Showing $\mathbb{R}$ with the standard topology union irrational subsets is normal.

Question

We are given that $\mathcal{T}$ is the standard topology on $\mathbb{R}$ and $\mathcal{O}=\{O\in\mathcal{P}(\mathbb{R}):O=U\cup A, U\in\mathcal{T}, A\subseteq\mathbb{R}\backslash\mathbb{Q}\}$. It can easily be shown that $(\mathbb{R},\mathcal{O})$ is a topology. Now we want to show that this space is normal, and as a hint, we are given that every metric space is normal.

So I assume that the idea is to show that $(\mathbb{R},\mathcal{O})$ is metrizable, by finding a suitable metric $d$ that induces the same topology $\mathcal{O}$. However, I have no idea where to start to find such a metric. Is this the correct approach?

score 4 · Accepted Answer · answered Jun 02 '20 at 15:51

4

That space is the Michael line. It is not metrizable, but it is paracompact and therefore normal. There is a proof of paracompactness at the link, but the result is quite easy using the fact that the usual topology on $\Bbb R$ is paracompact, so it’s worth trying on your own if you’ve done anything with paracompactness.

I gave a very different proof that the Michael line is normal in this answer to an earlier question.

answered Jun 02 '20 at 15:51

Brian M. Scott

631,399

I see, thank you! I will search a bit more on paracompact spaces and check your answer. Do you have any idea what the hint might be referring to? – Dylan Zammit Jun 02 '20 at 16:20
@DylanZammit: You’re welcome! No, I confess to being quite confused by the hint. – Brian M. Scott Jun 02 '20 at 16:39
The space is also monotonically normal and hence hereditarily normal. – Henno Brandsma Jun 02 '20 at 21:54

Showing $\mathbb{R}$ with the standard topology union irrational subsets is normal.

1 Answers1