5.24 Embeddings in Cubes
Definition 5.24.1 (Completely Regular).label Let $X$ be a topological space, then $X$ is completely regular if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in C(X; [0, 1])$ such that $f(x) = 1$ and $f|_{E} = 0$.
Definition 5.24.2 (Separation of Points and Closed Sets).label Let $X$ be a topological space and $\cf \subset C(X; [0, 1])$, then $\cf$ separates points and closed sets if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in \cf$ such that $f(x) \not\in \ol{f(E)}$.
Proposition 5.24.3.label Let $X$ be a $T_{1}$ space, then the following are equivalent:
- (1)
$X$ is completely regular.
- (2)
There exists $\cf \subset C(X; [0, 1])$ that separates points and closed sets.
Proof. (2) $\Rightarrow$ (1): Let $E \subset X$ closed and $x \in X \setminus E$, then there exists $f \in \cf$ such that $x \not\in \ol{f(E)}$. By Urysohn’s lemma, there exists $\phi \in C([0, 1]; [0, 1])$ such that $\phi(f(x)) = 1$ and $\phi(f(E)) = 0$.$\square$
Definition 5.24.4 (Embedding in Cube).label Let $X$ be a topological space, $\cf \subset C(X; [0, 1])$, and
then:
- (1)
$e \in C(X; [0, 1]^{\cf})$.
- (2)
If $\cf$ separates points, then $e$ is injective.
- (3)
If $X$ is $T_{1}$ and $\cf$ separates points and closed sets, then $e$ is an embedding.
The mapping $e$ is the mapping of $X$ into the cube $[0, 1]^{\cf}$ associated with $\cf$.
Proof, [Proposition 4.53, Fol99]. (1): By (U) of the product topology.
(3): Since $X$ is $T_{1}$, $e$ is injective by (2). Let $x \in X$ and $U \in \cn_{X}^{o}(x)$, then there exists $f \in \cf$ such that $f(x) \not\in \ol{f(U^c)}$. In which case, there exists $V \in \cn_{[0, 1]}^{o}(f(x))$ such that $V \cap f(U^{c}) = \emptyset$. Thus for any $y \in X$ with $\pi_{f}(e(y)) \in V$, $f(y) \not\in f(U^{c})$, so $y \in U$.$\square$
Proposition 5.24.5.label Let $X$ be a $T_{1}$ space, then the following are equivalent:
- (1)
$X$ is completely regular.
- (2)
There exists a uniformity $\fU$ on $X$ that induces the topology on $X$.
Proof. (1) $\Rightarrow$ (2): By Definition 6.1.17.
(2) $\Rightarrow$ (1): By Definition 5.24.4, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$.$\square$
Theorem 5.24.6 (Uryson Metrisation Theorem).label Let $X$ be a second countable regular space, then $X$ is metrisable.
Proof. By Proposition 5.9.3, $X$ is normal. Let $\cb \subset 2^{X}$ be a countable base for $X$, and let
By Urysohn’s Lemma, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF}\in C(X; [0, 1])$ such that $f|_{E} = 1$ and $f|_{F^c}= 0$. For any $x \in X$ and $U \in \cn^{o}_{X}(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E}\subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c}= 0$. Therefore
is a countable family of continuous functions that separate points and closed sets. By Definition 5.24.4, $X$ embeds into $[0, 1]^{\cf}$, which is metrisable by Proposition 8.1.3.$\square$
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