Some Point-set Topology: “Part 2: Separation Axioms”

The general structure of topological space is quite nice, but it is often not sufficient in practice. We often have to consider more certain types of topological spaces, that satisfy an additional separation axiom. In this section, we will present several such axioms, classified from less to more stringent. They basically tell us how efficient the open sets are in separating points or sets.

Definition 1.
Let (X,\tau) be a topological space.

  • (T_0) It is called T_0 (or Kolmogorov) if, for two points x and y in X with x\neq y, there exists an open set U such that x\in U and y \notin U or there exists an open set V such that x\notin V and y \in U.
  •  (T_1) It is called T_1 (or Fr\’echet) if, for for two points x and y in X with x\neq y, there exists an open set U such that x\in U and y \notin U and there exists an open set V such that x\notin V and y \in U.
  • (T_2) It is called T_2 (or Hausdorff) if, for two points x and y in X with x\neq y, there exists open sets U and V such that x \in U, y \in V, and U\cap V=\emptyset.
  •   (T_3) It is called T_3  if it is both T_1 and regular, that is to say, for any closed set C and x \notin C, there exist open sets U and V such that x \in U, C\subset V and U\cap V=\emptyset.
  • (T_{3\frac{1}{2}}) It is called (T_{3\frac{1}{2}}) (or Tychonoff) if it is both T_1 and completely regular, that is to say, for any closed set C and x \notin C, there exists a continuous function f:X\to [0,1], such that f(x)=0 and f\equiv 1 on C.
  • (T_4) It is called T_4 if it is both T_1 and normal, that is to say, for any closed sets C and D there exist open sets U and V such that C\subset U, D \subset V, and U\cap V=\emptyset.

The most important property is y far T_2: in T_2 spaces (which I will call Hausdorff spaces from now on), a convergent sequence has a unique limit. A topological space X is T_1 if, and only if, for all x\in X, \{x\} (the singleton containing x) is closed. A moment of reflexion shows that a T_{3\frac{1}{2}} topological space is T_3 (take U=f^{-1}([0,1/2[)) and V=f^{-1}(]1/2,1])). Clearly, a T_3  space is T_2, a T_2 space is T_1, and a T_1 space is T_0. It is also clear that a T_4 space is T_3, but it is not obvious that a T_4 space should be T_{3\frac{1}{2}}. This is indeed the case, and can be proved with the help of the following result.

Lemma 1. (Urysohn’s Lemma) If X is normal, if C and U are subsets of X with C closed, U open, and C\subset U, then there exists a continuous function f:X\to [0,1] such that f\equiv 0 on C and f\equiv 1 on X\setminus U.

This is very far from obvious. I have just learned about it, and I will try to convey my understanding of the proof (such as it is). Urysohn’s Lemma essentially tells us that we can separate closed sets using a continuous function. Let us try to do it on an example. We take as a topological space the segment [0,1], with its topology as a subspace of the real line \mathbb{R} equipped with the usual topology (the one I described at the beginning of the section on topological spaces). We take our closed set to be \{0\} and our open set to be [0,1[ (recall that we use the subspace topology). The function defined by f(x)=x then satisfies the three conditions we require: it is continuous, f(0)=0 and f(1)=1. To be more general, we have to define this function in such a way that an analogous definition could be found in a general normal space. It is reasonable (though I admit not obvious) to think that a definition of f(x) using a countable family of conditions on x would be suited to this task.

Let us consider the set
\displaystyle \mathbb{D}=\left\{\frac{j}{2^n}; n\in \mathbb{N}^*\mbox{ and } j\in\{1,\dots, 2^n-1\}\right\}.
We call it the set of dyadic numbers. Those are the numbers in ]0,1[ that can be written with a finite number of binary digits. For any r \in \mathbb{D}, we set U_r=[0,s[. Now (U_r)_{r\in \mathbb{D}} is a family of open sets of [0,1] such that:

  1. for all r\in \mathbb{D}, \{0\} \subset U_r;
  2. for all r\in \mathbb{D}, U_r\subset [0,1[;
  3. for all r and s in \mathbb{D} such that r<s, \overline{U}_r\subset U_s.

We can now define the function f using using the family (U_r)_{r\in \mathbb{D}}:
\displaystyle f(x)=\left\{\begin{array}{ll}  1 &\mbox{ if } x=1;\\  \inf\{r \in \mathbb{D}; x \in U_r\} &\mbox{ if } x\in [0,1[.\\  \end{array}\right.
This definition can be generalized.

Lemma 2.  Let X be a topological space, C a closed set and U an open set in X. Let us assume that (U_r)_{r \in \mathbb{D}} is a family of open sets in X satisfying:

  1.  for all r\in \mathbb{D}, C \subset U_r;
  2. for all r\in \mathbb{D}, \overline{U}_r\subset U;
  3. for all r and s in \mathbb{D} such that r<s, \overline{U}_r\subset U_s;
  4. U=\bigcup_{r\in \mathbb{D}}U_r.

Let us define the function f: X\to [0,1] by
\displaystyle f(x)=\left\{\begin{array}{ll}  1 &\mbox{ if } x\in X\setminus U;\\  \inf\{r \in \mathbb{D}; x \in U_r\} &\mbox{ if } x\in U.\\  \end{array}\right.
Then f is continous and f\equiv 0 on C.

Proof. Since C\subset U_{r} for all r \in \mathbb{D} and \inf \mathbb{D}=0, we have f\equiv 0 on C. Let us note (this will be used repeatedly ) that if x\in U_r, f(x) \le r, and if x\notin U_r, r\le f(x). It remains to prove that f is continuous. According to the statement at the end of the first section, it is enough to show that f is continuous at every point of X. Let x be a point of X and N a neighborhood of f(x) in [0,1].
Case 1: x\in X\setminus U. In that case f(x)=1. We recall that the intervals ]a-\varepsilon,a+\varepsilon[ (with a \in \mathbb{R} and \varepsilon>0) form a basis for the topological space \mathbb{R}. According to the definition of the subspace topology, there exists therefore \varepsilon>0 such that ]1-\varepsilon,1]\subset N. There also exists r\in \mathbb{D} such that r\in ]1-\varepsilon,1]. Then X\setminus \overline{U}_r is an open set, containing x. For all y \in U_r, we have f(y)> r according to the remark at the beginning, and thus f(y)\in N.
Case 2: x \in U with f(x)=0. With the same arguments than in the preceding case, there exists \varepsilon>0 such that [0,\varepsilon[\subset N. There exists r \in \mathbb{D} such that r\in [0,\varepsilon[. Then U_r is an open set containing x. For y\in U_r, we have f(y)\ge r, and therefore f(y)\in N.
Case 3: x\in U with f(x)>0. With the same arguments as before, there exists \varepsilon>0 such that ]f(x)-\varepsilon,f(x)+\varepsilon[\subset V. There exist also r and s in \mathbb{D} such that
\displaystyle f(x)-\varepsilon<r<f(x)<s<f(x)+\varepsilon.
Then U_s\setminus \overline{U}_r is an open set containing x such that f(y)\in N for all y\in U_s\setminus \overline{U}_r.
We have shown that in all cases there is a neighborhood N' of x, such that $f(N’)\subset N$. This shows that f is continuous at x and concludes the proof.

Let us turn to the proof of Urysohn’s Lemma. For n\in \mathbb{N}^*, we denote by \mathbb{D}_n the set
\displaystyle \left\{\frac{j}{2^n};j\in\{1,\dots,2^n-1\}\right\},
that is to say the set of numbers in ]0,1[ that have at most n binary digits. We have \mathbb{D}_n\subset\mathbb{D}_{n+1}  and \mathbb{D}=\bigcup_{n=1}^{\infty}\mathbb{D}_n. Let us fix an open set C and a closed set U in X. We will construct recursively a family of open set satisfying the hypotheses in Lemma 2.

Since X is normal, there exist two open sets U_{\frac{1}{2}} and V_1 such that C\subset U_{\frac{1}{2}}, X\setminus U \subset V_1 and U_{\frac{1}{2}}\cap V_1=\emptyset. In particular, U_{\frac{1}{2}} is contained in the closed set X\setminus V_1 and thus \overline{U}_{\frac{1}{2}}\subset X\setminus V_1\subset U. We have “constructed” the open set indexed by the element in \mathbb{D}_1=\{1/2\}.

Let now assume that, for some n\in\mathbb{N}^*, we have a family of open set (U_r)_{r\in \mathbb{D}_n} satisfying the hypotheses in Lemma 2. Let us now pick r \in \mathbb{D}_{n+1}\setminus \mathbb{D}_{n}.
Case 1: r=1/2^{n+1}. Then the closed sets C and X\setminus U_{\frac{1}{2^n}} are disjoint, and, proceeding as in the first step, we find an open set U_{\frac{1}{2^{n+1}}} such that
C\subset U_{\frac{1}{2^{n+1}}} and \overline{U}_{\frac{1}{2^{n+1}}}\subset U_{\frac{1}{2^n}}.
Case 2: r=(2j+1)/2^{n+1} for some j \in \{1,\dots, 2^n-3\}. Then the closed sets \overline{U}_{\frac{j}{2^n}}  and X\setminus U_{\frac{j+1}{2^n}} are disjoint, and, proceeding as before, we find an open set U_r such that \overline{U}_{\frac{j}{2^n}}\subset U_r and \overline{U}_r \subset U_{\frac{j+1}{2^n}}.
Case 3: r=(2^{n+1}-1)/2^{n+1}. Then the closed sets \overline{U}_{\frac{2^n-1}{2^n}} and X\setminus U are disjoint and, proceeding as before, we find an open set U_r such that \overline{U}_{\frac{2^n-1}{2^n}} \subset U_r and \overline{U}_r \subset U.

We can therefore build recursively a family of open sets (U_{r})_{r\in \mathbb{D}} satisfying the hypotheses of Lemma 2, and thus, by applying Lemma 2, a continuous function f:X\to [0,1] such that f\equiv 0 on C and f\equiv 1 on X\setminus U. This proves Urysohn’s Lemma.

We can now prove that a T_4 space is T_{3\frac{1}{2}} by applying Urysohn’s Lemma to the closed set \{x\} and the open set X\setminus C.

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