Metric and Hilbert Spaces

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

Last updated: 4 November 2014

Lecture 7: Connectedness and connected components

Homework questions

(1)	Let $X$ be a set. Show that the discrete topology on $X$ is a topology on $X\text{.}$
(2)	Let $(X,d)$ be a metric space. Show that the metric space topology on $X$ is a topology on $X\text{.}$
(3)	Let $(X,𝒯)$ be a topological space and let $Y\subseteq X\text{.}$ Show that the subspace topology on $Y$ is a topology on $Y\text{.}$
(4)	Let $(X,{𝒯}_X)$ and $(Y,{𝒯}_Y)$ be topological spaces. Show that the product topology on $X\times Y$ is a topology on $X\times Y\text{.}$

One could do these, one by one, directly, using proof machine. Alternatively, one can develop a few helpful definitions that, more or less, do them all in one fell swoop.

Let $X$ be a topological space with topology $𝒯\text{.}$

A base of the topology on $X$ is a collection $B\subseteq 𝒯$ such that if $U\in 𝒯$ then there exists $𝒮\subseteq B$ such that $$U=\bigcup _{A\in 𝒮}A\text{.}$$

Let $x\in X\text{.}$ A fundamental system of neighbourhoods of $x$ is a set $𝒮\subseteq 𝒩\left(x\right)$ such that if $N\in 𝒩\left(x\right)$ then there exists $W\in 𝒮$ such that $W\subseteq N\text{.}$

HW 1 Show that $B$ is a base of $𝒯$ 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 neighbourhoods of $x\text{.}$

HW 2 Let $(X,d)$ be a metric space. Show that the set of open balls is a base of the metric space topology on $X$ by showing that it satisfies the condition in HW 1.

HW 3 Let $(X,{𝒯}_X)$ and $(Y,{𝒯}_Y)$ be topological spaces. Show that $$B=\{U\times V\hspace{0.17em}|\hspace{0.17em}U\in {𝒯}_X\hspace{0.17em}\text{and}\hspace{0.17em}V\in {𝒯}_Y\}$$ is a base of the product topology on $X\times Y$ by showing that it satisfies the condition in HW 1.

Let $(X,𝒯)$ be a topological space.

A connected set is a subset $E\subseteq X$ such that there do not exist open sets $A$ and $B$ $\text{(}A,B\in 𝒯\text{)}$ with $$A\cap E\ne \varnothing \text{and}B\cap E\ne \varnothing \text{and}A\cup B\supseteq E\text{and}(A\cap B)\cap E=\varnothing \text{.}$$ Perhaps it is better to think of $E$ with the subspace topology ${𝒯}_E\text{.}$ Then $E$ is connected if there do not exist $U$ and $V$ open in $E$ $\text{(}U,V\in {𝒯}_E\text{)}$ such that $$U\ne \varnothing \text{and}V\ne \varnothing \text{and}U\cup V=E\text{and}U\cap V=\varnothing \text{.}$$

Let $(X,{𝒯}_X)$ and $(Y,{𝒯}_Y)$ be topological spaces. Let $f:X\to Y$ be a function.

The function $f:X\to Y$ is continuous if $f$ satisfies: if $V\in {𝒯}_Y$ then $f^{-1}\left(V\right)\in {𝒯}_X\text{.}$

Recall that, by definition, $$f^{-1}\left(V\right)=\{x\in X\hspace{0.17em}|\hspace{0.17em}f\left(x\right)\in V\}\text{.}$$

Recall that, by definition, a function $f:X\to Y$ is a subset $\mathrm{\Gamma }\subseteq X\times Y$ such that if $x\in X$ then there exists a unique $y\in Y$ such that $(x,y)\in \mathrm{\Gamma }\text{.}$ Use the notation $f\left(x\right)$ so that $$\mathrm{\Gamma }=\left\{(x,f\left(x\right))\hspace{0.17em}\right|\hspace{0.17em}x\in X\}\text{.}$$

Let $f:X\to Y$ be a continuous function. Let $E\subseteq X\text{.}$ If $E$ is connected
then $f\left(E\right)$ is connected.

		Proof.
	Assume $E$ is connected. To show: $f\left(E\right)$ is connected. Proof by contradiction. Assume $f\left(E\right)$ is not connected. Let $A$ and $B$ be open in $f\left(E\right)$ such that $$A\ne \varnothing \text{and}B\ne \varnothing \text{and}A\cup B\supseteq f\left(E\right)\text{and}A\cap B=\varnothing \text{.}$$ Let $C=f^{-1}\left(A\right)$ and $D=f^{-1}\left(B\right)\text{.}$ Then $$C\cup D=f^{-1}\left(A\right)\cup f^{-1}\left(B\right)=f^{-1}(A\cup B)\supseteq f^{-1}\left(f\left(E\right)\right)\supseteq E$$ and $$C\cap D=f^{-1}\left(A\right)\cap f^{-1}\left(B\right)=f^{-1}(A\cap B)=f^{-1}\left(\varnothing \right)=\varnothing$$ and $$\begin{array}{c}C\ne \varnothing \hspace{0.17em}\text{since}\hspace{0.17em}A\ne \varnothing \hspace{0.17em}\text{and}\hspace{0.17em}A\subseteq f\left(E\right),\\ D\ne \varnothing \hspace{0.17em}\text{since}\hspace{0.17em}B\ne \varnothing \hspace{0.17em}\text{and}\hspace{0.17em}B\subseteq f\left(E\right)\text{.}\end{array}$$ So $E$ is not connected. This is a contradiction. So $f\left(E\right)$ is connected. $\square$

Notes and References

These are a typed copy of Lecture 7 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 7, 2014.

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