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Urysohn metrization theorem

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Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

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topological homotopy theory

Theorem named after Urysohn.

Statement and proof

The proof can be divided into three parts. Recall that “second-countable” means having a countable base. All spaces considered are assumed to be T 1T_1 (points are closed), so that “regular” means regular Hausdorff.

  1. A regular space with a countable base is normal.

  2. A normal space XX is a completely regular space, i.e., given a point xXx \in X and an open set UXU \subseteq X containing xx, there exists a continuous function f:X[0,1]f: X \to [0, 1] such that f(x)=1f(x) = 1 and f(x)=0f(x) = 0 for xUx \notin U.

  3. A completely regular space XX with a countable base can be embedded in the Hilbert cube [0,1] [0, 1]^\mathbb{N}. Since [0,1] [0, 1]^\mathbb{N} is metrizable, so is XX.

The second assertion being proved at Urysohn lemma, we prove the first and third assertions.

Proposition

A regular space XX with a countable base is normal.

Proof

Let A,BA, B be disjoint closed sets of XX. The collection of open sets UU such that UAU \cap A is inhabited and U¯B=\widebar{U} \cap B = \emptyset is, by regularity, an open covering of AA. By second-countability, we may index it as U 1,U 2,U_1, U_2, \ldots. Similarly, there is a countable open covering V 1,V 2,V_1, V_2, \ldots of BB such that V k¯A=\widebar{V_k} \cap A = \emptyset.

Now form open sets Y nU n k=1 n¬V k¯Y_n \coloneqq U_n \cap \bigcap_{k=1}^n \neg \widebar{V_k} and Z n=V n k=1 n¬U k¯Z_n = V_n \cap \bigcap_{k=1}^n \neg \widebar{U_k}. It is clear that the Y nY_n cover AA and the Z nZ_n cover BB. Moreover, Y mZ n=Y_m \cap Z_n = \emptyset for all m,nm, n. For if mnm \geq n say, then

Y mZ nY mV nY mV n¯U m¬V n¯V n¯=.Y_m \cap Z_n \subseteq Y_m \cap V_n \subseteq Y_m \cap \widebar{V_n} \subseteq U_m \cap \neg \widebar{V_n} \cap \widebar{V_n} = \emptyset.

It follows that mY m\bigcup_m Y_m and nZ n\bigcup_n Z_n are disjoint open sets containing AA and BB, respectively. This completes the proof.

Proposition

A completely regular space XX with countable base can be embedded in [0,1] [0, 1]^\mathbb{N}.

Proof

XX has a countable base \mathcal{B}. The set S={(U,V)×:U¯V}S = \{(U, V) \in \mathcal{B} \times \mathcal{B}: \widebar{U} \subseteq V\} is countable. For each s=(U,V)Ss = (U, V) \in S there is by the Urysohn lemma a continuous map g s:X[0,1]g_s: X \to [0, 1] such that g sg_s is identically 11 on U¯\widebar{U} and identically 00 on ¬V\neg V. The map

g:X[0,1] S:x(sg s(x))g: X \to [0,1]^S: x \mapsto (s \mapsto g_s(x))

is a continuous injection, since if xyx \neq y, we can find a pair s=(U,V)Ss = (U, V) \in S with xUx \in U and yVy \notin V, so that g s(x)=1g_s(x) = 1 differs from g s(y)=0g_s(y) = 0, whence g(x)g(y)g(x) \neq g(y). The subspace topology on XX induced from the monomorphism gg is contained in the given topology of XX, simply by continuity of gg. On the other hand, if WW is an open neighborhood of xx in XX, there exists a smaller open neighborhood VV \in \mathcal{B} and s=(U,V)Ss = (U, V) \in S such that g s(x)=1g_s(x) = 1 and g sg_s is identically 00 outside VV. Provided that g s(x)g s(y)<1g_s(x) - g_s(y) \lt 1, we see g s(y)0g_s(y) \neq 0, so yVy \in V. In other words, letting π s:[0,1] S[0,1]\pi_s: [0, 1]^S \to [0, 1] be the evident projection, we have g s=π sgg_s = \pi_s \circ g, so that the inverse image under gg of the subbase element π s 1((0,1])\pi_s^{-1}((0, 1]) is g s 1((0,1])VWg_s^{-1}((0, 1]) \subseteq V \subseteq W. This shows that the subspace topology induced by gg contains the topology of XX. It follows that g:X[0,1] S[0,1] g: X \to [0, 1]^S \cong [0, 1]^\mathbb{N} is an embedding into the Hilbert cube.

Last revised on April 7, 2019 at 08:43:33. See the history of this page for a list of all contributions to it.