A case that shows a separable topological space is not normal

Question

Let ($X,\tau$) be a separable topological space. If $X$ has a closed subspace $D$, which is equipotent to $\mathbb{R}$ (exists a bijection between $D$ and $\mathbb{R}$), and the induced topology on $D$ is discrete, then $X$ is not normal.

Could someone give me some ideas on how to solve this problem? Thank you!

What is $R$? Do you mean $\Bbb R$? – Robert Shore Jul 03 '20 at 01:01 — Robert Shore, Jul 03 '20 at 01:01
Yes, thank you for pointing out, I have edited it. – cbyh Jul 03 '20 at 01:03 — cbyh, Jul 03 '20 at 01:03

score 0 · Answer 1 · answered Jul 03 '20 at 01:15

This is an application of Jones’ lemma. Henno Brandsma proved the lemma in this answer, and the special case that you want is treated in Dan Ma’s Topology Blog.

Before you look at those, though, here’s a HINT:

If $X$ is normal, for each non-trivial subset $A$ of $D$ there is a continuous $f_A:X\to[0,1]$ such that $f_A[A]=\{0\}$ and $f[D\setminus A]=\{1\}$.
If two continuous functions from $X$ to $[0,1]$ agree on a dense set, they are equal. How many functions are there from a countable set to $[0,1]$?

A case that shows a separable topological space is not normal

1 Answers1