(a) Let $\{U_j\}$ be a collection of open subsets of $X$ . Then for each $i$ , $\cap_j U_j\cap X_i=\cap_j(U_j\cap X_i)$ is open in $X_i$ if there are finitely many open sets, and $\cup_j U_j\cap X_i=\cup_j(U_j\cap X_i)$ is open in $X_i$ as well. Now we need to show that $X_i$ is a (closed) subspace of $X$ . From the definition of the coherent topology on $X$ it follows that the topology on $X_i$ is finer than the subspace topology on $X_i$ . We need to show that if $U_0$ is open in $X_i$ then there is some set $V\subseteq X$ such that $V\cap X_i=U_0$ and $V\cap X_n$ is open in $X_n$ for every $n$ . $X_n$ is a subspace of $X_{n+1}$ for every $n$ . For $k\ge 1$ let $U_k$ be a set open in $X_{i+k}$ such that $U_k\cap X_{i+k-1}=U_{k-1}$ . Note that for $k<k'$ , $U_{k'}\cap X_{i+k}=$ $U_{k'}\cap X_{i+k'-1}\cap . . .\cap X_{i+k}=U_k$ and for $k\ge k'$ , $U_{k'}\cap X_{i+k}=U_{k'}$ . Let $U=\cup_k U_k$ . Then, for $n\ge i$ , $U\cap X_n=$ $(\cup_k U_k)\cap X_n=$ $\cup_k(U_k\cap X_n)=$ $\cup_{k<n-i}U_k\cup_{k\ge n-i}U_{n-i}=U_{n-i}$ is open in $X_n$ . For $n<i$ , $U\cap X_n=$ $U\cap X_i\cap X_n=$ $U_0\cap X_n$ is open in $X_n$ . Therefore, the subspace topology is finer than the topology on $X_i$ , and, overall, we have that the topologies are equal. To show that $X_i$ is a closed subspace of $X$ we can take any point $x\in X-X_i$ and construct its neighborhood in $X$ disjoint from $X_i$ by taking the union of $X_n-X_i$ for all $n>i$ .(b) In the past we had the following fact: if a function is continuous on every space in a collection of subspaces covering the space and either all subspaces in the collection are open or all subspaces are closed and the collection is finite, then the function is continuous on the whole space. Here we have either the infinite union of closed subspaces or the infinite union of open subspaces $X_{n+1}-X_n$ union one closed subspace $X_1$ . If we knew that $X_1\subseteq U$ where $U$ is open in $X$ and there is some $n$ such that $U\subseteq X_n$ then we could apply the result stated above. However, it might be that every neighborhood of some point in $X_1$ intersects all spaces $X_n$ . For example, suppose $X_n=\{1, . . ., n\}$ and $U$ is open in $X_n$ iff for every $x\in U$ all $y>x$ also in $U$ . Note that every $X_n$ is closed in $X_{n+1}$ . Then $X$ is $\mathbb{Z}_+$ but there is no open set in $X$ such that it is contained in any $X_n$ . So we show the result directly.Let $V$ be open in $Y$ . Then for every $i\ge 1$ , $f^{-1}(V)\cap X_i=(f|X_i)^{-1}(V)$ which is open in $X_i$ . Therefore, $f^{-1}(V)$ is open in $X$ .(c) Let $A$ and $B$ be closed disjoint subsets of $X$ . Let $X_0=\emptyset$ (it is a closed subspace of $X_1$ ) and $f_0$ be defined on $A\cup B$ such that $f_0(A)=0$ and $f_0(B)=1$ . We assume that $f_0:A\cup B\to[0,1]$ . Note that both $A$ and $B$ are closed and open in $A\cup B$ and $f_0$ is continuous. For every $i\ge 1$ , given a continuous function $f_{i-1}:X_{i-1}\cup A\cup B\to[0,1]$ such that $f_{i-1}(A)=0$ and $f_{i-1}(B)=1$ , we extend it to a continuous function $f_i:X_i\cup A\cup B\to[0,1]$ which satisfies the same property: $f_i(A)=0$ and $f_i(B)=1$ . We cannot directly apply the Tietze Extension Theorem because $X_{i-1}\cup A\cup B$ may be not normal. However, $X_{i-1}$ is normal. So we can extend $f_{i-1}|X_{i-1}$ to a continuous function $g:X_i\to[0,1]$ as follows: first, we say that $g(A\cap X_i)=0$ and $g(B\cap X_i)=1$ , the function is still continuous ($A$ and $B$ are closed in $X$ , hence, their intersection with $X_i$ is closed in $X_i$ and the continuity follows by the pasting lemma of Section 18), then, we extend it from the closed subset $(A\cup B\cup X_{i-1})\cap X_i$ onto $X_i$ . Now, we define $f_i$ to be equal to $g$ on $X_i$ , $0$ on $A$ and $1$ on $B$ (it is still continuous by the pasting lemma). For every $x\in X-A-B$ there is the minimal $n_x$ such that $x\in X_{n_x}$ . We define $f(x)=f_{n_x}(x)$ . Note that $f_i(x)$ is not defined for $i<n_x$ and agrees with $f_{n_x}(x)$ for $i>n_x$ . For $x\in A$ we define $f(x)=0$ and for $x\in B$ we define $f(x)=1$ . Note, again, that this definition agrees with all $f_i$ . This implies that $f|X_i=f_i$ which is continuous, and by (b), $f$ is continuous. Since $f(A)=0$ and $f(B)=1$ , $f$ separates $A$ and $B$ . Hence, $X$ is normal.