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 $y \in U$.$\square$
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