TITLE: Compact topological space not having Countable Basis?
QUESTION [11 upvotes]: Does there exist a compact topological space not having countable basis?
I have constructed a product space from uncountably many unit intervals $[0,1]$, endowed with the product topology. Tychonoff's Theorem shows that this topological space is compact Hausdorff space, but I'm not sure how to prove that this space does not have any countable basis. Need help!
Any other suggestions on constructing such topological space (using Tychonoff's theorem) would be a huge help, too!

REPLY [6 votes]: Here's a very easy way to check that certain spaces cannot have a countable basis (though it will not detect all spaces that lack a countable basis).  Note that if $\mathcal{B}$ is a basis for a topological space $X$, then for every open set $U\subseteq X$ there exists a subset $\mathcal{A}\subseteq\mathcal{B}$ such that $U$ is the union of the elements of $\mathcal{A}$.  This means that there can be at most $2^{|\mathcal{B}|}$ different open subsets of $X$.  So if $X$ has a countable basis, it can have at most $2^{\aleph_0}$ open subsets.
This makes it easy to find compact spaces with no countable basis.  For instance, let $X$ be any compact $T_1$ space with more than $2^{\aleph_0}$ points (e.g., a product of copies of $[0,1]$, as long as you have enough copies that the cardinality of the product is greater than $2^{\aleph_0}$).  Then $X$ cannot have a countable basis, since for each $x\in X$ there is a distinct open set $X\setminus\{x\}$.