# Abstract Nonsense

## Thoughts about separation (Hausdorff)

In this post we go one step further and discuss Hausdorff spaces, a stronger property than either Kolomogorov or $T_1$.

Hausdorff ($T_2$): Let $X$ be a topological space, then we call $X$ Hausdorff (or $T_2$) if given any distinct $x,y\in X$ there exists open sets $O_x,O_y$ such that $x\in O_x,y\in O_y$ and $O_x\cap O_y=\varnothing$

Hausdorff spaces are the first real interesting spaces since we can say many important and profound things concerning mappings, etc.

It is clear first-off that every Hausdorff space is $T_1$. We begin by discussing some alternative definitions for when a space $X$ is Hausdorff.

Theorem: Let $X$ be a topological space, then $X$ is Hausdorff if and only if $\displaystyle \{x\}=\bigcap_{N\in\mathfrak{N}}\overline{N}$

Proof:

$\implies$: Clearly $\displaystyle x\in\bigcap_{N\in\mathfrak{N}}\overline{N}$. Now, let $y$ be distinct from $X$. By assumption there exists neighborhoods $O_x,O_y$ such that $O_x\cap O_y=\varnothing$. It is relatively easy to prove (see Random Questions 1) that since $O_x\cap O_y=\varnothing$ that $y\notin\overline{O_x}$. It follows then that $y\notin\displaystyle \bigcap_{N\in\mathfrak{N}}\overline{N}$. Since $y$ was arbitrary the conclusion follows.

$\Longleftarrow$: Conversely, let $x,y\in X$ be distinct. Since $\displaystyle \{x\}=\bigcap_{N\in\mathfrak{N}}\overline{N}$ we see there exists some neighborhood $N$ of $x$ such that $y\notin \overline{N}$. It follows that $N,\left(\overline{N}\right)'$ are sets that contain $x,y$ respectively but not the other.

The conclusion follows. $\blacksquare$

This next theorem ends up being very important later on.

Theorem: Let $X$ be a topological space. Define $\Delta\subseteq X\times X$ to be $\Delta=\left\{(x,x):x\in X\right\}$. Then $X$ is Hausdorff if and only if $\Delta$ is closed in $X\times X$.

Proof:

$\implies$: Suppose that $X$ is Hausdorff, then we prove that $\Delta\subseteq X\times X$ is closed by proving its compliment is open. So, let $(x,y)\in\Delta'$ then $x\ne y$. And since $X$ is Hausdorff there exists there exists disjoint open sets $O_x,O_y$ which contain $x$ and $y$ respectively. Clearly $(x,y)\in O_x\times O_y$ but since $O_x\cap O_y$ we see that $\Delta\cap\left(O_x\times O_y\right)=\varnothing$ and so $O_x\times O_y\subseteq \Delta '$. The conclusion follows.

$\Longleftarrow$: Conversely, suppose that $\Delta \subseteq X\times X$ is closed. Then, $\Delta'$ is open, and so given any $x\ne y$ we see that $(x,y)\in O_x\times O_y\subseteq\Delta'$. Clearly this implies though that $O_x\cap O_y=\varnothing$ (otherwise if they shared a point $z$ then $(z,z)\in \left(O_x\times O_y\right)\cap \Delta$) from where it follows that $O_x,O_y$ are disjoint open sets containing $x,y$ respectively. The conclusion follows. $\blacksquare$

Now we begin by showing that Haudorff spaces react well with compact subspaces.

Theorem: Let $X$ be a Hausdorff space and $E\subseteq X$ compact, then for any $x\in X-E$ there exists an open sets $O_E,O_x$ such that $E\subseteq O_E,x\in O_x$ and $O_E\cap O_x=\varnothing$.

Proof: For each $e\in E$ there exists disjoint corresponding sets $O_e,O_{x_e}$. Clearly then $\left\{O_{e}\right\}_{e\in E}$ is an open cover for $E$. But, by assumption it must have a finite subcover $\left\{O_{e_1},\cdots,O_{e_n}\right\}$. It clearly follows that $\displaystyle E\subseteq\bigcup_{k=1}^{n}O_{e_n}$ and $\displaystyle x\in\bigcap_{k=1}^{n}O_{x_{e_k}}$ and that these two open sets are disjoint. The conclusion follows. $\blacksquare$.

Corollary: A compact subspace of a Hausdorff space must be closed.

Another very nice property is the next.

Theorem: Let $X$ be a Hausdorff space and let $E,G\subseteq X$ be compact subspaces. Then there exists disjoint open sets $O_E,O_G$ such that $E\subseteq O_E,G\subseteq O_G$.

Proof: By the previous theorem for each $e\in E$ there exists some $O_e,O_{G_e}$ which are open, disjoint, and contain $e, X$ respectively. It is clear that $\left\{O_e\right\}_{e\in E}$ is an open cover for latex $E$ and so by it’s compactness there exists a finite subcover $\left\{O_{e_1},\cdots,O_{e_n}\right\}$ and so $\displaystyle O_E=\bigcap_{k=1}^{n}O_{e_n}$ and $\displaystyle O_G=\bigcap_{k=1}^{n}O_{G_{e_k}}$ are the necessary open sets.

An obvious question is “Is the product of Hausdorff spaces Hausdorff?”. The answer is…:

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty collection of Hausdorff spaces then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is Hausdorff under the product topology.

Proof:Let $\bold{x},\bold{y}\in X$ be distinct. Then, it follows that $\pi_{\ell}(\bold{x})\ne\pi_\ell(\bold{y})$ for some $\ell\in\mathcal{J}$. But, since $X_j$ is Hausdorff there exists disjoint open sets $O_x,O_y$ such that $\pi_\ell(\bold{x})\in O_x,\pi_\ell(\bold{y})\in O_y$. It clearly follows that $\displaystyle \prod_{j\in\mathcal{J}}E_j,\prod_{j\in\mathcal{J}}G_j$ where $\displaystyle E_j=\begin{cases} O_x & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell \end{cases}$,$G_j=\begin{cases} O_y & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell \end{cases}$ are disjoint open sets containing $\bold{x},\bold{y}$ respectively. The conclusion follows. $\blacksquare$

Hausdorffness is also hereditary (in the sense that subspaces inherit it).

Theorem: Let $X$ be a Hausdorff space and $E\subseteq X$ a subspace. Then, $E$ is Hausdorff.

Proof: Let $x,y\in E$ be distinct, then clearly they are distinct in $X$. Thus, by assumption we have that there exists disjoint open sets $O_x,O_y$ such that $x,y$ are in them respectively. Clearly then $x\in O_x\cap E,O_y\cap E$ which are open in $E$ and latex $\left(O_x\cap E\right)\cap\left(O_y\cap E\right)=\left(O_x\cap O_y\right)\cap E=\varnothing\cap E=\varnothing$. The conclusion follows. $\blacksquare$

Now, a nice theorem which wraps up the motivation of the very first question in the last thread is the following:

Theorem: Let $X$ be a topological space, then if $X$ is Hausdorff the limit of a sequence is unique.

Proof:

$\implies$: Suppose that $\left\{x_n\right\}_{n\in\mathbb{N}}$ be a sequence in $X$ such that $x_n\to x$. Now, let $y\in X-\{x\}$ by assumption there exists disjoint open sets $O_x,O_y$ which contain $x,y$ respectively. By assumption though we have that all but finitely many of latex $\{x_n\}_{n\in\mathbb{N}}$ are in $O_x$ and so at most finitely many are in $O_y$. Thus, $x_n\not\to y$. The conclusion follows. $\blacksquare$

We now discuss some nice qualities of Hausdorff spaces and maps.

Theorem: Let $\varphi:X\mapsto Y$ be continuous and suppose that $Y$ is Hausdorff, then $\Gamma_\varphi\subseteq X\times Y$ is closed.

Proof: By an earlier discussion we know that $\varphi\times \iota_Y:X\mapsto Y\mapsto Y\times Y$ given by $(x,y)\mapsto (\varphi(x),y)$ is continuous. But, it is relatively easy to see that $\Gamma_\varphi=\left(\varphi\times\iota_Y\right)^{-1}(\Delta)$ and since $Y$ is Hausdorff we know that $\Delta$ is closed we know by $\varphi\times \iota_Y$‘s continuity that $\Gamma_\varphi$ is closed. the conclusion follows. $\blacksquare$

We know discuss quite a fruitful subject. But, first a defition

Kernel: Let $\varphi:X\mapsto Y$ be a mapping from topological spaces $X,Y$. Define $\ker \varphi=\left\{(x,x')\in X\times X:\varphi(x)=\varphi(x')\right\}$.

Remark: For those familiar with set theory this can be interpreted as the quotient set $X/ \sim$ where $x\sim y\Leftrightarrow \varphi(x)=\varphi(y)$

Theorem: If $\varphi:X\mapsto Y$ is continuous and $Y$ Hausdorff, then $\ker \varphi$ is closed.

Proof: Once again, by previous argument we know that $\varphi\times\varphi:X\times X\mapsto Y\times Y$ given by $(x,x')\mapsto (\varphi(x),\varphi(x'))$ is continuous. But, it is relatively easy to see that $\ker\varphi=\left(\varphi\times\varphi\right)^{-1}\left(\Delta_Y\right)$. Now, since $Y$ is Hausdorff we know that $\Delta_Y$ is closed and the argument is finished by noting $\varphi\times\varphi$‘s continuity. $\blacksquare$

Theorem: Let $\varphi:X\mapsto Y$ be a open surjection and $\ker\varphi$ closed, then $Y$ is Hausdorff.

Proof: Define $\varphi\times\varphi:X\times X\mapsto Y\times Y$ by $(x,x')\mapsto (\varphi(x),\varphi(x'))$. Now, it was proved earlier that the product of open sujrective maps is openand surjective and since $\varphi$ is open it follows that $\varphi\times\varphi$ is open and surjective. So consider the following lemma

Lemma: $\left(\varphi\times\varphi\right)\left(\left[\ker\varphi\right]'\right)=\Delta_Y'$.

Proof: Let $(y,y')\in \left(\varphi\times\varphi\right)\left(\left[\ker\varphi\right]'\right)$, then since $(y,y')=\left(\varphi(x),\varphi(x')\right)$ for some $(x,x')\in \left[\ker\varphi\right]'$. But, this implies that $\varphi(x)\ne\varphi(x')$ and so $\left(\varphi(x),\varphi(x')\right)=(y,y')\in\Delta_Y'$.

Conversely, let $(y,y')\in\Delta_Y'$ then $y\ne y'$, but since $\varphi\times\varphi$ is surjective we know that $(y,y')=(\varphi(x),\varphi(x'))$ for some $x,x'\in X$. But this implies that $\varphi(x)\ne\varphi(x')$ and thus $(x,x')\in\left[\ker\varphi\right]'$ thus $\left(\varphi\times\varphi\right)(x,x')=\left(\varphi(x),\varphi(x')\right)=(y,y')\in\left(\varphi\times\varphi\right)\left(\left[\ker\phi\right]'\right)$.

The conclusion follows. $\blacksquare$

The theorem now follows since $\ker\varphi$ is closed we see that $\left[\ker\varphi\right]'$ is open and since $\varphi\times\varphi$ is open and $\left(\varphi\times\varphi\right)\left(\left[\ker\varphi\right]'\right)=\Delta_Y'$ we see that $\Delta_Y'$ is open. Thus, $\Delta_Y$ closed. The conclusion follows from an earlier problem. $\blacksquare$

From this we can easily prove a specific case of a future theorem.

Theorem: Let $\left\{X_1,X_2\right\}$ be two topological spaces. Then if $X=X_1\times X_2$ is Hausdorff under the product topology then so are $X_1$ and $X_2$.

Proof: We know that $\pi_1:X\mapsto X_1$ is a surjective open mapping. It just remains to show that $\ker\varphi$ is closed. But, $\ker\varphi=\left\{\left((x_1,x_2),(x_1',x_2')\right):x_1=x_1'\right\}$. Note though that since $X\times X\approx X_1\times X_2\times X_1\times X_2\approx X_2\times X_2\left(\times X_1\times X_1\right)$ we must merely prove that the set $\left\{\left(x_2,x_2'\left(x_1,x_1'\right)\right):x_1=x_1'\right\}$ is closed in $X_2\times X_2\times\left(X_1\times X_1\right)$. But it is clear that this is merely $X_2\times X_2\times \Delta_{X_1}$. But, the product of finitely many closed sets is closed and so it follows that $\left\{\left(x_2,x_2'\left(x_1,x_1'\right)\right):x_1=x_1'\right\}$ is closed in $X_2\times X_2\times\left(X_1\times X_1\right)$. But, from the previous observation $X_2\times X_2\times\left(X_1\times X_1\right)\approx \left(X_1\times X_2\right)\times\left(X_1\times X_2\right)$ it follows that $\left\{\left(\left(x_1,x_2\right),\left(x_1',x_2'\right)\right):x_1=x_1'\right\}=\ker\varphi$ is closed in $X\times X$. The conclusion follows.

The exact same process works for $X_2$.

$\blacksquare$

Corollary: It follows by induction and the homeomorphic-associative properties of the product topology that $X=X_1\times\cdots\times X_n$ is Hausdorff implies that $X_k,\text{ }1\leqslant k\leqslant n$ is Hausdorff.

We can combine these two theorems concerning kernels to get.

Theorem: Let $\varphi:X\mapsto Y$ be a quotient map (continuous open surjection) then $Y$ is Hausdorff if and only if $\ker\varphi$ is closed.

Proof: Follows from the combination of earlier theorems.

For the next theorem we need another definition.

Agreement set: Let $\varphi,\psi:X\mapsto Y$ be two mappings, then we define the agreement set $A(\varphi,\psi)$ to be $\left\{x\in X:\varphi(x)=\psi(x)\right\}$

Theorem: Let $\varphi,\psi:X\mapsto Y$ be two continuous mappings and $Y$ Hausdorff, then $A(\varphi,\psi)\subseteq X$ is closed.

Proof:

Lemma: Define $\varphi\times\psi:X\mapsto Y\times Y$ by $x\mapsto (\varphi(x),\psi(x))$ then $\varphi\times\psi$ is continuous.

Proof: It is readily verified that $\left(\varphi\times\psi\right)^{-1}\left(U\times V\right)=\varphi^{-1}(U)\cap\psi^{-1}(V)$ from where the conclusion follows immediately. $\blacksquare$.

Using this lemma we must merely note that $\left(\varphi\times \psi\right)^{-1}\left(\Delta_Y\right)=A(\varphi,\psi)$ from where it follows from $\varphi\times \psi$‘s continuity and $\Delta_Y$ being closed (since $Y$ is Hausdorff) that $A(\varphi,\psi)$ is closed. $\blacksquare$.

Using this we can prove a very powerful theorem.

Theorem: Let $\varphi,\psi:X\mapsto Y$ be two continuous mappings and $Y$ be Hausdorff. Then if $A(\varphi,\psi)$ is dense then $\varphi=\psi$.

Proof: By the previous problem we know that $A(\varphi,\psi)$ is  closed and thus $A(\varphi,\psi)=\overline{A(\varphi,\psi)}=X$. The conclusion follows. $\blacksquare$

Also, the next theorem is nice.

Latex: Let $\varphi:X\mapsto Y$ be continuous and $Y$ Hausdorff, then the set of fixed points of $\varphi$ is closed.

Proof: Clearly the set of fixed points of $\varphi$ is $A(\varphi,\iota_X)$ but these are both continuous mappings. The conclusion follows $\blacksquare$.

Remark: It follows from this that if you have a non-identity continuous function into a Hausdorff and you have some point $x\notin \overline{A(\varphi,x)}$ that there exists some neighborhood $N$ of $x$ such that $\varphi(x)\ne x,\text{ }\forall x\in N$.

We now discuss a partially discussed converse to the fact that the product of Hausdorff spaces is Hausdorff, namely:

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty collection of Hausdorff spaces, then if $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is Hausdorff then so is $X_\ell$ for each $\ell\in \mathcal{J}$.

Proof: Let $x,y\in X_\ell$ be distinct. So, let $\left\{x_j\right\}_{j\in\mathcal{J}-\{\ell\}}\text{ }(1)$ be a fixed but arbitrary set of points. It follows that $\displaystyle \bold{x}=\prod_{j\in\mathcal{J}}\{e_j\},\bold{y}=\prod_{j\in\mathcal{J}}\{g_j\}$ are distinct points in $X$ where $\displaystyle e_j=\begin{cases} x & \mbox{if} \quad j=\ell \\ x_j & \mbox{if} \quad j\ne \ell\end{cases}$ and $\displaystyle g_j=\begin{cases} y & \mbox{if} \quad j=\ell \\ x_j & \mbox{if} \quad j\ne \ell \end{cases}$. Thus, by $X$‘s assumed Hausdorffness there exists disjoint open sets $O_x,O_y$ such that $\bold{x}\in O_x,\bold{y}\in O_y$. It is clear then that $\pi_\ell(O_x),\pi_\ell(O_y)$ are open sets containing $x,y$ respectively. Now, suppose that $z\in \pi_\ell(O_x)\cap \pi_\ell(O_y)$ then we can see that $\displaystyle \bold{z}\in O_x\cap O_y$ where $\displaystyle \bold{z}=\prod_{j\in\mathcal{J}}\{c_j\}$ where $\displaystyle c_j=\begin{cases} z & \mbox{if} \quad j=\ell \\ x_j & \mbox{if} \quad j\ne\ell \end{cases}$ where $x_j$ is as in $(1)$. This of course contradicts that $O_x\cap O_y=\varnothing$. It follows that $\pi_\ell(O_x)\cap\pi_\ell(O_y)=\varnothing$ and the conclusion follows. $\blacksquare$

Before we move onto the next thing we need another definition.

Perfect map: Let $\varphi:X\mapsto Y$ be continuous closed and surjective. Then if $\varphi^{-1}(\{y\})$ is compact for every $y\in Y$ we call $\varphi$perfect map.

Theorem: Let $\varphi:X\mapsto Y$ be perfect, then if $X$ is Hausdorff so is $Y$.

Proof: Let $\varphi(x),\varphi(y)\in Y$ be distinct, then $\varphi^{-1}(\{x\}),\varphi^{-1}(\{y\})\in X$ are disjoint compact subsets of $X$. It follows from $X$‘s Hausdorffness that there exists disjoint open sets $O_x,O_y$ which contain $\varphi^{-1}(\{x\}),\varphi^{-1}(\{y\})$ respectively. Clearly then $X-O_x,Y-O_y$ are closed sets containing $\varphi^{-1}(\{y\}),\varphi^{-1}(\{x\})$ respectively. By virtue that $\varphi$ is perfect we see that $\varphi\left(X-O_x\right),\varphi\left(X-O_y\right)$ are closed sets containing $y,x$ respectively. Clearly then $Y-\varphi\left(X-O_x\right),Y-\varphi\left(X-O_y\right)$ are open sets containing $x,y$ respectively.

Now, let $\varphi(z)\in Y-\varphi\left(X-O_x\right)$. Then $z\notin X-O_x$ and so $z\in O_x$ and thus $z\notin O_y\implies z\in X-O_y$ and thus $\varphi(z)\in \varphi\left(X-O_y\right)\implies \varphi(z)\notin Y-\varphi\left(X-O_y\right)$. Using the exact same argument we see that $\varphi(z)\in Y-\varphi\left(X-O_y\right)\implies \varphi(z)\notin Y-\varphi\left(X-O_x\right)$. It follows that $\left(Y-\varphi\left(X-O_x\right)\right)\cap\left(Y-\varphi\left(X-O_y\right)\right)=\varnothing$. The conclusion follows. $\blacksquare$

We lastly show that not every $T_1$ space is Hausdorff.

Example: Let $\mathbb{N}$ be given the cofinite topology (sets are open if their compliments are finite). Then clearly given any distinct $x,y\in \mathbb{N}$ we see that $\mathbb{N}-\{y\},\mathbb{N}-\{x\}$ are open sets which contain $x,y$ and don’t contain $y,x$ respectively. But, it is not possible to have disjoint open sets. To see this, let $O$ be open, then $O'$ is finite and so any disjoint set $V$ must be finite. But, this implies that $V'$ is infinite contradicting $V$‘s openness.

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March 2, 2010 -

## 2 Comments »

1. […] An (-dimensional continuous) atlas for consists of a set of (-dimensional) charts such that . A Hausdorff second countable (i.e. having a countable basis) topological space which admits a continuous […]

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2. […] let, for functions , be the agreement set . Note that is closed, indeed since is Hausdorff we know that the diagonal is closed in . Now, by the definition of the product topology, the map given […]

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