set $A_\alpha$ may be different so that we cannot prove that $x$ lies in the closure of some particular $A_\alpha$ (consider the example of the previous exercise).

8 (a) and (b) $\subseteq$ in both cases. Using 6(a), $\overline{\cap_\alpha A_\alpha}\subseteq \overline{A_\alpha}$ for every $\alpha$ , and therefore, the left hand side is a subset of the right hand side. They do not equal in general, as there might be some common limiting or interior/limiting points which are not in the intersection and not limiting points of the intersection. For example, the intersection can be empty, while there is a common limiting point: $(0,1)$ and $(1,2)$ is an example.

(c) Using 6(a,b): $\={A}-\={B}=(\overline{A-B}\cup\overline{A\cap B})-\={B}=$ $\overline{A-B}-\={B}$ . This implies $\={A}-\={B}\subseteq\overline{A-B}$ .
Alternatively, assume that $x$ is in $\={A}-\={B}$ , i.e. every neighborhood of $x$ intersects $A$ but there is some neighborhood $V$ that does not intersect $B$ . Suppose there is some neighborhood $U$ that does not intersects $A-B$ , then the neighborhood $U\cap V$ does not intersects $A-B$ and it does not intersects $B$ , i.e. it does not intersects $A$ , contradiction. So, $x\in\overline{A-B}$ but $x\notin\={B}$ . The other direction is immediate using 6(a).
The inclusion is strict and the same example $A=(0,1)$ and $B=(1,2)$ demonstrates the fact: $\={A}-\={B}=\overline{A-B}-\={B}=[0,1)\subset\overline{A-B}=[0,1]$ .

9 $(x,y)\in\overline{A\times B}$ iff for every basis set $U\times V$ of $X\times Y$ , i.e. $U$ is open in $X$ and $V$ is open in $Y$ , such that $(x,y)\in U\times V$ the intersection $U\times V\cap A\times B\neq\emptyset$ iff for every neighborhood $U$ of $x$ and every neighborhood $V$ of $y$ : $U\cap A\neq\emptyset\neq V\cap B$ iff $x\in\={A}$ and $y\in\={B}$ iff $(x,y)\in\={A}\times\={B}$ .

10 If $x<y$ then if there is $c$ such that $x<c<y$ then neighborhoods $(-\infty, c)$ and $(c, +\infty)$ are disjoint, otherwise $(-\infty, y)$ and $(x, +\infty)$ are disjoint.

11 If two points have equal $x$ -coordinates, i.e. the points are $(x,y)$ and $(x,z)$ , then let $U$ be a neighborhood of $x$ , and the disjoint neighborhoods of the points are given by product of $U$ and disjoint neighborhoods of $y$ and $z$ . Similar in the case they have equal $y$ -coordinates. Otherwise, the disjoint neighborhoods are products of disjoint neighborhoods of the projections of the points onto $X$ and $Y$ .

12 The disjoint neighborhoods of two points in the subspace $Y$ are the intersections of $Y$ and two disjoint neighborhoods of the points in the space $X$ .

13 Cool! $\bigtriangleup$ is closed in $X\times X$ iff for every $x\neq y$ there are two open sets $U$ containing $x$ and $V$ containing $y$ such that for no point $(z,z)\in U\times V$ iff any pair of different points have disjoint neighborhoods.

14 For any point and any its neighborhood there is only finite number of points in the sequence that may be not in the neighborhood, so the set of “limits” is $\mathbb{R}$ .

15 $\forall x$ : $\{x\}$ is closed iff $\forall x$ : $X-\{x\}$ is open iff $\forall x\neq y$ : there is open $U$ such that $x\notin U$ and $y\in U$ .

16 (a) and (b) The closure is the union of the set and its limiting points. So we describe only limiting points.

Topology:	$\mathcal{T}_1=\mathbb{R}$	$\mathcal{T}_2=\mathbb{R}_K$	$\mathcal{T}_3=\mathbb{R}_{fc}$	$\mathcal{T}_4=\mathbb{R}_u$	$\mathcal{T}_5=\mathbb{R}_{-\infty}$
$K'$	$\{0\}$	$\emptyset$	$\mathbb{R}$	$\emptyset$	$\={\mathbb{R}}_+$
Hausdorff	+	+	-	+	-
$T_1$	+	+	+	+	-

17 In the lower limit topology $\={A}=[0,\sqrt{2})$ and $\={B}=[\sqrt{2},3)$ . In $\mathcal{C}$ $\={A}=[0,\sqrt{2}]$ and $\={B}=[\sqrt{2}, 3)$ . The difference for the set $A$ is exactly the one described in the solution for Exercise 8 of §13: set $[\sqrt{2},2)$ is open in the lower limit topology, but not open in $\mathcal{C}$ (every open set in $\mathcal{C}$ containing an irrational point has a rational point below it).

18 $A'=\{(0,1)\}$ as every neighborhood of $(0,1)$ is $(0,a)<(x,y)<(b,c)$ for some $b>0$ and contains a point $(1/n,0)$ for sufficiently large $n$ . $B'=(1,0)$ for similar reason. $C'=D'=(0,1]\times\{0\}\cup[0,1)\times\{1\}$ . $E'=\{0.5\}\times[0,1]$ .

19 (a) $x\in\={A}\Leftrightarrow\forall$ open $U:x\in U\Rightarrow$ $U\cap A\neq\emptyset\Leftrightarrow$ $\exists$ open $V:x\in V\subseteq A$ or $\forall$ open $U:x\in U\Rightarrow$ $U\cap A\neq\emptyset\neq U\cap(X-A)\Leftrightarrow$ $x\in int(A)\cup Bd(A)$ . Also, the two cases are disjoint, so that $int(A)\cap Bd(A)=\emptyset$ .

(b) $Bd(A)=\emptyset\Leftrightarrow$ $\forall x:\exists$ open $U$ s.t. $x\in U\subseteq A$ or $x\in U\subseteq (X-A)\Leftrightarrow$ $A,(X-A)\in\mathcal{T}$ .

(d) No, $U\subseteq\={U}$ is open, therefore, $U\subseteq Int(\={U})$ , but, for example, $\matbb{R}-\{0\}\subset Int(\overline{\matbb{R}-\{0\}})=\mathbb{R}$ .

Set	Bd	Int
A	A	$\emptyset$
B	$\{0\}\times\mathbb{R}$ $\cup$ $\mathbb{R}_+\times\{0\}$	B
C	$\mathbb{R}_-\times\{0\}$ $\cup$ $\{0\}\times\mathbb{R}$	$\mathbb{R}_+\times\mathbb{R}$
D	$\mathbb{R}\times\mathbb{R}$	$\emptyset$
E	$\{(x,y):\|x\|=\|y\|$ or $x^2-y^2=1\}$	$\{(x,y):0<x^2-y^2<1\}$
F	$\{(x,y):(x\neq 0$ and $y=1/x)$ or $x=0\}$	$\{(x,y):x\neq 0$ and $y<1/x\}$

21 We use the following notation here: we define four operators as follows, $iA=Int(A)$ , $cA=\={A}$ , $bA=Bd(A)$ , $xA=X-A$ . So that, for example, $xicA=X-Int(\={A})$ and $cA=iA\cup bA$ (19(a)). First, several statements:

$i$ and $c$ are monotone: if $A\subseteq B$ then $i(A)\subseteq i(B)$ and $c(A)\subseteq c(B)$ . Indeed, every open set contained in $A$ is contained in $B$ as well, and every closed set containing $B$ contains $A$ as well.
$ccA=cA$ , $xxA=A$ , $iiA=iA$ , but $bbA\neq bA$ in general. Indeed, $cA$ is a closed set containing $cA\Rightarrow$ $cA\subseteq ccA\subseteq cA$ . $xxA=X-(X-A)=A$ . $iA$ is an open set contained in $iA\Rightarrow$ $iA\subseteq iiA\subseteq iA$ . $bb(\mathbb{Q})=b(\mathbb{R})=\emptyset$ .
$bA=bxA$ and $X=iA\cup bA\cup ixA$ is a partion of $X$ (the sets are disjoint). This follows from the definition of the boundary, 19(a) and the fact that $X=A\cup xA\subseteq cA\cup cxA$ .
$iA=xcxA$ , $xcA=ixA$ , $cxA=xiA$ . Using 19(a) and properties 2 and 3, $xcxA=x(ixA\cup bxA)=x(xiA)=iA$ . Then, using this and property 2, $xcA=xcxxA=ixA$ , and $cxA=xxcxA=xiA$ .
$iA\subseteq icA$ , $ciA\subseteq cA$ . Indeed, $iA\subseteq cA$ and is open. $iA\subseteq cA$ and $cA$ is closed. Both can be strict, as, for example, for $A=\{0\}\cup[1,2)\cup(2,+\infty)$ .
$icA$ , $ciA$ and $A$ are, in general, not comparable by inclusion. Consider the following set: $A=[0,1]\cap\mathbb{Q}\cup(2,+\infty)$ . Then $icA=(0,1)\cup(2,+\infty)$ and $ciA=[2,+\infty)$ , and each of the following points belongs to one set only: $0$ , $1/\sqrt{2}$ and $2$ .
$icicA=icA$ , $ciciA=ciA$ . By properties 2 and 5, $cicA\subseteq ccA=cA$ , but $icA\subseteq cicA$ , therefore, $icA=iicA\subseteq icicA\subseteq icA$ . Now, using this and properties 2 and 4, $ciciA=$ $cicixxA=$ $xicicxA=$ $xicxA=$ $cixxA=$ $ciA$ .

The property 4 shows that even if we take $Int$ as an additional operation we are not going to get more sets. The property 2 shows that in a sequence of operations of closure and complementation we can substitute any two operations of the same type by one operation of the same type. Therefore, we may construct only two sequences of sets: starting with two initial sets $A$ and $xA$ we apply the sequence of operations $. . .xcxcxc$ . Now, properties 4 and 2 are especially useful for moving operators $x$ to the right (by switching it with another operator on the right: $xc=ix$ and $xi=cx$ ) and for canceling two $x$ 's in a row ( $xxA=A$ ). The first few members of the sequence starting with $A$ are $A$ , $cA$ , $ixA$ , $cixA$ , $icA$ , $cicA$ , $icixA$ , $cicixA$ , and the second sequence starts with $xA$ , $cxA$ , $iA$ , $ciA$ , $icxA$ , $cicxA$ , $iciA$ , $ciciA$ . Note that, by 7, in each sequence the 8th member equals the 4th member, therefore, there can be at most 7 distinct sets in each sequence. Note also, that each set starting from the second one is open or closed. Let $A$ be neither open nor closed. This ensures that no other set in the sequence equals $A$ or $xA$ . The following table splits the sets in the sequences in 4 different groups according to whether a set is a sequence of $i$ and $c$ starting at $A$ or $xA$ , and whether it is closed or open.

	$A$	$xA$
open	$iA, icA, iciA$	$ixA, icxA, icixA$
closed	$cA, ciA, cicA$	$cxA, cixA, cicxA$

The sets in different groups can easily be made different. For example, closed sets and open sets are different in the standard topology on the real line. Also, since interior is preserved, we can have two different points in $iA$ and $ixA$ such that each one will be in every set in one column but not in any set in the other column. So, the challenge is to try to make all sets within one group different. For example, to make $ciA$ and $cicA$ different $A$ must have a limiting point such that it is an interior point of $cA$ but not an interior point of $A$ . The set we used to illustrate property 6 is an example of how this can be done. But that example will not work as for that set $A$ : $cA=cicA$ . We also need to add some isolated points to the set. Now, the same should work for the complement $xA$ as well: to create an isolated point for $xA$ we need a "punctured" subset in $A$ . Here is an example:

	$A$	$xA$
	$\{-1\}\cup([0,1]\cap\mathbb{Q})\cup(1,2)\cup(2,+\infty)$	$(-\infty,-1)\cup(-1,0)\cup([0,1]-\mathbb{Q})\cup\{2\}$
$c$	$\{-1\}\cup[0,+\infty)$	$(-\infty,1]\cup\{2\}$
$xc$	$(-\infty,-1)\cup(-1,0)$	$(1,2)\cup(2,+\infty)$
$cxc$	$(-\infty,0]$	$[1,+\infty)$
$xcxc$	$(0,+\infty)$	$(-\infty,1)$
$cxcxc$	$[0,+\infty)$	$(-\infty,1]$
$xcxcxc$	$(-\infty,0)$	$(1,+\infty)$

In both cases the next closure equals the 4th member of the sequence. So, there are maximum of 14 sets that can be obtained from a given set by taking closure and complement (and interior, as the latter can be expressed in terms of the former two) operations.

Munkres, Section 18 Continuous Functions

1 Let $V$ be open and $f$ be continuous in the $\epsilon-\delta$ definition. Then, $x\in f^{-1}(V)$ implies $f(x)=y\in V$ and $\exists$ $\delta>0$ and $\epsilon>0$ s.t. $f((x-\epsilon,x+\epsilon))\subseteq(y-\delta,y+\delta)\subseteq V$ implies $\exists\epsilon:(x-\epsilon,x+\epsilon)\subseteq f^{-1}(V)$ . Therefore, $f^{-1}(V)$ is open.

2 No, for example, $f$ can be a constant function.

3 (a) Both mean that “any set open in $\mathcal{T}$ is open in $\mathcal{T}'$ ”. (b) Follows from (a).

4 Let $V=f(U)$ . Then $V$ is open in $X\times\{y_0\}$ iff $U\times\{y_0\}$ is open in $X\times\{y_0\}$ iff $U$ is open in $X$ . Similarly for $g$ .

5 In both cases $f(x)=(x-a)/(b-a)$ works as a homeomorphism.

6 For example, $f(x)=x$ if $x\in\mathbb{Q}$ or $0$ otherwise.

7 (a) If $V$ is open in $\mathbb{R}$ and $x\in f^{-1}(V)$ then $y=f(x)\in V$ and for some $\delta>0$ $(y-\delta,y+\delta)\subseteq V$ . Therefore, $\exists\epsilon>0$ s.t. $f([x,x+\epsilon))\subseteq(y-\delta,y+\delta)\subseteq V$ and $[x,x+\epsilon)\subseteq f^{-1}(V)$ . Thus, $f^{-1}(V)$ is open in $\mathbb{R}_l$ .

(b) From $\mathbb{R}$ to $\mathbb{R}_l$ only constant functions: for any $[a,b)$ the inverse image has to be both open and closed, and there are only two sets in $\mathbb{R}$ , namely the empty set and $\mathbb{R}$ , such that they are both open and closed. Now, from $\mathbb{R}_l$ to $\mathbb{R}_l$ . $f:\mathbb{R}_l\rightarrow\mathbb{R}_l$ is continuous iff for any $x$ and $\delta>0$ there exists $\epsilon>0$ such that $f([x,x+\epsilon))\subseteq[f(x),f(x)+\delta)$ . This can be proved exactly the same way as it was proved one direction in the text (Example 1), and the other direction in the Exercise 1. Now, what does this $\epsilon-\delta$ definition of continuity means... $f$ must be continuous from the right and non-decreasing.

8 (a) $\{x|f(x)>g(x)\}=\cup_{y\in Y}\{x|f(x)>y>g(x)\}\cup$ $\cup_{y<y'\in Y:(y,y')=\emptyset}\{x|f(x)>y,g(x)<y'\}$ is open.

(b) $h(x)=f(x)$ for $x$ such that $f(x)\le g(x)$ , and $h(x)=g(x)$ for $x$ such that $f(x)\ge g(x)$ , both sets are closed by (a), and $h$ restricted on them is continuous. Using the pasting theorem, $h$ is continuous.

9 (a) The pasting lemma applied several times.

(b) $f(x)=0$ on $[1/(n+1);1/n]$ , $f(x)=1$ on $(-\infty,0]$ are continuous, but $f$ on $(-\infty,1]$ is not. The reason is that the inverse image of a closed interval is a union of the closed inverse images, but if the union is not finite then it may not be closed, as for $f^{-1}(\{0\})$ in the example above.

(c) Let $B$ be a closed subset of $Y$ . Then $A=f^{-1}(B)=\cup_\alpha(f|A_\alpha)^{-1}(B)$ (this follows from the fact that $\cup_\alpha A_\alpha=X$ ). Suppose $x\notin A$ . There is a neighborhood $U$ of $x$ such that it intersects only a finite number of sets in the collection: $A_1, . . ., A_n$ . For each $i=1, . . ., n$ : $(f|A_i)^{-1}(B)=S_i$ is closed in A_i and, therefore, closed in $X$ (as $A_i$ is closed). Moreover, $x\notin S_i$ . Hence, there is a neighborhood $U_i$ of $x$ such that $U_i\cap S_i=\emptyset$ . The intersection $U'=U\cap\cap_i U_i$ is a neighborhood of $x$ such that $U'\cap A=\emptyset$ . We conclude that $A$ is closed.

10 The inverse image of any open set is the union of the inverse images of basis elements contained in the set, each of them as a product of two open sets has the inverse image equal to the product of the inverse images of the two open sets, and since $f$ and $g$ are continuous, the inverse images of the two open sets are open and their product is open.

11 $x\in h^{-1}(A)$ , $A$ is open in $Z$ implies $F(x,y_0)\in A$ and there is a basis neighborhood $U\times V$ in $X\times Y$ containing $x\times y_0$ such that it lies in $F^{-1}(A)$ . It follows that $h(U)\subseteq A$ , and, therefore, $h^{-1}(A)$ is open.

12 (a) It is symmetric, so we check for $x$ only. If $y_0=0$ then $F(x,y_0)=0$ and continuous, otherwise it is given by the expression and is continuous as well. (b) $g(x)=0.5$ if $x\neq 0$ and $0$ for $x=0$ . (c) $F^{-1}(\{0.5\})=\{(x,x)|x\neq 0\}$ is not closed.

13 Let $f(x)$ and $g(x)$ be two continuous functions on $\={A}$ such that they agree on $A$ . Let $C=\{x\in\={A}|f(x)=g(x)\}$ . In particular, $A\subseteq C$ . According to Theorem 18.4 $(f,g)$ is continuous, and $C=(f,g)^{-1}(\bigtriangleup)$ , where $\bigtriangleup=\{(y,y)|y\in Y\}$ . According to Exercise 17.13 $\bigtriangleup$ is closed in $Y$ iff $Y$ is Hausdorff. Since it is closed, $C$ is closed. Therefore, $\={A}\subseteq C\subseteq \={A}$ . The requirement “if it may be extended” is needed, of course, because not every continuous function can be extended onto the closure of the domain: for example, $1/x$ defined on $\mathbb{R}_+$ cannot be extended onto $\overline{\mathbb{R}}_+$ .

Munkres, Section 19 The Product Topology

1 Each of the product is in the corresponding topology, and every basis set (as it was defined in the text) of each topology is a union of basis sets as defined in Theorem 19.2.

2 $\sqcap_\alpha U_\alpha\cap\sqcap_\alpha A_\alpha=\sqcap_\alpha(U_\alpha\cap A_\alpha)$ so that for a given set every element in the set has a basis neighborhood in the topology of the product iff every element in the set has a basis neightborhood in the topology of the product as a subspace.

3 If two points in the product are different, then they have at least one different coordinate, we can separate their values in that space by two open sets and take the product of all other complete spaces and these neighborhoods — the result will be two disjoint subsets of the product open under either topology.

4 The homeomorphism is $f((x_1,…,x_{n-1}),x_n)=f(x_1,…,x_n)$ . Let $A_m$ equals the product of the first $m$ spaces. If a set is open in $A_n$ then every its element has a basis neighborhood in $A_n$ which can be represented as the product of open sets, and the product of the first $n-1$ of these sets is open in $A_{n-1}$ . Similarly, vice versa.

5 The first part of the theorem relies on the fact that the inverse image of the projection is open, which is true in both topologies. Therefore, the proof still holds, and if a function on the product space is continuous in either topology, then each component of the function is continuous.

6 Suppose that the sequence converges in the product topology. Let $U$ be an open neighborhood of $\pi_\alpha(x)$ in $X_\alpha$ . Then the product of $U$ and all other spaces gives an open neighborhood of $x$ in the product and all members of the sequence starting from some $N$ lies in the neighborhood. This works for the box topology as well. The other direction. Suppose, that the projections of a sequence converge. Take any neighborhood of $x$ , it contains a basis set containing $x$ . This basis set is a product of open sets in each space, and if we are in the product topology, only finite number of this sets are different from the corresponding space. For every open set in each space there is a number $N_\alpha$ such that starting from this number all projections of elements of the sequence into this space fall into the open set. For all open sets that equal the space we choose $N_\alpha=1$ . For the product topology, there is only finite number of $N$ ’s that are greater than $1$ , and we can take the maximum of all $N$ ’s. All elements of the sequence starting from the maximum of $N$ ’s fall into the open set in the product. For the box topology this direction may fail: the could be no greatest $N_\alpha$ . Some examples are in Exercise 4(b) of §20 (note that in all examples every projection converges to 0).

7 Using Exercise 6, in the product topology a sequence converges to a limiting point iff every projection converges to the projection of the limiting point. For $(x_1,x_2, . . . )$ in $\mathbb{R}^\omega$ define a sequence of points $y_k$ such that the first $k$ members are equivalent to $x$ and all others are zero. Then, clearly the projection onto any coordinate $n$ converges to $x_n$ , and, therefore, $\{y_n\}$ converges to $x$ , i.e. $x$ is limiting in the product topology, and $\overline{\mathbb{R}^\infty}=\mathbb{R}^\omega$ (see also Example 7 of §23). In the box topology, if a point $(x_1,x_2, . . . )$ is outside the $\mathbb{R}^\infty$ then there is a sequence $\{n_k\}$ such that $x_{n_k}\neq 0$ for all $k$ , and for each $k$ there is a neighborhood of $x_{n_k}$ that does not contain $0$ . The product of all such intervals and all other spaces is open in the box topology and does not intersect $\mathbb{R}^\infty$ . Therefore, in the box topology $\mathbb{R}^\omega-\mathbb{R}^\infty$ is open, and $\overline{\mathbb{R}^\infty}=\mathbb{R}^\infty$ .

8 Obviously, $h$ is bijective (as a product of bijective functions). Also note, that the reverse function has the same form with different coefficients. So, we need only to prove that $h$ is continuous. Any point $y=h(x)$ in an open set $U$ in the product has open basis neighborhood $B$ in the set, which is a product of open sets $B_n$ in $\mathbb{R}$ . The preimage $A_n=\{x_n|a_n x_n + b_n\in B_n\}$ of each such set is open in $R$ as well, and the preimage of $R$ is $R$ . Therefore, any point $x$ in $h^{-1}(U)$ has an open neighborhood in this set, and therefore, the set is open, $h$ is continuous. This works for both product and box topologies.

9 Given the definition of the product on page 113, the product is not empty if the choice axiom holds. Vice versa, if for any collection of sets the product is not empty, then any element in the product is the choice function.

10 (a) There exists at least one topology such that every function is continuous: namely, the discrete topology. The intersection of all such topologies is the coarsest topology such that every function is continuous in it. Another way to see this is as follows. Suppose there are two such topologies that are not comparable. Then in each topology there exists an open set which is not open in the other topology. Let us take an intersection of these two topologies. It is coarser then both topologies, and the inverse image $f_\alpha^{-1}$ of any open set in $X_\alpha$ is open, as it must be open in both topologies.

(b) Every set in $\mathcal{S}$ must be open in $\mathcal{T}$ . So, every topology such that every given function is continuous in it must contain this collection of sets, and their arbitrary unions of finite intersections. The coarsest such topology is generated by $\mathcal{S}$ as a subbasis.

(c) Let $h_\alpha=f_\alpha\circ g$ . Then if $g$ is continuous, every $h_\alpha$ is continuous. If each $h_\alpha$ is continuous, then for every $U$ open in $A$ , $U$ is the union of the finite intersection of some subbasis elements in $\mathcal{S}$ , and for each such subbasis set there is a function $f_\beta$ and an open set $U_\beta$ in $X_\beta$ such that this subbasis set equals to $f_\beta^{-1}(U_\beta)$ . Therefore, $g^{-1}(U)$ is the union of the finite intersections of $h_\beta^{-1}(U_\beta)$ and, therefore, it is open in $Y$ .

(d) Let $U\in\mathcal{T}$ and $\mathbf{x}\in f(U)$ . Take $a\in U$ such that $\mathbf{x}=f(a)$ . Since $U$ is open there is a basis element $V$ in $\mathcal{T}$ such that $a\in V\subseteq U$ and $V=\cap_{i=1, . . ., n} f_{\alpha_i}^{-1}(U_{\alpha_i})$ where $U_{\alpha_i}$ is open in $X_{\alpha_i}$ (all $\alpha$ ’s are without loss of generality assumed to be distinct). For all other $\alpha$ ’s assume $U_\alpha=X_\alpha$ . Then, $V=f^{-1}(\sqcap_\alpha U_\alpha)$ and $f(V)=f(A)\cap\sqcap_\alpha U_\alpha$ . This set is open in the subspace topology (as only finite number of $U$ ’s are distinct from their spaces) and $\mathbf{x}\in f(V)\subseteq f(U)$ . Therefore, $f(U)$ is open in the subspace topology.

Munkres, Section 20 The Metric Topology

1 (a) The basis elements are squares turned by 45 degrees (right angle “rhombuses”). It is a distance by inequality on page 122. The equivalence can be shown the same way as for the “squares” and “balls”. Every such rhombus is contained in the square with the same size. Every such rhombus of size $\epsilon$ contains a square with size $\epsilon/n$ . In other words, it follows from $\rho\le d'\le n\rho$ .

(b) It follows from $\rho\le d'\le n^{1/p}\rho$ .

2 Let $d((x,y),(a,b))=1$ if $x\neq a$ and $\={d}(y,b)$ otherwise. Any ball under this metric is either a vertical interval open in the dictionary order topology or the whole space. For any open set in the dictionary order topology and any its point there is an open vertical interval centered at the point and contained in the set. There is an open ball in this interval that is centered at the point.

3 (a) $d^{-1}((-\infty, a))=\cup_x B_d(x,a)$ and $d^{-1}((b, \infty))=\cup_x \{y:d(x,y)>b\}$ . While the first union of sets is obviously open, for any $x$ and $y$ in the second set $B_d(y,d(x,y)-b)$ is in the same set, so that it is open as well.

(b) If $d:X'\times X'\rightarrow\mathbb{R}$ is continuous, then for any fixed $x\in X'$ $d'(y):X'\rightarrow\mathbb{R}=d(x,y)$ is continuous (Exercise 11 of §18). Therefore, $B_d(x,r)=\{y|d'(y)<r\}=d'^{-1}((-\infty,r))$ must be open in the topology of $X'$ .

4 (a) The finer is the topology of the range, the “harder” it is for a function to be continuous. So, for each function we may specify the coarsest topology out of all three given such that it is not continuous in it. For $f$ it is the uniform topology, so that $f$ is continuous in the product topology only, and for $g$ and $h$ it is the box topology (they are continuous in both other topologies). $((-1,1),(-1/4,1/4), . . . ,(-1/k^2,1/k^2), . . .)$ is open in the box topology, but all three functions give the inverse image equal to $\{0\}$ . For the uniform topology the ball $B(x,\epsilon)$ has an open inverse image for $g$ and $h$ only. At the same time, as it is easy to see, all three are continuous in the product topology (for any basis element only finite number of open sets will define some range for the size of the ball centered at any point in the inverse image, or, alternatively, apply Theorem 19.6).

(b) The finer is the topology, the “harder” it is for a sequence to converge. If $w$ converges in a metric space, it converges to $(0,0, . . . )$ (any other point will give a non-zero distance to the members of the sequence starting from some index). In the product topology (metricized) it converges as $\={d}(w_k,0)/k=1/k$ , and, therefore, any ball centered at 0 has all elements of the sequence starting from some index. But in the uniform topology the distance always equals 1. So, $w$ converges in the product topology only. $x$ converges to $0$ in the uniform topology, but does not converge in the box topology, as, for example, there are no points of the sequence in $(-1/2,1/2)\times(-1/3,1/3)\times . . .$ (actually, this show that it does not converge to $0$ or any point in this open set, but it obviously does not converge to any other point as well). $y$ converges to $0$ in the uniform topology, but not in the box topology (the same set shows this). $z$ converges to $0$ in all three topologies, as it clearly converges in the finest one — the box topology.

5 The finer is the topology, the more closed sets it has, therefore, the closure under the finer topology is a subset of the closure under the coarser topology. Exercise 7 of §19 shows that the closure of $\mathbb{R}^\infty$ under the box topology is the set itself, and under the product topology is the whole space. So here the answer can be either one of those or anything in between. Let $X\in\mathbb{R}^\omega$ be the set of all sequences of real numbers that converge to $0$ . Note, that $\mathbb{R}^\infty\subseteq X$ . If a sequence $y$ of real numbers does not converge to 0 then there is a positive $\epsilon$ and an infinite subsequence such that each point in the subsequence is at least $\epsilon$ from $0$ . This implies that the uniform ball around $y$ of the size, for example, $\epsilon/2$ does not contain any point in $X$ . Therefore, $X$ is closed and all points outside $X$ are not in the closure of $\mathbb{R}^\infty$ . On the other hand, for any $x\in X$ and $\epsilon>0$ : $B_{\overline{\rho}}(x,\epsilon)$ has a point in $\mathbb{R}^\infty$ . So, the closure is the set of all sequences of real numbers converging to zero.

6 (a) $(x_1+\epsilon/2, x_2+2\epsilon/3, x_3+3\epsilon/4, . . .)$ is in $U$ but not in the ball. (b) A point in (a) has no ball centered around it that is in $U$ . (c) The distance between any point in $U(x,\delta)$ and $x$ is less than $\delta<\epsilon$ . So the RHS is in the LHS. Any point in the ball is such that the distance to $x$ is $\delta<\epsilon$ , and, therefore, it is in $U(x,(\epsilon+\delta)/2)$ .

7 $h^{-1}(y)=\sqcap(y_i/a_i-b_i/a_i)$ , so if we find the conditions on $a$ ’s and $b$ ’s such that $h$ maps open sets to open sets we can apply this conditions to $1/a$ ’s and $-b/a$ ’s to find the conditions under which $h$ is continuous. The combination of these two sets of conditions will give us the conditions on $h$ such that it is homeomorphism. Let $U$ be open, then it is a union of balls. $h(B_{\overline{\rho}}(x,\epsilon))=\cup_{\delta<\epsilon}h(U(x,\delta))=\cup_{\delta<\epsilon}U(ax+b,a\delta)=B_{\overline{\rho}}(ax+b,a\epsilon)$ . So, it is a homeomorphism. One could try to apply the following logic to prove that it is a continuous function. Since we have already shown (in the exercise 8 of §19) that it is a continuous mapping in both coarser product and finer box topologies, it must be continuous in the uniform topology. The problem here is that we change the topology for both the domain and the range. In this case if a function is continuous when both its domain and range are in a coarser topology and it is continuous when they both are in a finer topology we cannot conclude that it is continuous in the given topology. For example, $x+1, x<0$ and $x, x\ge 0$ is a continuous function when both the domain and the range are in the finite complement topology and, of course, in the discrete topology, but it not continuous in the standard topology.

8 (a) Let $M(x)=\{y:|x_i-y_i|^2<\epsilon^2/2^i\}$ . Then it is open in the box topology, and $M(x)\subset B_{l^2}(x,\epsilon)\subset B_{\overline{\rho}}(x,\epsilon)$ .

(b) The finer is the topology in the space, the finer it is in the subspace. So, given (a) and the fact that uniform topology is finer then the product topology, to show they are different in the subspace we need only to find an open set in the box topology of the subspace that is not open in the $l^2$ -topology, and so on. The set we used in 4(b) which is an infinite product of intervals is open in the box topology, and its intersection $A$ with $\mathbb{R}^\infty$ is open in the subspace. $(0,0, . . .)\in A$ , but for all $\epsilon>0$ , $B_{l^2}(0,\epsilon)\cap\mathbb{R}^\infty$ has points not in the intersection. Now, $B_{l^2}(0,\epsilon)$ is open in $l^2$ -topology, but for any $\delta>0$ there is a point $(\delta/2, . . . , \delta/2, 0, 0, . . .)$ such that it is in $\mathbb{R}^\infty$ and within the ball of size $\delta$ centered at 0 in the uniform topology, but not in $B_{l^2}(0,\epsilon)$ . Finally, for any ball $B_{\overline{\rho}}(0,\epsilon)$ , $0<\epsilon<1$ , there is a sequence $(\delta/2,\delta,3\delta/2, . . .,k\delta/2, 0, 0, . . .)$ such that it lies within $\mathbb{R}^\infty$ and $\delta$ -ball centered at 0 in the product topology, but not in $B_{\overline{\rho}}(0,\epsilon)$ (if $k$ was chosen sufficiently large).

(c) (Hilbert cube) The box topology is strictly finer than the $l^2$ -topology: $\sqcap[0,1/2^n)$ is open in it, and in any $B_{l^2}(0,\epsilon)$ there is a sequence $x_n=const\cdot\epsilon/n^2$ that is not in the set. The all three other topologies are equivalent on the cube. To show this we show that the product topology is finer than the $l^2$ topology. Intuitively, this happens because every unbounded set in the basis element of the product topology is now bounded by a sequence such that it converges in squares. For any $\epsilon>0$ consider the ball $B_{l^2}(x,\epsilon)$ . Let us take an open set in the product topology of the form $\sqcap_{i\le n}(x_i-\epsilon_i,x_i+\epsilon_i)\sqcap_{i>n}[0,1/i)$ . Then the square of the distance between $x$ and any point in the set in the $l^2$ topology is no greater than $\sum_{i\le n}\epsilon_i^2\sum_{i>n}(1/i^2)$ which can easily be made less than $\epsilon^2$ for sufficiently large $n$ and sufficiently small $\epsilon_i$ 's.

9 (a) Obviously. (b) $0\le \|ax+by\|^2=a^2\|x\|^2+b^2\|y\|^2+2abx\cdot y$ . For the opposite value of $b$ we get $a^2\|x\|^2+b^2\|y\|^2-2abx\cdot y\ge 0$ . Therefore, $|2abx\cdot y|\le a^2\|x\|^2+b^2\|y\|^2$ . Now take the values in the hint, and it is done. (c) $\|x+y\|^2\le \|x\|^2+\|y\|^2+2|x\cdot y|\le (\|x\|+\|y\|)^2$ and both are positive. (d) $d(x,y)\ge 0$ and is $0$ iff $x=y$ . The triangle inequality holds.

10 (a) Partial sums are bounded by the product of norms. (b) $cx$ is obvious, and $x+y$ follows from 9(c) (partial sums are bounded). (c) It is. The sum converges by (b), nonnegative, and zero iff $x=y$ , the triangle inequality holds by 9(c) (using for partial sums).

11 On $\={\mathbb{R}}_+$ it's well-defined, nonnegative, zero for same points only, symmetric, monotonically increasing and bounded. To show the triangle inequality we use the hint. First, $f(x+y)-f(y)$ is strictly decreasing in $y$ as $f(x)=1-1/(1+x)$ is strictly concave. Therefore, $f(a+b)-f(b)\le f(a)-f(0)=f(a)$ . $d'(x,y)+d'(y,z)\ge f(d(x,y)+d(y,z))\ge d'(x,z)$ . Since $f$ and $f^{-1}$ are continuous, $d'$ is continuous in $d$ and vice versa, which means the topologies are the same (Exercise 3(b)).

Munkres, Section 21 The Metric Topology (continued)

1 Let $B_A(x,r)\subseteq A$ be a ball centered at $x\in A$ given by the metric restricted on $A\times A$ . $A$ with the restricted metric has topology generated by $B_A(x,r)$ for all $x\in A$ . $A$ as a subspace has a basis element $C(x,r)=A\cap B(x,r)=\{y\in A|d(x,y)<r\}$ for all $x\in X$ . For every $x\in A$ : $B_A(x,r)=B(x,r)\cap A=C(x,r)$ . So the topology of $A$ generated by the restricted metric is coarser then the topology of $A$ as a subspace of $X$ . Consider any non-empty set $C(x,r)$ for $x\notin A$ . Let $y\in C(x,r)\subseteq A$ and $r'=r-d(x,y)$ . Then $B_A(y,r')\subseteq C(x,r)$ . Therefore, $C(x,r)$ is open under the restricted metric.

2 By the properties of metric it is injective (therefore, there exists bijective $f^{-1}$ from $f(X)$ onto $X$ ). The inverse image of any open ball in $f(X)$ as a subspace of $Y$ (which, by Exercise 1, has the restricted metric of $Y$ ) is an open ball in $X$ . Similar for $f^{-1}$ .

3 (a) It is well-defined as the product is finite. All properties including the triangle inequality are obviously satisfied (similar to the proof on page 122).

(b) $\={d}_i$ is a metric on $X_i$ (Theorem 20.1), and, therefore, $\={d}_i/i$ is a metric. It is well-defined and all the metric properties hold (similar to Theorem 20.5).

4 For $x\in\mathbb{R}_l$ : $[x,x+1/n)$ . For $x=(a,b)$ in the ordered square $[0,1]\times[0,1]$ : if $b\in(0,1)$ then $((a,b-\epsilon/n),(a,b+\epsilon/n))$ , if $b=0,a>0$ then $((a-\epsilon/n,1),(a,1/n))$ , if $b=0,a=0$ then $[(0,0),(0,1/n))$ , if $b=1,a<1$ then $((a,1-1/n),(a+\epsilon/n,0))$ , and, finally, if $b=1,a=1$ then $((1,1-1/n),(1,1)]$ .

5 Exercise 6 of §19 shows that $(x_n,y_n)\rightarrow(x,y)$ and we may now apply Theorem 21.3 and Lemma 21.4.

6 For every $x<1$ : $f_n(x)\rightarrow 0$ , but for any $\epsilon>0$ and points $x<1$ sufficiently close to $1$ : $f_n(x)>\epsilon$ (something like $(1-\delta)^n\ge 1-n\delta$ , which can be proved by induction, might help).

7 Once the notion of the uniform metric on $\mathbb{R}^X$ is clarified, the rest is an easy implication. By definition,
$\bar{\rho}(f,g)=\sup_{x\in X}\{\min\{|f(x)-g(x)|,1\}\}$ (we need the minimum to bound the metric so that it is well-defined).
Now, it is clear that if $f_n$ converges uniformly to $f$ then starting from some $N$ : $f_n\in B_{\bar{\rho}}(f,\epsilon)$ . And vice versa.

8 $f$ is continuous, by Theorem 21.6, therefore, there is an open neighborhood $U$ of $x$ such that $f(U)\subseteq B_{d_Y}(f(x),\epsilon/2)$ . Let $N$ be such that for $n>N$ : $x_n\in U$ and $d_Y(f_n(x),f(x))<\epsilon/2$ for all $x\in X$ . Then, by the triangle inequality, $d_Y(f_n(x_n),f(x))\le d_Y(f_n(x_n),f(x_n))+d_Y(f(x_n),f(x))<\epsilon$ for $n>N$ .

9 (a) It’s unimodal, the mode at $1/n$ , $f(0)\rightarrow 0$ , and for $x>0$ : $f(x)\rightarrow 0$ as well. (b) The value $f(1/n)=1$ .

10 The functions are continuous by Lemma 21.4 and as the composition of continuous functions. Then we may apply Theorem 18.1.

11 (a) It converges to the supremum of the set $\{s_n\}$ , as can be easily shown as the sequence is non-decreasing.

(b) For sufficiently large $N$ both partial sums are sufficiently close to their limits so that the linear combination of the partial sums sufficiently close to the linear combination of the limits.

(c) Both sequences of partial sums in the hint are non-decreasing and bound, so that, by (a), they converge, and by (b), their linear combination also converges.

(d) By (c), $s_n$ converges point-wise to some function $s$ . Now, for $k>n$ : $|s_k(x)-s_n(x)|\le \sum_{i=n+1 . . k}|f_i(x)|\le r_n$ . Therefore, $|s(x)-s_n(x)|\le r_n$ uniformly for all $x$ (if that was not true than for some sufficiently large $k$ the previous inequality would not hold as well). Now, note that for any $\epsilon>0$ we can find $N$ such that for $n>N$ : $r_n<\epsilon$ .

12 (a) If a sequence converges to $(x,y)$ then $x$ -coordinate converges to $x$ and $y$ -coordinate converges to $y$ and starting from some point they are within a ball of size $\delta=\epsilon/2$ of their limits. Using the hint, the sum is within a ball of size $\epsilon$ .

(b) Similar to (a) but taking another $\delta$ from the hint.

(c) The inverse image of $(a,b)$ is (we may consider finite intervals only): $(1/b,1/a)$ if they have the same sign, $(-\infty,1/a)\cup(1/b,+\infty)$ if they have different signs, $(1/b,+\infty)$ if $a=0$ , and $(-\infty,1/a)$ if $b=0$ .

(d) The operation of taking the opposite element is continuous. So, the composition of the sum with it is continuous. Also, the composition of the multiplication with the operation of taking reciprocals is continuous on $\mathbb{R}\times(\mathbb{R}-\{0\})$ .

Munkres, Section 22* The Quotient Topology

1 $\{a\}$ , $\{b\}$ and their union are open, $\{c\}$ and other proper subsets of $\{a,b,c\}$ are closed.

2 (a) We can reformulate as follows: if a continuous function has a continuous right inverse then it is a quotient map. $p$ has a right inverse, therefore, by exercise 5(a) of §2, it is surjective. Since it is also continuous, all we need to show is that it maps open saturated sets to open sets. Let $A=p^{-1}(B)$ be open. Now, $f^{-1}(A)=f^{-1}(p^{-1}(B))=B$ is open as $f$ is continuous.

(b) The inclusion map is a continuous right inverse for the retraction which is continuous by definition.

3 Instead of $\pi_1$ consider the following function: let $f:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\{0\}$ be the retraction $f(x,y)=(x,0)$ .Then it is a quotient map. There is an obvious homeomorphism $h$ between $\mathbb{R}\times\{0\}$ and $\mathbb{R}$ . Moreover, $\pi_1=h\circ f$ . Therefore, $\pi_1$ is a quotient map as well. Now, if $\pi_1$ were open or closed, so would be $f$ . But $f(\{(x,y)|y>0\})$ is not open and $f(\{(x,1/x)\})$ is not closed.

4 (a) $\mathbb{R}$ . Let $f(x\times y)=(x+y^2,0)$ . It is a retraction, therefore, a quotient map (see 2(b)). There is an obvious homeomorphism between $\mathbb{R}\times\{0\}$ and $X^*$ .

(b) $\mathbb{R}_+$ . Similar to (a).

5 $U$ open in $A$ , $A$ open, imply $U$ is open in $X$ , $p(U)\subseteq p(A)$ is open in $X$ .

6 (a) Remember, that set $K$ is closed in $\mathbb{R}_K$ (not in $\mathbb{R}$ ). Therefore, every one-point set in $Y$ is closed, and it is a $T_1$ -topological space. At the same time, there is no neighborhood of $0$ in $\mathbb{R}$ that does not contain points in $K$ . Therefore, any neighborhood of $0$ in $\mathbb{R}_K$ intersects any neighrborhood of some points in $K$ .

(b) (It is important here that $p$ is not an open quotient map.) $Y$ is not Hausdorff, therefore, the diagonal of $Y\times Y$ is not closed (every neighborhood of $(K,0)$ contains $(0,0)$ : §17, Hausdorff Spaces). At the same time, its preimage is the diagonal in $\mathbb{R}_K\times\mathbb{R}_K$ union $K\times K$ , and it is closed as $\mathbb{R}_K$ is Hausdorff (the diagonal is closed) and $K$ is closed in $\mathbb{R}_K$ .

Supplementary Exercises*: Topological Groups

1 If it is a topological group then $f(x,y)=x\cdot y^{-1}$ is continuous as a composition of continuous functions. If $f$ is continuous then it is continuous in both variables (§18), and, in particular, $f(1,y)$ and $f(x,f(1,y))$ are continuous.

2 (a) (b) (c) The spaces are Hausdorff, the operations are continuous (§21, exercise 12).

(d) It is Hausdorff (as a subspace). If $x=e^{i\phi}$ and $y=e^{i\theta}$ then $x\cdot y=e^{i(\phi+\theta)}$ . For a given open $U$ the set $\{(x,y)|x\cdot y^{-1}\in U\}$ is open (needs some work).

(e) It is Hausdorff (as a subspace). The operation of multiplication can be represented as a product of compositions of addition and multiplication. This should help.

3 Let $f(x,y)=x\cdot y^{-1}$ . $\={H}$ is a subgroup iff $f(\={H}\times\={H})\subseteq \={H}$ , which is true as $f(\={H}\times\={H})=f(\overline{H\times H})\subseteq\overline{f(H\times H)}\subseteq\={H}$ ( $f$ is continnuous and also Theorem 19.5). Now, if $H$ (or $\={H}$ ) is a subgroup, then the restriction of $f$ on $H\times H\rightarrow H$ is continuous.

4 They are bijective and continuous. Their inverses are continuous as well. For every $x$ and $y$ : $f_{y\cdot x^{-1}}(x)=y$ or $g_{x^{-1}\cdot y}(x)=y$ .

5 (a) $xH=yH$ iff $\alpha xH=\alpha yH$ , therefore, it is a bijection. Now, $(f_\alpha|H)(xH)=\alpha xH$ , therefore, $(f_\alpha|H)\circ p=p\circ f_\alpha$ where $p$ is the quotient map. Using Theorem 22.2, since $p\circ f_\alpha$ is a quotient (a composition of a homeomorphism and a quotient), it induces a homeomorphism $f_\alpha|H$ . Moreover, $(f_{y\cdot x^{-1}}|H)(xH)=yH$ .

(b) $H$ is closed, $f_\alpha$ is a homeomorphism, therefore, $xH=f_x(H)$ is closed in $G$ .

(d) Using (b), $G|H$ is a $T_1$ -space. Now, let $f(x,y)=x\cdot y^{-1}$ . Using exercise 1, we need only to show that $f|H$ is continuous. Similar to (a), using the fact that the subgroup is normal, $(f|H)(xH,yH)=f(x,y)H$ (indeed, for this to hold we need $f_\alpha(H)=g_\alpha(H)$ ), therefore, $(f|H)\circ (p\times p)=p\circ f(x,y)$ . $p$ and $f$ are continuous, moreover, $p$ is an open quotient so that $p\times p$ is an open quotient as well, and using Theorem 22.2, $p\circ f(x,y)$ induces a continuous function $f|H$ .

6 $[0,1)$ with addition by modulus. Or, just $(S^1,\cdot)$ .

7 (a) $x\cdot y$ and $x\cdot y^{-1}$ are continuous, therefore, there is a neighborhood $V'$ of $e$ such that $V'\cdot V'\subseteq U$ and $V$ such that $V\cdot V^{-1}\subseteq V'$ . Therefore, $(V\cdot V^{-1})\cdot(V\cdot V^{-1})\subseteq U$ .

(b) We need a symmetric neighborhood of $e$ such that for no $u^{-1},v\in V$ : $u^{-1}\cdot x=v\cdot y$ or $x\cdot y^{-1}=u\cdot v$ . In other words, we need a symmetric neighborhood $V$ of $e$ such that $V\cdot V$ does not contain $x\cdot y^{-1}$ . $G$ is a $T_1$ -space, therefore, $\{x\cdot y^{-1}\}$ is closed, and there is a neighborhood $U$ of $e$ such that it does not contain $\{x\cdot y^{-1}\}$ . Using (a), we construct $V$ .

(c) We need a symmetric neighborhood of $e$ such that for no $u^{-1},v\in V$ and $y\in A$ : $u^{-1}\cdot y=v\cdot x$ or $y\cdot x^{-1}=u\cdot v$ . In other words, $V\cdot V$ does not intersect $A\cdot x^{-1}$ , which is closed ( $g_{x^{-1}}$ is a homeomorphism). Let $U$ be a neighborhood of $e$ such that it does not intersect $A\cdot x^{-1}$ and $V$ be a symmetric neighborhood of $e$ such that $V\cdot V\subseteq U$ (see (a)).

(d) $H$ is closed, therefore, $G|H$ is a $T_1$ -space (5(b)) and $xH$ is closed. A closed subset of $G|H$ is an image of a closed saturated subset of $G$ . Let $A$ be a saturated closed subset of $G$ that does not intersect $xH$ . Using (c), let $V$ be a symmetric neighborhood of $e$ such that $V\cdot A$ does not intersect $V\cdot x$ . Then, since $A$ is saturated, $V\cdot A$ does not intersect $V\cdot xH$ . Suppose, $zH\in p(V\cdot A)\cap p(V\cdot xH)$ . Then for some elements $z=v\cdot a\cdot h$ , $z=v'\cdot x\cdot h'$ which contradicts $V\cdot A\cap V\cdot xH=\emptyset$ . Using 5(c), $p(V\cdot A)$ is an open neighborhood of $p(A)$ disjoint from $p(V\cdot xH)$ , an open neighborhood of $xH$ .

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bara

2012/08/31 at 9:09 am (UTC -5) Link to this comment


nice work
1. Vadim
  
  2012/10/13 at 3:37 pm (UTC -5) Link to this comment
  
  
  Thanks!
tesfalem

2012/11/28 at 12:52 am (UTC -5) Link to this comment


the solution for (a,b) is homeomorphic to (0,1) is not clear. so please help me mean the verification.
1. Vadim
  
  2012/11/28 at 1:02 am (UTC -5) Link to this comment
  
  
  That will be easier for me to find the problem you are asking about if you provide the section and problem number. Thank you.

Topology: product, uniform, and box topology converging sequences

2012/04/02 at 6:58 pm (UTC -5) Link to this comment


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Amazon Student Account

2000 Munkres
Topology: Solutions > Chapter 2 Topological Spaces and Continuous Functions

Munkres, Section 12 Topological Spaces

Munkres, Section 13 Basis for a Topology

Munkres, Section 14 The Order Topology

Munkres, Section 15 The Product Topology on XxY

Munkres, Section 16 The Subspace Topology

Munkres, Section 17 Closed Sets and Limit Points

Munkres, Section 18 Continuous Functions

Munkres, Section 19 The Product Topology

Munkres, Section 20 The Metric Topology

Munkres, Section 21 The Metric Topology (continued)

Munkres, Section 22* The Quotient Topology

Supplementary Exercises*: Topological Groups

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bara

Vadim

tesfalem

Vadim

Topology: product, uniform, and box topology converging sequences

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Buy Me Coffee

Smiling coffee

Cappuccino

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Like dbFin on Facebook

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TransRu

Amazon Student Account

2000 MunkresTopology: Solutions > Chapter 2 Topological Spaces and Continuous Functions

Munkres, Section 12 Topological Spaces

Munkres, Section 13 Basis for a Topology

Munkres, Section 14 The Order Topology

Munkres, Section 15 The Product Topology on XxY

Munkres, Section 16 The Subspace Topology

Munkres, Section 17 Closed Sets and Limit Points

Munkres, Section 18 Continuous Functions

Munkres, Section 19 The Product Topology

Munkres, Section 20 The Metric Topology

Munkres, Section 21 The Metric Topology (continued)

Munkres, Section 22* The Quotient Topology

Supplementary Exercises*: Topological Groups

4 comments

1 ping

bara

Vadim

tesfalem

Vadim

Topology: product, uniform, and box topology converging sequences

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Content

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2000 Munkres
Topology: Solutions > Chapter 2 Topological Spaces and Continuous Functions