Abstract Nonsense

Crushing one theorem at a time

Thoughts about separation (Regularity and Complete Regularity)

So up until this point we’ve discussed spaces which are Kolomogorov, T_1, or Hausdorff. We now discuss the next property in the progression.

Regular (T_3): Let X be a topological space. We call X regular if it’s T_1 and given any closed subset E of X and any point x\in X-E there exists disjoint open neighborhoods of both.

Remark: It is quite common to call a space regular if it only has the second property mentioned and T_3 if is also T_1. We, obviously, do not have that sentiment.

It clearly follows that every regular space is Hausdorff since its T_1ness guarantees that  points are closed sets.

We first discuss the most functional aspect of regular spaces.

Theorem: Let X be a topological space, then X is regular if and only if given any point x\in X and any neighborhood N of x there exists some neighborhood U of x such that \overline{U}\subseteq N.


\implies: Suppose that X is regular and let x\in X be arbitrary. For any neighborhood N of x we have that X-N is a closed set not containing x. By assumption then there exists disjoint open neighborhoods U_x,U_{X-N} of each. We claim that \overline{U_x}\subseteq N.

To see this, let x\notin N then x\in U_{X-N} and since U_x\cap U_{X-N}=\varnothing it follows by an earlier problem that x\notin \overline{U_{x}}. The conclusion follows.

\Longleftarrow: Let latex E\subseteq X be closed and x\in X-E. Since they are disjoint we have that E' is an open set containing x and by assumption there exists some neighborhood U of x such that \overline{U}\subseteq E'. So, it follows quite readily that U,\left(\overline{U}\right)' are the sets we seek.


From this we get the following almost corollary.

Theorem: Let X be regular and x,y\in X be distinct. Then there exists neighborhoods U,V of x,y respectively such that \overline{U}\cap\overline{V}=\varnothing.

Proof: Since X is Hausdorff there exists disjoint open neighborhoods O_x,O_y of x,y respectively. By the previous problem these neighborhoods contain neighborhoods U,V such that \overline{U}\subseteq O_x,\overline{V}\subseteq O_y. These are clearly the neighborhoods we desired. \blacksquare

The following is the next step in the obvious progression which shows that in some sense compact sets act very similarly to points.

Theorem: Let X be regular and E,F\subseteq X be disjoint closed subsets such that E is compact. There exists disjoint open sets O_E,O_F such that E\subseteq O_E,F\subseteq O_F.

Proof: Clearly for each e\in E we clearly have that e\in X-F and by the regularity of X there exists disjoint open sets O_e,O_{F_e} such that e\in O_e,F\subseteq O_{F_e}. Clearly the class \left\{O_{e}\right\}_{e\in E} is an open cover of E and thus by E‘s compactness it has a finite subcover \left\{O_{e_1},\cdots,O_{e_n}\right\}. Clearly if \left\{O_{F_{e_1}},\cdots,O_{F_{e_n}}\right\} are the corresponding open sets containing F that E\subseteq\displaystyle \bigcup_{j=1}^{n}O_{e_j} and \displaystyle F\subseteq \bigcap_{j=1}^{n}O_{F_{e_j}} and both are disjoint. The conclusion follows. \blacksquare

We note that regularity is hereditary.

Theorem: Let X be regular and E a subspace of X then E is regular.

Proof: Let G\subseteq E be closed and x\in E-G. Since G is closed we have that G=E\cap C for some closed set C in X. Clearly x\notin C otherwise x\in C\text{ }x\in E\implies x\in C\cap E=G. Thus, by X‘s regularity there exists open sets O_x,O_C such that x\in O_x,\text{ }C\subseteq O_C and O_x\cap O_C. Clearly then E\cap O_x,E\cap O_C are disjoint sets which contain x and G respectively. The conclusion follows. \blacksquare

Next is the obvious.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of regular spaces, then \displaystyle X=\prod_{j\in\mathcal{J}}X_j is regular.

Proof: We actually need a little lemma first which we should have proved in the product topology section, namely:

Lemma: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a collection of topological spaces and \displaystyle X=\prod_{j\in\mathcal{J}}X_j. Then, given any class \left\{E_j\right\}_{j\in\mathcal{J}} such that E_\ell\subseteq X_\ell,\text{ }\forall \ell\in \mathcal{J} then \displaystyle \overline{\prod_{j\in\mathcal{J}}E_j}=\prod_{j\in\mathcal{J}}\overline{E_j}.


Let \displaystyle \bold{x}\in \prod_{j\in\mathcal{J}}\overline{E_j} and let N be any neighborhood of x. By assumption \pi_j(\bold{x})\in\overline{E_j} for each j\in\mathcal{J} and so there exists some point e_j \in E_j in \prod_{j}(N). It follows that \displaystyle \prod_{j\in\mathcal{J}}\{e_j\}\in \overline{\prod_{j\in\mathcal{J}}E_j}\cap N. Since N was arbitrary it follows that \bold{x} is an adherent point of \displaystyle \prod_{j\in\mathcal{J}}E_j and so \displaystyle \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}.

Conversely, let \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}. Let \ell\in \mathcal{J} be arbitrary, and let O_\ell be an open set containing \pi_\ell(\bold{x}). Since \pi_\ell^{-1}(O_\ell) is open in X it contains a point \displaystyle \bold{y}\in\prod_{j\in\mathcal{J}}E_j. Then \pi_\ell(\bold{y})\in O_\ell \cap E_\ell. The conclusion follows. \blacksquare

Let \bold{x}\in X be arbitrary and let N be any basic neighborhood of \bold{x}. Let j_1,\cdots,j_n be the finite number of indices such that \pi_{j_k}(\bold{x})\ne X_{j_k}. It is clear that \pi_{j_k}(N) is a neighborhood of \pi_{j_k}(\bold{x}) and by X_{j_k}‘s regularity there exists some U_{j_k}\subseteq \pi_{j_k}(N) such that \overline{U_{j_k}}\subseteq \pi_{j_k}(N). Clearly then \displaystyle \prod_{j\in\mathcal{J}}G_j where \displaystyle G_j=\begin{cases} U_{j_k} & \mbox{if} \quad j=j_k,\text{ }1\leqslant k\leqslant n \\ X_j & \mbox{if} \quad j\ne j_k\text{ }1\leqslant k\leqslant n\end{cases}. This is clearly a neighborhood of \bold{x} and \displaystyle \overline{\prod_{j\in\mathcal{J}}G_j}=\prod_{j\in\mathcal{J}}\overline{G_j}\subseteq N since \overline{U_{j_k}}\subseteq \pi_{j_k}(N). The conclusion follows. \blacksquare

We now prove a partial converse.

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty collection of topological spaces and \displaystyle X=\prod_{j\in\mathcal{J}}X_j be under the product topology. If X is regular then so is X_\ell for any \ell\in\mathcal{J}

Proof: Let F\subset X_\ell be closed and x\in X_\ell-F. Clearly then \displaystyle \bold{F}=\prod_{j\in\mathcal{J}}E_j where \displaystyle \begin{cases} F & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell \end{cases} is a  closed subset of X  and \displaystyle \bold{x}\in X-\bold{F} where \displaystyle \bold{x}=\prod_{j\in\mathcal{J}}\{e_j\} with \displaystyle e_j=\begin{cases} x & \mbox{if} \quad j=\ell \\ x_j & \mbox{if} \quad j\ne \ell \end{cases} where \left\{x_j\right\}_{j\in\mathcal{J}-\{\ell\}} is a fixed but arbitrary set in X-\bold{F}. Thus, by X‘s regularity there exists disjoint open sets O_{\bold{x}},O_{\bold{F}} such that \bold{x}\in O_{\bold{x}} and \bold{F}\subseteq O_{\bold{F}}. Clearly then x\in\pi_\ell(O_{\bold{x}}) and F\subseteq \pi_\ell(O_{\bold{F}}). It thus remains to show that they are disjoint.

Thus, assume that z\in \pi_\ell(O_{\bold{x}})\cap \pi_\ell(O_{\bold{F}}) then \displaystyle \bold{z}\in O_{\bold{x}}\cap O_{\bold{F}} where \displaystyle \bold{z}=\prod_{j\in\mathcal{J}}\{y_j\} where \displaystyle y_j=\begin{cases} z & \mbox{if} \quad j=\ell\\ x_j & \mbox{if} \quad j\ne\ell \end{cases} where x_j is as before. This is of course a contradiction. The conclusion follows.

Theorem: Let X be a regular space and \varphi:X\mapsto Y be perfect. Then, Y is regular.

Proof: Let F\subseteq Y be closed and \varphi(x)\in X-F. It follows from \varphi‘s perfection that \varphi^{-1}\left(\varphi(x)\right),\varphi^{-1}\left(F\right) are disjoint closed subsets of X such that \varphi^{-1}(\varphi(x)) is compact. By an earlier theorem there exists open sets O_x,O_F such that \varphi^{-1}(\varphi(x))\subseteq O_x,\varphi^{-1}\left(F\right)\subseteq O_F. Clearly then X-O_x,X-O_F are disjoint closed sets containing \varphi^{-1}\left(F\right),\varphi^{-1}(\varphi(x)) respectively. Once again, by \varphi‘s perfection we see that \varphi\left(X-O_x\right),\varphi\left(X-O_F\right) are closed sets containing F,\varphi(x) respectively. Thus, Y-\varphi\left(X-O_X\right),Y-\varphi\left(X-O_F\right) are open sets containing \varphi(x),F respectively. All that remains is to show that they are disjoint.

So, let \varphi(z)\in Y-\varphi\left(X-O_x\right) then z\notin X-O_x\implies z\in O_x and so z\notin O_F and so z\in X-O_F and thus \varphi(z)\in \varphi\left(X-O_x\right) and thus \varphi(z)\notin Y-\varphi\left(X-O_x\right). A similar argument shows that \varphi(z)\in Y-\varphi\left(X-O_F\right)\implies \varphi(z)\notin Y-\varphi\left(X-O_x\right). The conclusion follows. \blacksquare

Now, of course you might ask for an example a Hausdorff space which is not regular. Unfortunately, they are not as easy to come by. We outline such a topological space.

Example: Let \mathbb{R} be such that O\subseteq\mathbb{R} is open if  O=U-C where U is open in the usual topology and C is countable.

Our next example is a much more analysis motivated concept.

Tychonoff (completely regular T_\pi): Let X be a topological space, then it is called Tychonoff (completely regular/T_\pi) if it is T_1 and given any closed F\subseteq X and any x\in X-F there exists some \varphi\in \mathcal{C}\left[X,[0,1]\right] such that \varphi(x)=0 and \varphi(F)=1

It is clear that every Tychonoff space is regular since given any closed F\subseteq X, x\in X-F, and the corresponding \varphi\in\mathcal{C}\left[X,\mathbb{R}\right] the sets \varphi^{-1}\left(\left([0,\frac{1}{2}\right)\right),\varphi^{1}\left(\left(\frac{1}{2},1\right]\right) are disjoint open sets containing x,F respectively.

Theorem: Let X be Tychonoff and E a subspace of X. Then, E is Tychonoff.

Proof: Let F\subseteq E be closed and e\in X-F. Since E is a subspace of X we have that F=E\cap G for some closed G in X. Clearly e\notin G otherwise e\in G\cap E=F. Thus, e\in X-G and by X‘s complete regularity there exists some \varphi:X\mapsto [0,1] which is continuous and \varphi(x)=0,\varphi(G)=1. Thus, \varphi\mid E:E\mapsto [0,1] is clearly continuous and \varphi(x)=0,\varphi(F)=1. The conclusion follows. \blacksquare

Theorem: Let \left\{X_j\right\}_{j\in\mathcal{J}} then \displaystyle X=\prod_{j\in\mathcal{J}}X_j is Tychonoff under the product topology if  X_\ell is Tychonoff for each \ell\in\mathcal{J}.

Proof: Let \bold{E}\subseteq X be closed and \bold{x}\in X-\bold{E}. We may clearly choose a basic open set \mathcal{U} such that \bold{x}\in\mathcal{U}\subseteq X-\bold{E}. Let j_1,cdots,j_n be the finitely many indices such that \pi_{j_k}(\bold{U})\ne X_{j_k}. Since by assumption each of X_{j_1},\cdots,X_{j_n} are Tychonoff we may find some function \varphi_{k}:X_{j_k}\mapsto [0,1] such that \varphi\left(\pi_{j_k}(\bold{x})\right)=1,\text{ }\varphi\left(X_{j_k}-\pi_{j_k}(\bold{U})\right)=0. Let \psi_k=\varphi_k(\pi_k(\bold{y})) then \psi_k:X\mapsto [0,1] continuously and vanishes outside of \pi_{j_k}^{-1}\left(\pi_{j_k}\left(\bold{U}\right)\right). So, let \varphi:X\mapsto [0,1] be given by \varphi(\bold{y})=\psi_{1}(\bold{y})\cdots \psi_n(\bold{y}). Then this is a continuous mapping of X into [0,1] such that \varphi(\bold{x})=1\cdots 1=1 and \varphi\left(X-\bold{U}\right)=0. The conclusion follows.

\implies: Let


March 9, 2010 - Posted by | General Topology, Topology | , , , , , ,

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