\chapter{Continuity: When Nearby Points Stay Together}

\begin{thm}
\label{continuity-equivs}
Let $X$ and $Y$ be topological spaces and let $f: X\to Y$ be a function. Then the following
are equivalent:
\begin{enumerate}
\item The function $f$ is continuous.
\item For every closed set $K$ in $Y$, the inverse image $f^{-1}(K)$ is closed in $X$.
\item For every limit point $p$ of a set $A$ in $X$, the image $f(p)$ belongs to
$\overline{f(A)}$.
\item For every $x\in X$ and open set $V$ containing $f(x)$, there is
an open set $U$ containing $x$ such that $f(U) \subset V$.
\end{enumerate}
\end{thm}

\oneperpage

\begin{thm}
Let $X, Y$ be topological spaces and $y_{0} \in Y$.  
The constant map $f:X\rightarrow Y$ defined by $f(x)=y_{0}$ is continuous.
\end{thm}

\oneperpage

\begin{thm}
Let $X \subset Y$ be topological spaces.  The inclusion map $i:X \rightarrow Y$ defined by $i(x)=x$ is continuous.
\end{thm}

\oneperpage

\begin{thm}
Let $f:X\rightarrow Y$ be a continuous map between topological spaces, and let $A$ be a subset of $X$.  
Then the restriction map $f |_{A}: A \rightarrow Y$ defined by $f|_{A}(a)=f(a)$ is continuous.
\end{thm}

\oneperpage

\begin{thm}    
A function $f:\R_{std}\to \R_{std}$ is continuous
if and only if for every point $x$ in $\R$ and $\varep >0$, there is a
$\delta >0$ such that for every $y\in \R$  with $d(x,y) < \delta$, then 
$d(f(x),f(y)) <\varep$.
\end{thm}

\oneperpage

\begin{thm}    
Let $X$ be a $1^{\text{st}}$ countable space and $Y$ be a 
topological space. Then a function $f:X\to Y$ is continuous if and only
if for each convergent sequence $x_n\to x$ in $X$, $f(x_n)$ converges to $f(x)$ in $Y$.
  \begin{inst} %%%
     $1^{\text{st}}$ countability is used in the reverse direction.
  \end{inst}
\end{thm}

\oneperpage

\begin{thm}
Let $X$ be a space with a dense set $D$, and let $Y$ be Hausdorff.  Let $f:X \rightarrow Y$ and $g:X \rightarrow Y$ be continuous functions such that for every $d$ in $D$, $f(d)=g(d)$.  Then for all $x$ in $X$, $f(x)=g(x)$.
\end{thm}

\oneperpage

\begin{thm}
\label{continuous-dense}
The cardinality of the set of continuous functions from $\R$ to $\R$ is the same as the cardinality of $\R$.
    
\end{thm}

\oneperpage

\begin{thm}
If $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are continuous, then
their composition $g\circ f :X\rightarrow Z$ is continuous.
\end{thm}

\oneperpage

\begin{thm}[pasting lemma]
Let $X=A \cup B$, where $A,B$ are closed in $X$.  Let $f:A\rightarrow Y$
and $g:B\rightarrow Y$ be continuous functions that agree on $A\cap B$.
Then the function $h:A\cup B \rightarrow Y$
such that $h=f$ on $A$ and $h=g$ on $B$ is continuous.
\end{thm}

\oneperpage

\begin{thm}[pasting lemma]
Let $X=A \cup B$, where $A,B$ are open in $X$.  Let $f:A\rightarrow Y$
and $g:B\rightarrow Y$ be continuous functions which agree on $A\cap B$.
Then the function $h:A\cup B \rightarrow Y$
such that $h=f$ on $A$ and $h=g$ on $B$ is continuous.
\end{thm}

\oneperpage

\begin{exercise}
%Is the pasting lemma true when $A$ and $B$ are both open?  What if $A$ and $B$ are arbitrary sets?
Is the pasting lemma true when $A$ and $B$ in the preceding theorems are arbitrary sets?
\end{exercise}

\oneperpage

\begin{thm}
Let $f:X\to Y$ be a function and let $\mathcal B$ be a basis for $Y$. Then $f$ is continuous if and only if for every open set $B$ in $\mathcal B$, $f^{-1}(B)$ is open in $X$.  
\end{thm}

\oneperpage

\begin{thm}
Let $f:X\to Y$ be a function and let $\mathcal B$ be a subbasis for $Y$. Then $f$ is continuous if and only if for every open set $B$ in $\mathcal B$, $f^{-1}(B)$ is open in $X$.  
\end{thm}

\oneperpage

\begin{thm}
\label{continuity-compactness}
If $X$ is compact, and $f:X\rightarrow Y$ is continuous and surjective, 
then $Y$ is compact.
\end{thm}

\oneperpage

\begin{thm}     
If $X$ is Lindel\"of and $f: X\to Y$ is continuous and surjective, then $Y$ is Lindel\"of.
\end{thm}

\oneperpage

\begin{thm}
If $X$ is countably compact and $f:X\rightarrow Y$ is continuous and surjective, 
then $Y$ is countably compact.
\end{thm}

\oneperpage

\begin{thm}     
Let $D$ be a dense set of a topological space $X$ and let $f:X\to Y$
be continuous and surjective. Then $f(D)$ is dense in $Y$.
\end{thm}

\oneperpage

\begin{cor}     
Let $X$ be a separable space and let $f:X\to Y$
be continuous and surjective.  Then $Y$ is separable.
\end{cor}

\oneperpage

\begin{exercise}
\begin{enumerate}
\item Find an open function that is not continuous.
\item Find a closed function that is not continuous.
\item Find a continuous function that is neither open nor closed.
\item Find a continuous function that is open but not closed.
\item Find a continuous function that is closed but not open. 
\end{enumerate}
\end{exercise}

\oneperpage

\begin{thm}
If $X$ is normal and $f:X\rightarrow Y$ is continuous, surjective, and closed, 
then $Y$ is normal.
\end{thm}

\oneperpage

\begin{thm}
If $\{B_\alpha \}_{\alpha \in \lambda}$ is a basis for $X$ and $f:X\rightarrow Y$ is continuous, surjective, and open, 
then $\{f(B_\alpha) \}_{\alpha \in \lambda}$ is a basis for $Y$.
\end{thm}

\oneperpage

\begin{cor}
If $X$ is $2^{nd}$ countable and $f:X\rightarrow Y$ is continuous, surjective, and open, 
then $Y$ is $2^{nd}$ countable.
\end{cor}

\oneperpage

\begin{thm}    
Let $X$ be compact and $Y$ be Hausdorff.  Then any 
continuous function $f:X\to Y$ is closed.
\end{thm}

\oneperpage

\begin{thm}
\label{compact-Hausdorff-2ndcountable}
Let $X$ be compact and $2^{nd}$ countable and let $Y$ be Hausdorff.  
If $f:X\rightarrow Y$ is continuous and surjective, then $Y$ is $2^{nd}$ countable.
\end{thm}

\oneperpage

\begin{thm}
Being homeomorphic is an equivalence relation on topological spaces.
\end{thm}

\oneperpage

\begin{exercise}
Let $a$ and $b$ be points in $\R^1$ with $a<b$. Show that $(a,b)$ with the subspace topology from $\R^1_{std}$ is homeomorphic to $\R^1_{std}$.
\end{exercise}

\oneperpage

\begin{thm}     
If $f:X\to Y$ is continuous, the following are equivalent:
\item{a)} $f$ is a homeomorphism.
\item{b)} $f$ is a closed bijection.
\item{c)} $f$ is an open bijection.
\end{thm}

\oneperpage

\begin{thm}  
\label{compact-to-hausdorff-homeo}  
Suppose $f:X\to Y$ is a continuous bijection where $X$ is compact and $Y$ is Hausdorff.  Then $f$ is a homeomorphism.
\end{thm}

\oneperpage

\begin{exercise}
Construct some examples to show why the compactness and Hausdorff assumptions in the previous theorem are necessary.
\end{exercise}

\oneperpage

\begin{cor}
Let  $X$ be a compact space and let $Y$ be Hausdorff. If $f:X\to Y$ is a continuous, injective map, then $f$ is an embedding.
\end{cor}

\oneperpage

\begin{thm}   
Let $X$ and $Y$ be topological spaces. The projection maps $\pi_X, \pi_Y$ on $X\times Y$ are continuous, surjective, and open.
%In fact, if $X$ and $Y$ are nonempty, then $\pi_X$ and $\pi_Y$ are quotient maps.  
\end{thm}

\oneperpage

\begin{thm}   
Let $X$ and $Y$ be topological spaces. The product topology on $X\times Y$ is the coarsest topology on $X\times Y$ that makes the projection maps $\pi_X, \pi_Y$ on $X\times Y$ continuous.
%In fact, if $X$ and $Y$ are nonempty, then $\pi_X$ and $\pi_Y$ are quotient maps.  
\end{thm}

\oneperpage

\begin{exercise}
Find an example of $X$ and $Y$ that shows that the projection map $\pi_X : X\times Y\to X$ is not necessarily a closed map.
 \end{exercise}

\oneperpage

\begin{thm}
Let $X$ and $Y$ be topological spaces.  
For every $y\in Y$, the subspace $X \times \{y\}$ of $X \times Y$ is homeomorphic to $X$.
\end{thm}

\oneperpage

\begin{thm} 
\label{universal-prop-product}  
Let $X$, $Y$, and $Z$ be topological spaces.  A function $g: Z\to X\times Y$
is continuous if and only if $\pi_X \circ g$ and $\pi_Y \circ g$ are both continuous.
\end{thm}

\oneperpage

\begin{exercise}
What about maps {\em out of} 
a product space, i.e., $f:X \times Y \rightarrow Z$?
Do you think $f$ is continuous if it is continuous in each coordinate?
%No!  There are strange multivariable functions which do not work.
% Try $f(x,y)=xy/(x^2+y^2)$ with $f(0,0)=0$.
\end{exercise}

\oneperpage

\begin{thm} 
Let $\prod_{\alpha\in\lambda}X_\alpha$ be the product of topological spaces $\{ X_\alpha\}_{\alpha\in\lambda}$. The projection map $\pi_\beta : \prod_{\alpha\in\lambda}
X_\alpha \to X_\beta$ is a continuous, surjective, and open map.
\end{thm}

\oneperpage

\begin{thm}
\label{product-coarsest}
The product topology is the coarsest (smallest) topology on $\prod_{\alpha\in\lambda}  X_\alpha$ that makes each projection map continuous.
\end{thm}

\oneperpage

\begin{thm}
 \label{univ-map-product}  
Let $\prod_{\alpha\in\lambda}X_\alpha$ be the product of topological spaces $\{ X_\alpha\}_{\alpha\in\lambda}$ and
let $Z$ be a topological space.  A function $g: Z\to \prod_{\alpha\in\lambda}X_\alpha$
 is continuous if and  only if $\pi_\beta \circ g$ is continuous for
 each $\beta$ in $\lambda$.
 \end{thm}

\oneperpage

\begin{exercise}
Let $\R^\omega$ be the countably infinite product of $\R$ with itself.
Let $f:\R \rightarrow \R^\omega$ be defined by
$ f(x):= (x, x, x, ...)$.
Then $f$ is continuous if $\R^\omega$ is given the product topology, 
but not if given the box topology.  (This strange result once again shows why the box topology would be a poor choice as the standard topology for infinite products.)
\end{exercise}

\oneperpage

\begin{thm}    
The Cantor set is homeomorphic to the product $\prod_{n\in \N} \{ 0,1\}$
where $\{0,1\}$ has the discrete topology.
\end{thm}

\oneperpage

\begin{exercise}
\label{exercise:cylinder}
The cylinder $C$ from our example above did not need to be embedded in $\R^{3}$ to be defined; 
it could have been defined as an identification space of $X =[0,1] \times [0,1]$, using the partition 
whose sets are either singletons or pairs:
$$C^{*} = \bigg\{ \{(x,y)\} : x \in (0,1), y \in [0,1] \bigg\} \cup \bigg\{ \{ (0,y) \cup (1,y) \} : y \in [0,1]  \bigg\}.$$
What is the identification map $f:X \rightarrow C^{*}$?  What is a basis for the topology on $C^{*}$?
\end{exercise}

\oneperpage

\begin{exercise}
A \textbf{M\"obius band} is obtained by taking a strip of paper $X$ and gluing two opposite sides with a  ``twist''.  Sometimes this gluing is notated by drawing $X$ with arrows on two parallel sides that point in opposite directions.
Construct a M\"obius band explicitly as an identification space of $X=[0,8] \times [0,1]$.
\end{exercise}

\oneperpage

\begin{exercise}
\label{exercise:torus-identification}
A \textbf{torus} is the surface of a doughnut.  Construct a torus explicitly as:
\begin{enumerate}
\item an identification space of $C$, the cylinder.
% from Exercise \ref{exercise:cylinder}.
\item an identification space of $X=[0,1] \times [0,1]$.
\item an identification space of $\R^{2}$.
\end{enumerate}
\end{exercise}

\oneperpage

\begin{exercise}
Describe the $2$-dimensional sphere (the boundary of a $3$-dimensional ball in $\R^{3}$)
as an identification space of two discs in $\R^2$ by drawing a figure.
%The projective spaces $\R P^n$. 
%Explanations will be given in class.
\end{exercise}

\oneperpage

\begin{thm}
The quotient topology actually defines a topology.
\end{thm}

\oneperpage

\begin{thm}
Let $X$ be a topological space, $Y$ be a set, and 
$f:X \rightarrow Y$ be a surjective map.  The quotient topology
on $Y$ is the finest (largest) topology that makes
$f$ continuous. 
\end{thm}

\oneperpage

\begin{thm}
Let $X$ and $Y$ be topological spaces.
A surjective, continuous map $f:X\rightarrow Y$ that is an open
map is a quotient map.
\end{thm}

\oneperpage

\begin{thm}
Let $X$ and $Y$ be topological spaces.
A surjective, continuous map $f:X\rightarrow Y$ that is a closed map 
is a quotient map.
\end{thm}

\oneperpage

\begin{exercise}
Show with examples that not all quotient maps are open maps, and not all quotient maps are closed maps.
\end{exercise}

\oneperpage

\begin{exercise}
Is $\pi:\R^{2}\rightarrow \R$ defined by $\pi(x,y)=x$ a quotient map?
\end{exercise}

\oneperpage

\begin{thm}
\label{map-out-of-quotient}
Let $f:X\rightarrow Y$ be a quotient map.  Then a map $g:Y\rightarrow Z$
is continuous if and only if $g \circ f$ is continuous.
\end{thm}

\oneperpage

\begin{exercise}
Let the cylinders $C^{*}$ and $C$ be defined as at the beginning of this Section.
% in Exercise \ref{exercise:cylinder}.  
Prove that $C^{*}$ is homeomorphic to $C$ by constructing a map
$h:C^{*}\rightarrow C$ and showing it is a continuous bijection from a compact space into a Hausdorff space.
\end{exercise}

\oneperpage

\begin{exercise}
Suppose $X$ is a subspace of
$\R^n$ for some $n$.  View $\R^n$ as a subset of $\R^{n+1}$ in the
usual way (that is, $\R^n$ is the space of the first $n$
coordinates of $\R^{n+1}$ where the final coordinate is $0$). Choose a point $x_0\in\R^{n+1}-\R^n$. 
Let $C$ be the subspace of $R^{n+1}$ consisting of the union of
all the line segments from $x_0$ to points in $X$. Show that $C$ is homeomorphic to the cone over $X$ as defined above, thus justifying the name "cone." 
\end{exercise}

\oneperpage

\begin{lem}    
\label{nested-urysohn}
Let $A$ and $B$ be disjoint closed sets in a 
normal space $X$.  Then for each rational $r\in[0,1]$,
there exists an open set $U_r$ such that $A\subset U_0$, $B\subset
(X-U_1)$, and for $r<s$, $\overline{U_r} \subseteq U_s$.
\end{lem}

\oneperpage

\begin{urysohn}\index{Urysohn's lemma} 
A topological space $X$ is normal if and only if for each pair of disjoint closed sets
$A$ and $B$ in $X$, there exists a continuous function $f: X\to [0,1]$ such
that $A\subset f^{-1}(0)$ and $B\subset f^{-1}(1)$.
\end{urysohn}

\oneperpage

\begin{lem}
Let $X$ be a normal space, and let $A$ be a closed subset of $X$.  Let $f:A\rightarrow [0,1]$ be a continuous function and let $r\in (0,1)$.
Then there exist disjoint open sets $U_r$ and $V_r$
such that $f^{-1}([0,r)) \subset U_r$ and 
$f^{-1}((r,1]) \subset V_r$.  Or equivalently, there exists an open set $U_r$ such that $f^{-1}([0,r)) \subset U_r$ and $\ol{U_r} \cap f^{-1}((r,1]) = \emptyset$.
\end{lem}

\oneperpage

\begin{Tietze}
\index{Tietze extension theorem}
A space $X$ is normal if and only if for every closed set $A \subset X$ and continuous function $f:A \to [0,1]$,
there exists a continuous function $F:X \to [0,1]$ such that $F(x)=f(x)$ for every $x \in A$.  
%A space $X$ is normal if and only if every continuous function $f$
%from a closed set $A$ in $X$ into $[0,1]$ can be extended to a
%continuous function $F:X\to [0,1]$.  ($F$ extends $f$ means for each point
%$x$ in $A$, $F(x) = f(x)$.) 
\end{Tietze}

\oneperpage

\begin{thm}
A space $X$ is normal if and only if for every closed set $A \subset X$ and continuous function $f:A \to (0,1)$,
there exists a continuous function $F:X \to (0,1)$ such that $F(x)=f(x)$ for every $x \in A$.  
\end{thm}

\oneperpage

\begin{thm}
A space $X$ is normal if and only if for every closed set $A \subset X$ and continuous function $f:A \to [0,1)$,
there exists a continuous function $F:X \to ([0,1)$ such that $F(x)=f(x)$ for every $x \in A$.  
\end{thm}

\oneperpage

\begin{thm}
A space $X$ is normal if and only if for every closed set $A \subset X$ and continuous function $f:A \to [0,1] \times [0,1]$,
there exists a continuous function $F:X \to [0,1] \times [0,1]$ such that $F(x)=f(x)$ for every $x \in A$.  
\end{thm}

\oneperpage

\begin{thm}
A space $X$ is normal if and only if for every closed set $A \subset X$ and continuous function $f:A \to \prod_{\alpha\in\lambda}  [0,1]_\alpha$, where each $[0,1]_\alpha$ is a copy of $[0,1]$ in the usual topology,
there exists a continuous function $F:X \to \prod_{\alpha\in\lambda}  [0,1]_\alpha$ such that $F(x)=f(x)$ for every $x \in A$.  
\end{thm}

\oneperpage

\begin{thm}
Let $X$ be a normal space and let $A$ be a closed subspace of $X$ homeomorphic to $[0,1]$ with the usual topology. Then there exists a continuous function $r:X \to A$ such that for every $x \in A$, $r(x)=x$.
\end{thm}

\oneperpage

\begin{thm}
Let $X$ be a normal space and let $A$ be a closed subspace of $X$ homeomorphic to $S^1$ with the usual topology. Then there exists an open set $U$ containing $A$ and a continuous function $r:U \to A$ such that for every $x \in A$, $r(x)=x$.
\end{thm}

\oneperpage

\begin{exercise}
Think of (many) other possible alternatives to $S^1$ in the preceding theorem that would allow you to draw the same conclusion. 
\end{exercise}

\oneperpage

\begin{thm}
Suppose $X$ is perfectly normal. Then for each pair of disjoint closed sets
$A$ and $B$ in $X$, there exists a continuous function $f: X\to [0,1]$ such
that $A= f^{-1}(0)$ and $B= f^{-1}(1)$.
\end{thm}

\oneperpage

\begin{thm}
Every perfectly normal space is completely normal.
\end{thm}

\oneperpage

\begin{thm}    
 Let $X$ be a normal, $T_1$ space.
Then $X$ is homeomorphic to a subspace of $\prod_{\alpha\in\lambda}  [0,1]_\alpha$ for some $\lambda$, where each factor is the unit interval with the standard topology.
 \end{thm}

\oneperpage

\begin{scholium}    A space $X$ is completely regular and $T_1$ if and only if
 $X$ can be embedded in  $\prod_{\alpha\in\lambda}  [0,1]_\alpha$ for some $\lambda$, where each factor is the unit interval with the standard topology.
\end{scholium}

\oneperpage

\begin{thm}%(Existence of Partitions of Unity)
Given a locally finite open cover  
$\{ U_\alpha \}_{\alpha\in \lambda}$ of a normal, $T_1$ space
$X$, there is a collection of corresponding continuous functions 
$\phi_\alpha:X \rightarrow [0,1]$ such that 
(i) each $\phi_\alpha$ is zero outside $U_\alpha$, and 
(ii) the $\phi_\alpha$ pointwise add to 1.  The collection $\{\phi_{\alpha}\}_{\alpha\in\lambda}$ is called a \textbf{partition of unity}.
\end{thm}

\oneperpage

