Hausdorff and separable spaces

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

Last updates: 4 March 2014

Hausdorff spaces

Let $X$ be a set. The diagonal map is

$\Delta :X\to X\times X\text{given by}\Delta \left(x\right)=(x,x)$.

The image of the diagonal map $\Delta \left(X\right)$ is the graph of the identify function ${\mathrm{id}}_X:X\to X$.
A Hausdorff space is a topolgical space such that $\Delta \left(X\right)$ is closed in $X\times X$, where $X\times X$ has the product topology.

Let $X$ be a topological space. The following are equivalent.

(H)	If $x,y\in X$ and $x\ne y$ then there exist $U\in 𝒩\left(x\right)$ and $V\in 𝒩\left(y\right)$ such that $U\cap V=\varnothing \text{.}$
$\left({\text{H}}^{\text{I}}\right)$	If $x\in X$ then $\bigcap _{N\in 𝒩\left(x\right)}\bar{N}=\left\{x\right\}\text{.}$
$\left({\text{H}}^{\text{II}}\right)$	If $\Delta :X\to X\times X$ is the diagonal map then $\Delta \left(X\right)$ is closed in $X\times X\text{.}$
$\left({\text{H}}^{\text{III}}\right)$	If $I$ is a set and $\Delta :X\to \prod _{k\in I}{X}_k,$ where $X_k=X$ for $k\in I,$ is the diagonal map then $\Delta \left(X\right)$ is closed in $\prod _{k\in I}{X}_k\text{.}$
$\left({\text{H}}^{\text{IV}}\right)$	If $𝒢$ is a filter on $X$ then $𝒢$ has at most one limit point.
$\left({\text{H}}^{\text{V}}\right)$	If $𝒥$ is a filter on $X$ and $x$ is a limit point of $𝒥$ then $x$ is the only cluster point of $𝒥\text{.}$

		Proof.
	$\left({\text{H}}^{\text{III}}\right)\Rightarrow \left({\text{H}}^{\text{II}}\right)\text{:}$ ${\text{H}}^{\text{II}}$ is a special case of ${\text{H}}^{\text{III}}\text{.}$ $\left({\text{H}}^{\text{II}}\right)\Rightarrow \left(\text{H}\right)\text{:}$ Assume $x,y\in X$ and $x\ne y\text{.}$ Then $(x,y)\in X\times X$ and $(x,y)\notin \Delta \left(X\right)\text{.}$ Thus, by $\left({\text{H}}^{\text{II}}\right),$ $(x,y)\notin \overline{\Delta \left(X\right)}\text{.}$ So $(x,y)$ is not a close point of $\Delta \left(X\right)\text{.}$ So there exists a neighborhood $Z\in 𝒩\left((x,y)\right)$ such that $Z\cap \Delta \left(X\right)=\varnothing \text{.}$ By the definition of the product topology, there exist $U\in 𝒩\left(x\right)$ and $V\in 𝒩\left(y\right)$ such that $(U\times V)\cap \Delta \left(X\right)=\varnothing \text{.}$ So $U\cap B=\varnothing \text{.}$ $\left(\text{H}\right)\Rightarrow \left({\text{H}}^{\text{III}}\right)\text{:}$ Assume (H). To show: $\Delta \left(X\right)$ is closed in $\prod _{k\in I}{X}_k,$ where $X_k=X\text{.}$ To show: If $x\in \prod _{k\in I}{X}_k$ and $x\notin \Delta \left(X\right)$ then $x$ is not a close point of $\Delta \left(X\right)\text{.}$ Assume $x=\left(x_k\right)\in \prod _{k\in I}{X}_k$ and $x\notin \Delta \left(X\right)\text{.}$ To show: There exists $W\in 𝒩\left(x\right)$ such that $W\cap \Delta \left(X\right)=\varnothing \text{.}$ Let $i,j\in I$ be such that $x_i\ne x_j\text{.}$ Let $V_i\in 𝒩\left(x_i\right)$ and $V_j\in 𝒩\left(x_j\right)$ such that $V_i\cap V_j=\varnothing \text{.}$ Then $W=V_i\times V_j\times \prod _{k\ne i,j}{X}_k\in 𝒩\left(x\right)$ and $W\cap \Delta \left(X\right)=\varnothing \text{.}$ So $x$ is not a close point of $\Delta \left(X\right)\text{.}$ So $\Delta \left(X\right)$ is closed in $\prod _{k\in I}{X}_k\text{.}$ $\left(\text{H}\right)\Rightarrow \left({\text{H}}^{\text{I}}\right)\text{:}$ Assume (H). To show: If $x\in X$ then $\bigcap _{N\in 𝒩\left(x\right)}\bar{N}=\left\{x\right\}\text{.}$ Assume $x\in X\text{.}$ To show: If $y\in X$ and $y\notin \left\{x\right\}$ then $y\notin \bigcap _{N\in 𝒩\left(x\right)}\bar{N}\text{.}$ Assume $y\in X$ and $y\notin \left\{x\right\}\text{.}$ To show: There exists $U\in 𝒩\left(x\right)$ such that $y\notin \bar{U}\text{.}$ By (H), since $y\ne x,$ there exist $U\in 𝒩\left(x\right)$ and $V\in 𝒩\left(y\right)$ such that $U\cap V=\varnothing \text{.}$ So $y\notin \bar{U}\text{.}$ So $y\notin \bigcap _{N\in 𝒩\left(x\right)}\bar{N}\text{.}$ $\left({\text{H}}^{\text{I}}\right)\Rightarrow \left({\text{H}}^{\text{V}}\right)\text{:}$ Assume $\left({\text{H}}^{\text{I}}\right)\text{.}$ Assume $𝒥$ is a filter on $X$ and $x$ is a limit point of $𝒥\text{.}$ So $𝒥\supseteq 𝒩\left(x\right)\text{.}$ Let $y\in X$ with $y\ne x\text{.}$ To show: $y$ is not a cluster point of $𝒥\text{.}$ To show: $y\notin \bigcap _{M\in 𝒥}\bar{M}\text{.}$ To show: There exists $M\in 𝒥$ such that $y\notin \bar{M}\text{.}$ By $\left(H^I\right),$ $y\notin \bigcap _{N\in 𝒩\left(x\right)}\bar{N}\text{.}$ So there exists $M\in 𝒩\left(x\right)$ such that $y\notin \bar{M}\text{.}$ Since $𝒥\supseteq 𝒩\left(x\right),$ then $M\in 𝒥\text{.}$ So $y$ is not a cluster point of $𝒥\text{.}$ $\left({\text{H}}^{\text{V}}\right)\Rightarrow \left({\text{H}}^{\text{IV}}\right)\text{:}$ Assume $\left({\text{H}}^{\text{V}}\right)\text{.}$ Let $𝒢$ be a filter on $X$ and let $x$ be a limit point of $𝒢\text{.}$ Let $y\in X$ with $y\ne x\text{.}$ To show: $y$ is not a limit point of $𝒢\text{.}$ If $y$ is a limit point of $𝒢$ then $y$ is a cluster point of $𝒢,$ which is a contradiction to $\left({\text{H}}^{\text{V}}\right)\text{.}$ So $y$ is not a limit point of $𝒢\text{.}$ $\left({\text{H}}^{\text{IV}}\right)\Rightarrow \left(\text{H}\right)\text{:}$ Assume not (H). Let $x,y\in X$ with $x\ne y$ such that there do not exist $U\in 𝒩\left(x\right)$ and $V\in 𝒩\left(y\right)$ with $U\cap V=\varnothing \text{.}$ Let $𝒥$ be the filter generated by $$\{U\cap V\hspace{0.17em}|\hspace{0.17em}U\in 𝒩\left(x\right),V\in 𝒩\left(y\right)\}\text{.}$$ Then $x$ and $y$ are both limit points of $𝒥\text{.}$ So $\left({\text{H}}^{\text{IV}}\right)$ does not hold. $\square$

HW: Show that metric spaces are Hausdorff.

Separable spaces

Remove separability, this must have to do with normed linear spaces. See [Bou, Top, Ch. 1 Sec 1 Ex 7].

Separability appears in [BR] Chapter 2 Exercises 22 and 23, and in [Ru] Chapter 4 Exercises 2, 3, 4 and 18. A uniform space is almost a metric space: By [Bou??] the separable Hausdorff uniform spaces are exactly the separable metric spaces.

A topological space $X$ is separable if it has a countable base, or?? if it has a countable dense set.

Hausdorff separable spaces are separable uniform spaces and metric spaces???

HW: Give an example of a metric space that is not separable. See [Ru] Chapter 4 Exercises 3 and 18. The key examples are ${\ell }^p$ is separable if $p\ne \infty$ and ${\ell }^{\infty }$ is not separable.

HW: Show that a Hilbert space is separable if and only if it contains a maximal orthonormal system which is at most countable. (See [Ru, Ch. 4 Ex. 4]).

HW: Give an example of a uniform space that is not separable.

HW: Show that ${ℝ}^k$ is separable.

Notes and References

These notes follow Bourbaki [Bou, Ch.I §8 no.1]. The condition that $\Delta$ is closed is the condition used in algebraic geometry for a separated scheme (see [Hartshorne, Ch. II Sec 4] and Macdonald (1.11) in [Carter-Segal-Macdonald, LMS Lecture Notes]).

By Theorem 1.1e, Hausdorff spaces are spaces such that limits are unique, when they exist.

For the filters and topology pages: Define base of a topology and base of a filter? Or just go by topology generated by and filter generated by? Then put other stuff in exercises.

The treatment of metric spaces and completion follows [BR] Chapter 2 Exercise ??.

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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