Hausdorff

 

Def 1

Let $(X, \mathscr{T})$ be a topological space. $X$ is Hausdorff if

$\forall a, b \; (a \neq b) \in X$,  there exists $U, V \in \mathscr{T}$  s.t. 

$a \in U$,     $b \in V$,     $U \cap V = \varnothing$

 

 

 

Thm 1

 $X$ is metrizable  $\Rightarrow$  $X$ is Hausdorff

 

  Let $\mathscr{T}$ be a topology of $X$

  $\forall a, b \; (a \neq b) \in X$

 

  Let $r = d(a, b)$  $(>0)$

  $B_{\frac{r}{3}}(a), B_{\frac{r}{3}}(b) \in \mathscr{T}$

  It is clearly $a \in B_{\frac{r}{3}}(a)$  and  $b \in B_{\frac{r}{3}}(b)$.

 

 

  Suppose : $B_{\frac{r}{3}}(a) \cap B_{\frac{r}{3}}(b) \neq \varnothing$

  So there exists $y$  s.t.

$y \in B_{\frac{r}{3}}(a) \cap B_{\frac{r}{3}}(b)$

 

  Therefore

  $r = d(a, b) \leq d(a, y) + d(y, b)$

                          $< \frac{r}{3} + \frac{r}{3} = \frac{2r}{3}$

  Contradiction

 

  $\therefore$  $B_{\frac{r}{3}}(a) \cap B_{\frac{r}{3}}(b) = \varnothing$

  $\therefore$  $X$ is Hausdorff

 

 

 

Cor 1

Let $X$ be a set  and  $\left|X \right| \geq 2$ . Then

$(X, \text{discrete topology})$ is Hausdorff

 

# Prop 1

 

 

 

Cor 2

Let $X$ be a set  and  $\left|X \right| \geq 2$ . Then

$(X, \text{indiscrete topology})$ is not Hausdorff

 

# So $(X, \text{indiscrete topology})$ is not metrizable

 

 

 

Prop 1

Let $X$ be an infinite set. Then

$(X, \text{finite complement topology})$ is not Hausdorff

 

# So $(X, \text{finite complement topology})$ is not metrizable

 

  $\forall U, V \; (\neq \varnothing) \in \mathscr{T}$

 

 

  Suppose : $U \cap V = \varnothing$

  So $U \subseteq X \setminus V$.

  Since $U, V \in \mathscr{T}$ and $X$ is an infinite set,

  $U$ is infinite set  and  $X \setminus V$ is finite set.

  Contradiction

 

  $\therefore$  $U \cap V \neq \varnothing$

  $\therefore$  $(X, \text{finite complement topology})$ is not Hausdorff

 

 

 

Prop 2

$(\mathbb{R}, \text{usual})$ is Hausdorff

 

  $\forall a, b \; (a \neq b) \in \mathbb{R}$

  WLOG, WMA

  $a < b$

 

  Let $d = b - a$

  Let $U = (a - \dfrac{d}{2}, a + \dfrac{d}{2})$  and  $V = (b - \dfrac{d}{2}, b + \dfrac{d}{2})$

  $\therefore$  $a \in U$  and  $b \in V$

 

  Since $(b - \dfrac{d}{2}) - (a + \dfrac{d}{2}) = 0$,

  $U \cap V = \varnothing$

 

  $\therefore$  $(\mathbb{R}, \text{usual})$ is Hausdorff