1

I am trying to show that given a set of topological spaces $(X_i, \tau_i)$, if $X = \prod_{i \in \mathbb{N}} X_i$ with the product topology is separable, then every space $X_i$ is separable. The other direction I didn't have any problems with.

Usually to prove such things I've seen that the homeomorphism between $X_1$ to $ X_1 \times \prod_{i=2}^{\infty} \left \{ x_i \right \}$ for some $(x_i)_{i \in \mathbb{N}} \in X$ is used together with hereditary topological properties. But since separability is only hereditary to open subsets I don't think this approach works here.

Sumanta
  • 9,534
paxtibimarce
  • 645
  • 5
  • 13
  • @SumantaDas So you're saying if $A$ is the dense subset in $X$, I should use $\pi_i (A)$? – paxtibimarce Oct 12 '21 at 12:11
  • 2
    Yup! Let $\mathcal D$ be a countable dense subset of $X$, and $\pi_i$ be the projection on the $i$-th component. Take any open subset $U_i$ of $X_i$, by continuity of $\pi_i$, the set $\pi_i^{-1}(U_i)$ is open in $X$, and let ${d_n}\in \mathcal D\cap \pi_i^{-1}(U_i)$. Then, $\pi_i$ sends ${d_n}$ to $d_i\in \pi_i(\mathcal D)\cap U_i$. – Sumanta Oct 12 '21 at 12:19
  • 1
    Projections being continuous is all you need. – Henno Brandsma Oct 12 '21 at 13:10
  • 1
    @SumantaDas openness is irrelevant, continuity is the only thing that matters here. – Henno Brandsma Oct 12 '21 at 13:18

1 Answers1

2

You are right that being separable is not hereditary in general. However it is well known that a continuous image of a separable space is separable. Observe that each $X_i$ is a continuous image of $X$ via the projection $\pi_i:X\to X_i$. Note that this works for any product, not only countable.

freakish
  • 42,851