Suppose that $(X,\mathcal{T})$ is a topological space. If we know that $X$ is compact can we assume that is also second countable ??
Because X is compact we have that $X=\cup_{i=1}^{n}V_{i}$, where $V_{i}$ are open sets. So for every $U\subset X$ we have that $U\subset \cup_{i=1}^{n}V_{i}$. Can we assume that by choosing some of the $V_{i}$ we will have that $U=\cup_{i\in J}V_{i}$, $J$ is countable.
In order for $(X,\mathcal{T})$ to have a countable topology base ??
By Tychonoff's theorem, the arbitrary product of compact spaces is compact. Take $X=\{0,1\}$ with the discrete topology and let $Y$ be the direct product of an uncountable collection of copies of $X$. Then $Y$ is compact, but $Y$ is not first countable (let alone second countable).
If $X$ is any set, the cofinite topology on $X$ is compact and any base for it has size $|X|$.
The $\epsilon$-order topology on the ordinal $\omega_1+1$ is compact Hausdorff but not 2nd-countable. It has an uncountable subset of isolated points: $\{x+1:x\in \omega_1\}\cup \{0\}.$
It is not even 1st-countable: The point $ \omega_1$ has uncountable character.
