I need an example for a compact, Hausdorff, separable space which is not first-countable. I tried to look for it for some time without success...
Asked
Active
Viewed 2,111 times
7
Martin Sleziak
- 53,687
RyArazi
- 379
-
14Check this site out. – David Mitra Jun 29 '14 at 10:44
-
$\beta \omega$ satisfies the first three properties clearly. It takes a little work to show it is not first countable though. – Forever Mozart Jun 29 '14 at 10:49
-
@David Mitra: This is a fantastic site. You should write this as an answer so that more people know about it. – Moishe Kohan Jun 30 '14 at 00:46
1 Answers
11
The space $I^I$ (i.e., product of $\mathfrak c$-many copies of the unit interval $I=[0,1]$ is a compact Hausdorff space.
It is not first-countable, see here: Uncountable Cartesian product of closed interval
It is separable by Hewitt-Marczewski-Pondiczery theorem, see here: On the product of $\mathfrak c$-many separable spaces
As pointed out in a comment, we could also prove separability by directly showing that polynomials with rational coefficients form a countable dense subset. See also this answer for a similar approach in a slightly different space.
Martin Sleziak
- 53,687
-
No big theorem is needed to show this space is separable. For example, polyomial functions with rational coefficients do the job. Given $\epsilon > 0$ and finitely many points $a_i,b_i$ in $I$, there exists a rational polynomial $p$ with $p(a_i) \approx_\epsilon b_i$. – Mike F Apr 17 '18 at 13:44