Topological spaces and Continuous functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 4 March 2014

Topological spaces

A topological space is a set $X$ with a specification of the open subsets of $X$ where it is required that

(a)   $\varnothing$ is open and $X$ is open,

(b)   Unions of open sets are open,

(c)   Finite intersections of open sets are open.

In other words, a topology on $X$ is a set $𝒯$ of subsets of $X$ such that

(a)   $\varnothing \in 𝒯$ and $X\in 𝒯$,

(b)   If $𝒮\subseteq 𝒯$ then ${\displaystyle \left(\bigcup _{U\in 𝒮}U\right)\in 𝒯}$,

(c)   If $n\in {\mathbb{Z}}_{\ge 0}$ and $U_1,U_2,\dots ,U_n\in 𝒯$ then $U_1\cap U_2\cap \dots \cap U_n\in 𝒯.$

A topological space is a set $X$ with a topology $𝒯$ on $X$.

Let $X$ be a set and let $𝒯$ be a topology on $X$.

• An open set is a set in $𝒯$.
• A closed set is a subset $E$ of $X$ such that the complement $E^c$ of $E$ is open.
• A connected set is a subset $E\subseteq X$ such that there do not exist open sets $A$ and $B$ with $$A\cap E\ne \varnothing ,B\cap E\ne \varnothing ,A\cup B\supseteq E,\text{and}(A\cap B)\cap E=\varnothing .$$
• A compact set is a subset $E\subseteq X$ such that every open cover of $E$ contains a finite subcover.
  More precisely, a compact set is a subset $E\subseteq X$ such that

  if $𝒮\subseteq 𝒯$ and ${\displaystyle \bigcup _{U\in 𝒮}U\supseteq E}$

  then there exists $n\in {\mathbb{Z}}_{>0}$ and $U_1,\dots ,U_n\in 𝒮$ such that $(U_1\cup U_2\cup \dots \cup U_n)\supseteq E.$

HW: Let $X$ be a topological space. Let $𝒞$ be the collection of closed sets of $X\text{.}$ Show that $𝒞$ satisfies:

(Ca)	If $𝒮\subseteq 𝒞$ then $\bigcap _{C\in 𝒮}C\in 𝒞\text{.}$
(Cb)	If $\ell \in {\mathbb{Z}}_{>0}$ and $C_1,C_2,\dots ,C_{\ell }\in 𝒞$ then $C_1\cup \cdots \cup C_{\ell }\in 𝒞\text{.}$

HW: Let $X$ be a set and let $𝒞$ be a collection of subsets of $X$ which satisfies (Ca) and (Cb).

(a)	Show that $$𝒯=\{A\subseteq X\hspace{0.17em}|\hspace{0.17em}{A}^c\in 𝒞\}$$ is a topology on $X$ such that $𝒞$ is the set of closed sets in $X\text{.}$
(b)	Show that $𝒯$ is the unique topology on $X$ such that $𝒞$ is the set of closed sets in $X\text{.}$

Continuous Functions

Continuous functions are for comparing topological spaces.

Let $X$ and $Y$ be topological spaces.

• A function $f:X\to Y$ is continuous if it satisfies the condition

  if $V$ is an open subset of $Y$     then     $f^{-1}\left(V\right)$ is an open subset of $X$.

• Let $X$ and $Y$ be topological spaces. An isomorphism, or homeomorphism, is a continuous function $f:X\to Y$ such that the inverse function $f^{-1}:Y\to X$ exists and is continuous.
• A subspace of $X$ is a subset $E$ of $X$ with the topology given by making the open sets be the sets $${\iota }^{-1}\left(V\right)\text{such that}V\text{is an open subset of}X,$$ where $\iota :E\to X$ is the inclusion.

Let $f:X\to Y$ be a continuous function and let $E\subseteq X$.

(a)   If $E$ is connected then $f\left(E\right)$ is connected.

(b)   If $E$ is compact then $f\left(E\right)$ is compact.

		Proof of (a):
	Proof by contradiction. Assume $f\left(E\right)$ is not connected. Let $A$ and $B$ be open in $f\left(E\right)$ such that $A\ne \varnothing$ and $B\ne \varnothing$ and $A\cup B\supseteq f\left(E\right)$ and $A\cap B=\varnothing$. Then let $C=f^{-1}\left(A\right)$ and $D=f^{-1}\left(B\right)$. Then $C\cup D=f^{-1}\left(A\right)\cup f^{-1}\left(B\right)=f^{-1}\left(A\cup B\right)\supseteq f^{-1}\left(f\right(E\left)\right)\supseteq E$, and $C\cap D=f^{-1}\left(A\right)\cap f^{-1}\left(B\right)=f^{-1}(A\cap B)=f^{-1}\left(\varnothing \right)=\varnothing$. Now $C\ne \varnothing$ since $A\ne \varnothing$ and $A\subseteq f\left(E\right)$, and $D\ne \varnothing$ since $B\ne \varnothing$ and $B\subseteq f\left(E\right)$. So $E$ is not connected. This is a contradiction. So $f\left(E\right)$ is connected. $\square$

		Proof of (b):
	Assume $f:X\to Y$ is continuous and $E$ is compact. To show: If $𝒮$ is an open cover of $f\left(E\right)$ then it has a finite subcover. To show: If $𝒴$ is the topology on $Y$, $𝒮\subseteq 𝒴$, and ${\displaystyle \left(\bigcup _{V\in 𝒮}V\right)}\supseteq f\left(E\right)$ then there exists $n\in {\mathbb{Z}}_{>0}$ and $V_1,\dots ,V_n\in 𝒮$ such that $V_1\cup \dots \cup V_n\supseteq f\left(E\right)$. Assume $𝒴$ is the topology on $Y$ and $𝒮\subseteq 𝒴$ and ${\displaystyle \left(\bigcup _{V\in 𝒮}V\right)}\supseteq f\left(E\right)$. Then ${\displaystyle \left(\bigcup _{V\in 𝒮}{f}^{-1}\left(V\right)\right)}\supseteq E$. Let $𝒯=\left\{f^{-1}\right(V\left)\right|V\in 𝒮\}$. Since $f$ is continuous, $f^{-1}\left(V\right)$ is open. Thus $𝒯\subseteq 𝒳$ where $𝒳$ is the topology of $X$, and also ${\displaystyle \left(\bigcup _{f^{-1}\left(V\right)\in 𝒯}{f}^{-1}\left(V\right)\right)}\supseteq E$. Since $E$ is compact there exists $n\in {\mathbb{Z}}_{>0}$ and $f^{-1}\left(V_1\right),\dots ,f^{-1}\left(V_n\right)\in 𝒯$ such that $\left(f^{-1}\right(V_1)\cup \dots \cup f^{-1}(V_n\left)\right)\supseteq E$. So $(V_1\cup \dots \cup V_n)\supseteq f\left(E\right)$. Thus $𝒮$ contains a finite subcover of $f\left(E\right)$. $\square$

Let $Y$ be a topological space. Let $y\in Y$.

• A neighbourhood of $y$ is a subset $N\subseteq Y$ such that there exists an open set $U$ of $Y$ with $y\in U\subseteq N$.
• The neighbourhood filter of $y$ is

  $𝒩\left(y\right)=\left\{\text{neighborhoods of}y\right\}$.

Let $X$ and $Y$ be topological spaces. Let $a\in X$.

• A function $f:X\to Y$ is continuous at $a$ if it satisfies the condition

  if $V$ is a neighborhood of $f\left(a\right)$ in $Y$     then     $f^{-1}\left(V\right)$ is a neighborhood of $a$ in $X$.

  Here $f^{-1}\left(V\right)=\left\{x\in X\right|f\left(x\right)\in V\}.$

HW: Let $X$ and $Y$ be topological spaces and let $f:X\to Y$ be a function. Show that $f:X\to Y$ is continuous if and only if $$f\text{satisfies:}\text{if}a\in X\text{then}f\text{is continuous at}a.$$

HW: Let $X$ be a topological space. For $x\in X$ let $𝒩\left(x\right)$ be the set of neighborhoods of $x\text{.}$ Show that the collections $𝒩\left(X\right)$ satisfy:

$\left({\text{<mi>V</mi>}}_{\text{I}}\right)$	If $A\subseteq X$ and there exists $N\in 𝒩\left(x\right)$ such that $A\supseteq N$ then $A\in 𝒩\left(x\right)\text{.}$
$\left({\text{V}}_{\text{II}}\right)$	If $\ell \in {\mathbb{Z}}_{>0}$ and $N_1,N_2,\dots ,N_{\ell }\in 𝒩\left(x\right)$ then $N_1\cap N_2\cap \dots \cap N_{\ell }\in 𝒩\left(x\right)\text{.}$
$\left({\text{V}}_{\text{III}}\right)$	If $N\in 𝒩\left(x\right)$ then $x\in N\text{.}$
$\left({\text{V}}_{\text{IV}}\right)$	If $N\in 𝒩\left(x\right)$ then there exists $W\in 𝒩\left(x\right)$ such that if $y\in W$ then $N\in 𝒩\left(y\right)\text{.}$

HW: Let $X$ be a set with a collection of subsets of $X,$ $𝒩\left(x\right)$ for each $x\in X,$ such that the $𝒩\left(x\right)$ satisfy $\left({\text{V}}_{\text{I}}\right),\left({\text{V}}_{\text{II}}\right),\left({\text{V}}_{\text{III}}\right),\left({\text{V}}_{\text{IV}}\right)\text{.}$ Let $$𝒯=\{A\subseteq X\hspace{0.17em}|\hspace{0.17em}\text{if}\hspace{0.17em}x\in A\hspace{0.17em}\text{then}\hspace{0.17em}A\in 𝒩\left(x\right)\}\text{.}$$

(a)	Show that $𝒯$ is a topology on $X\text{.}$
(b)	Show that the $𝒩\left(x\right),$ $x\in X,$ are the neighborhood filters for the topology $𝒯\text{.}$
(c)	Show that $𝒯$ is the unique topology on $X$ such that $𝒩\left(x\right),$ $x\in X,$ are the neighborhood filters for $𝒯\text{.}$

Let $I$ be a set and let $X_i,$ $i\in I,$ be topological spaces.

• The product space $\prod _{i\in I}{X}_i$ is the set $\prod _{i\in I}{X}_i$ with topology generated by the sets $${\text{pr}}_i\left(U_i\right)\hspace{0.17em}\text{with}\hspace{0.17em}{U}_i\hspace{0.17em}\text{open in}\hspace{0.17em}{X}_i,$$ where ${\text{pr}}_i:\prod _{i\in I}{X}_i\to X_i$ is the projection (onto the $i^{\text{th}}$ coordinate).

HW: Show that the topology on $\prod _{i\in I}{X}_i$ is the weakest topology such that the projections ${\text{pr}}_i:\prod _{i\in I}{X}_i\to X_i,$ for $i\in I,$ are continuous.

Examples

(1) Let $X$ be a set. The discrete topology on $X$ is the topology such that every subset of $X$ is open.

(2) A metric space is a set $X$ with a function $d:X\times X\to {ℝ}_{\ge 0}$ such that

(a)   If $x\in X$ then $d(x,x)=0$,

(b)   If $x,y\in X$ and $d(x,y)=0$, then $x=y$,

(c)   If $x,y,z\in X$ then $d(x,z)\le d(x,y)+d(y,z)$.

Let $X$ be a metric space. Let $x\in X$ and let $\epsilon \in {ℝ}_{>0}$. The ball of radius $\epsilon$ at $x$ is the set $$B_{\epsilon }\left(x\right)=\{p\in X\left|d\right(x,y)<\epsilon \}.$$ Let $X$ be a metric space. The metric space topology on $X$ is the topology generated by the sets $$B_{\epsilon }\left(x\right),\text{for}x\in X\text{and}\epsilon \in {ℝ}_{>0}.$$

HW: Show that the metric space topology is a topology on $X$.

HW: Let $X=\{0,1\}$ and let $𝒯=\{\varnothing ,\{0\},X\}$. Show that $𝒯$ is a topology on $X$ and that there does not exist a metric $d:X\times X\to {ℝ}_{\ge 0}$ on $X$ such that $𝒯$ is the metric space topology on $X$.

Generating topologies

Let $X$ be a topological space. Let $x\in X\text{.}$

• A base of the topology on $X$ is a collection $ℬ$ of open sets of $X$ such that if $U$ is open in $X$ then there exists $𝒮\subseteq ℬ$ such that $U=\bigcup _{A\in ℬ}A\text{.}$
• A fundamental system of neighborhoods of $X$ is a set $𝒮$ of neighborhoods of $X$ such that if $N\in 𝒩\left(x\right)$ then there exists $W\in 𝒮$ such that $W\subseteq N\text{.}$

HW: Show that $ℬ$ is a base of the topology on $X$ if and only if $ℬ$ satisfies if $x\in X$ then $\{V\in ℬ\hspace{0.17em}|\hspace{0.17em}x\in V\}$ is a fundamental system of neighborhoods of $x\text{.}$

This is found in [Bou, Ch I §1.3 Proposition 3].

HW: Show that the topology on $\prod _{i\in I}{X}_i$ has base $$\left\{\prod _{i\in I}{A}_i\hspace{0.17em}\right|\hspace{0.17em}\begin{array}{c}{A}_i\hspace{0.17em}\text{is open in}\hspace{0.17em}{X}_i,\\ A_i=X_i\hspace{0.17em}\text{for all but a finite number of}\hspace{0.17em}i\in I\end{array}\}\text{.}$$

HW: Let $x\in \prod _{i\in I}{X}_i\text{.}$ Show that $$\left\{\prod _{i\in I}{A}_i\hspace{0.17em}\right|\hspace{0.17em}\begin{array}{c}{A}_i\hspace{0.17em}\text{is open in}\hspace{0.17em}{X}_i,\\ A_i=X_i\hspace{0.17em}\text{for all but a finite number of}\hspace{0.17em}i\in I,\\ x\in \prod _{i\in I}{A}_i\end{array}\}$$ is a fundamental system of neighborhoods of $x\text{.}$

Homework

1. $f^{-1}(C\cup D)=f^{-1}\left(C\right)\cup f^{-1}\left(D\right)$.
2. $f^{-1}(C\cap D)=f^{-1}\left(C\right)\cap f^{-1}\left(D\right)$.
3. $f^{-1}\left(f\right(E\left)\right)\supseteq E$.
4. Give an example where $f^{-1}\left(f\right(E\left)\right)\ne E$.
5. If $E$ is compact then $E$ is closed.

Notes and References

This material is found in almost every textbook on introductory topology or mathematical analysis. One comprehensive reference is Bourbaki -- see [Bou, Ch I § 1 nos. 1,2,4] for open sets, closed sets and neighborhoods, [Bou, Ch I § 3 no. 1] for subspaces, [Bou Ch I § 4 no. 1] for products, [Bou Ch I § 9 no. 1 Axiom C'''] and [Bou, Ch. I § 9 no. 4] for compact sets, [Bou, Ch. I § 11 nos. 1,2] for connected sets, and [Bou, Ch. I § 2 no. 1] for continuous functions.

The most important theorem in the theory of continuous functions is the following. This theorem will be proved in the section on Limits in Topological spaces.

Let $X$ and $Y$ be topological spaces and let $a\in X$. A function

$f:X\to Y$ is continuous at $a$    if and only if

References

[Bou] N. Bourbaki, General Topology, Springer-Verlag, 1989. MR1726779.

[BR] W. Rudin, Principles of mathematical analysis, Third edition, International Series in Pure and Applied Mathematics, McGraw-Hill 1976. MR0385023.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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