Section 18: Continuous Functions

A continuous function $f:X\rightarrow Y$ (relative to the topologies on $X$ and $Y$ ) is a function such that the preimage (the inverse image) of every open set (or, equivalently, every basis or subbasis element) of $Y$ is open in $X$ .

A function $f$ is continuous at a point $x$ if for each neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$ .

Equivalent definitions

$f:X\rightarrow Y$ is continuous iff
$f$ is continuous at every point iff
$f(\overline{A})\subset\overline{f(A)}$ for every subset $A\subset X$ iff
$f^{-1}(B)$ is closed for every closed subset $B\subset Y$ .

Just for fun we prove each statement from each other.

If $V\subset Y$ is open and $f^{-1}(V)$ is nonempty, then
$2\Rightarrow1$ let $x\in f^{-1}(V)$ , then there is an open set $U\subset X$ such that $x\in U$ and $f(U)\subset V$ , hence, $x\in U\subset f^{-1}(V)$ .

$3\Rightarrow1$ $f(\overline{f^{-1}(Y-V)})\subset\overline{f(f^{-1}(Y-V))}$ $\subset\overline{Y-V}=Y-V$ , so that $\overline{X-f^{-1}(V)}=X-f^{-1}(V)$ .

$4\Rightarrow1$ $f^{-1}(V)=X-f^{-1}(Y-V)$ .
If $V\subset Y$ is a neighborhood of $f(x)$ , then
$1\Rightarrow2$ $f^{-1}(V)$ is a neighborhood of $x$ and $f(f^{-1}(V))\subset V$ .

$3\Rightarrow2$ $f(\overline{f^{-1}(Y-V)})\subset\overline{f(f^{-1}(Y-V))}$ $\subset\overline{Y-V}=Y-V$ , so that $x\in X-f^{-1}(Y-V)=X-\overline{f^{-1}(Y-V)}$ and $f(X-\overline{f^{-1}(Y-V)})\subset V$ .

$4\Rightarrow2$ $X-f^{-1}(Y-V)$ is a neighborhood of $x$ and $f(X-f^{-1}(Y-V))\subset V$ .
If $x\in\overline{A}$ and $V$ is a neighborhood of $f(x)$ , then
$1\Rightarrow3$ let $y\in f^{-1}(V)\cap A$ , hence, $f(y)\in V\cap f(A)$ .

$2\Rightarrow3$ there is an open set $U\subset X$ such that $x\in U$ and $f(U)\subset V$ , so let $y\in U\cap A$ , then, $f(y)\in V\cap f(A)$ .

$4\Rightarrow3$ $x\in X-f^{-1}(Y-V)$ , so let $y\in(X-f^{-1}(Y-V))\cap A$ , then $f(y)\in V\cap f(A)$ .
If $B\subset Y$ is closed, then
$1\Rightarrow4$ $f^{-1}(B)=X-f^{-1}(Y-B)$ .

$2\Rightarrow4$ if $x\notin f^{-1}(B)$ , then there is an open set $U\subset X$ such that $x\in U$ and $f(U)\subset Y-B$ , hence, $x\in U\subset X-f^{-1}(B)$ .

$3\Rightarrow4$ $f(\overline{f^{-1}(B)})\subset\overline{f(f^{-1}(B))}$ $\subset\overline{B}=B$ , hence, $\overline{f^{-1}(B)}=f^{-1}(B)$ .

Properties

Constructing continuous functions

Suppose $f:X\rightarrow Y$ is continuous.

$f|A$ from a subspace $A\subset X$ to $Y$ is continuous.
$f:X\rightarrow W$ is continuous if $W$ is a subspace of $Y$ containing $f(X)$ or $Y$ is a subspace of $W$ .
If $g:Y\rightarrow Z$ is also continuous, $g\circ f$ is continuous.
If $g:X\rightarrow Y$ is also continuous, and $Y$ is ordered, then $\min\{f,g\}$ is continuous.

Extending the domain

Local definition of continuity: $f$ is continuous iff $f|U_{\alpha}$ is continuous for each $\alpha$ where $\{U_{\alpha}\}$ is an arbitrary collection of open subsets of $X$ such that $\cup_{\alpha}U_{\alpha}=X$ .
The pasting lemma: $f$ is continuous iff $f|S_{i}$ is continuous for each $i$ where $\{S_{i}\}$ is a finite collection of closed subsets of $X$ such that $\cup_{i}S_{i}=X$ .
- If the collection is arbitrary but locally-finite then the result still holds.
Extending the domain to its closure: if $A\subset X$ , $f:A\rightarrow Y$ is continuous and $Y$ is Hausdorff then there is no more than one way to extend $f$ continuously on $\overline{A}$ .

Continuity and product spaces

$f:X\rightarrow Y_{1}\times...\times Y_{n}$ is continuous iff $f_{i}=\pi_{i}\circ f:X\rightarrow Y_{i}$ is continuous for $i=1,...,n$ ( $f_{i}$ are called coordinate functions of $f$ ).
Let $f:X\rightarrow Y$ and $g:V\rightarrow W$ be continuous, then $(f\times g)(x,v)=(f(x),g(v))$ is a continuous function from $X\times V$ to $Y\times W$ .
If $f:X\times Y\rightarrow Z$ is continuous then it is continuous in each variable separately, but not vice versa.

$f:X\times Y\rightarrow Z$ is said to be continuous in variable $x$ if for every $c\in Y$ , $f(x,c):X\rightarrow Z$ is continuous.

Homeomorphism

A homeomorphism of $X$ with $Y$ (both are topological spaces) is a bijective function such that $U$ is open in $X$ iff $f(U)$ is open in $Y$ iff both $f$ and $f^{-1}$ are continuous.

An imbedding of $X$ in $Y$ is a homeomorphism of $X$ with a subspace of $Y$ .

Subpages

Section 18: Continuous Functions

Section 18: Continuous Functions

Equivalent definitions

Properties

Constructing continuous functions

Extending the domain

Continuity and product spaces

Homeomorphism