Let $(X,\tau)$ a Normal topological space, $\beta \subseteq \tau$ a
basis of $\tau$. Then, there exists a embedding between $X$ and $[0,1]^J$, with $J \subseteq \beta \times \beta$.
My attempt was:
For any $\mathcal{F} \subseteq C(X,[0,1])$ a subset of the family of continuous functions from $(X,\tau)$ to $([0,1], \sigma)$ with $\sigma$ the usual topology, the function
$F:(X,\tau)$ $\rightarrow$ $([0,1]^{\mathcal{F}},\tau_{P})$ , $F(x)=\big( f(x) \big)_{f \in \mathcal{F}}$, with $\tau_P$ = the product topology
$X$ normal implies Tychonoff; $X$ and $Y$ are $T_1$ (points are closed)
Then F is a embedding.
maybe exists $G: [0,1]^J \rightarrow [0,1]^{\mathcal{F}}$ surjective and countinuous?
then $F \circ G$
Only hints please, not all the question, thankyou <3
Best Answer
You're not using normality or the special form of the index set that was hinted at, so your attempt is not correct for meta-reasons alone.
Try to use Urysohn functions for all pairs $(B, B')$ from the base with $\overline{B} \subseteq B'$ (for appropriate closed sets).
Do you know any theorems to recognise embeddings into a cube $[0,1]^J$? There is a quite famous one, also used to prove facts on Tychonoff spaces..