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 19: Hausdorff, normal and path connected

A Hausdorff topological space is a topological space $(X,𝒯)$ such that if $x_1,x_2\in X$ and $x_1\ne x_2$ then there exist $U_1,U_2\in 𝒯$ such that $x_1\in U_1,$ $x_2\in U_2$ and $U_1\cap U_2=\varnothing \text{.}$

A normal topological space is a topological space $(X,𝒯)$ such that if $C_1,C_2$ are closed sets in $X$ and $C_1\cap C_2=\varnothing$ then there exist $U_1,U_2\in 𝒯$ such that $C_1\subseteq U_1,$ $C_2\subseteq U_2$ and $U_1\cap U_2=\varnothing \text{.}$

HW: If $X$ is a Hausdorff topological space and $A\subseteq X$ then $$A\hspace{0.17em}\text{is cover compact}\Rightarrow A\hspace{0.17em}\text{is closed.}$$

HW: If $X$ is a compact Hausdorff topological space then $X$ is normal.

A topological space $(X,𝒯)$ is path connected if $X$ satisfies: if $p,q\in X$ then there exists a continuous function $f:[0,1]\to X$ with $f\left(0\right)=p$ and $f\left(1\right)=q\text{.}$

HW: Show that if $X$ is path connected then $X$ is connected.

HW: Show that the graph of $$f\left(x\right)=\{\begin{array}{ll}\text{sin}\left(\frac{1}{x}\right),& \text{if}\hspace{0.17em}x\in (0,1],\\ 0,& \text{if}\hspace{0.17em}x=0\end{array}$$ is a connected set which is not path connected.

One point compactification

A topological space $X$ is locally compact if $X$ is Hausdorff and if $x\in X$ then there exists $N\in 𝒩\left(x\right)$ such that $N$ is compact.

HW: Show that $ℝ$ is locally compact but $ℝ$ is not compact.

Let $X$ be locally compact. The one point compactification of $X$ is $$X\prime =X\cup \left\{\omega \right\}$$ with topology $$𝒯\prime =𝒯\cup \{(X-K)\cup \left\{\omega \right\}\hspace{0.17em}|\hspace{0.17em}K\subseteq X\hspace{0.17em}\text{and}\hspace{0.17em}K\hspace{0.17em}\text{is compact}\}\text{.}$$

Compactness and closedness

HW: Let $(X,d)$ be a metric space and let $Y\subseteq X\text{.}$

(a)	Show that if $X$ is compact and $Y$ is closed then $Y$ is compact.
(b)	Show that if $Y$ is compact then $Y$ is closed.

Closed and bounded $\nRightarrow$ compact.

Let $X=C([0,1];ℝ)$ with metric given by $$d(f,g)=\text{sup}\left\{\mid f\left(x\right)-g\left(x\right)\mid \hspace{0.17em}\right|\hspace{0.17em}x\in [0,1]\}\text{.}$$ Let $E=\overline{B(0,1)}=\{f\in X\hspace{0.17em}|\hspace{0.17em}d(f,0)\le 1\}\text{.}$
Since $d$ is continuous then $E$ is closed.
Since $E\subseteq B(0,2)$ then $E$ is bounded.
Let $f_n:[0,1]\to ℝ$ be given by $f_n\left(x\right)=x^n\text{.}$
The pointwise limit of $f_1,f_2,\dots$ is $f:[0,1]\to ℝ$ given by $$f\left(x\right)=\{\begin{array}{ll}0,& \text{if}\hspace{0.17em}x\ne 1,\\ 1,& \text{if}\hspace{0.17em}x=1\text{.}\end{array}$$ Since $\Vert f_n-f\Vert =1$ for $n\in {\mathbb{Z}}_{>0},$ $$\underset{n\to \infty }{\text{lim}}d(f_n,f)=\underset{n\to \infty }{\text{lim}}\Vert f_n-f\Vert =\underset{n\to \infty }{\text{lim}}1=1\text{.}$$ So $f_1,f_2,\dots$ does not have a convergent subsequence.

Examples of completions

The completion of $\mathbb{Q}$ is $ℝ\text{.}$

The completion of $\mathbb{Q}$ is ${\mathbb{Q}}_p\text{.}$

The completion of $ℂ\left[X\right]$ is $ℂ\left[\left[X\right]\right]\text{.}$

The completion of $ℂ\left(X\right)$ is $ℂ\left(\left(X\right)\right)\text{.}$

Notes and References

These are a typed copy of Lecture 19 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 28, 2014.

page history