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...

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    $\begingroup$ Check this site out. $\endgroup$ – David Mitra Jun 29 '14 at 10:44
  • $\begingroup$ $\beta \omega$ satisfies the first three properties clearly. It takes a little work to show it is not first countable though. $\endgroup$ – Forever Mozart Jun 29 '14 at 10:49
  • $\begingroup$ @David Mitra: This is a fantastic site. You should write this as an answer so that more people know about it. $\endgroup$ – Moishe Kohan Jun 30 '14 at 0:46

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.

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  • $\begingroup$ 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 $p(a_i) \approx_\epsilon b_i$. $\endgroup$ – Mike F Apr 17 '18 at 13:44

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