Abstract Nonsense

Crushing one theorem at a time

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 \varphiperfect 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 - Posted by | General Topology, Topology | , , ,

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 […]

    Pingback by Topological Manifolds (pt. I) « Abstract Nonsense | August 30, 2012 | Reply

  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 […]

    Pingback by Holomorphic Maps and Functions (Pt. III) « Abstract Nonsense | October 4, 2012 | Reply


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