Is there an example of a compact Hausdorff space that is not metrizable?
I was thinking maybe the space of continuous functions $f: X \rightarrow Y$ between topological spaces $X, Y$, might work, but I'm sure I'm missing some conditions.
[Math] A compact Hausdorff space that is not metrizable
compactnessexamples-counterexamplesgeneral-topology
Best Answer
I like the double arrow space, as a classical example:
Let $X = [0,1] \times \{0, 1\}$, where $X$ has the lexicographical ordering $(x,i) < (y,j)$ iff $x < y$ or ( $x = y$ and $i=0, j=1$). Then $X$ in the order topology is separable ($\mathbb{Q} \times \{0, 1\}$ is countable and dense), compact, hereditarily normal and perfectly normal and first countable, but its square is not hereditarily normal (it contains the square of the Sorgenfrey line, which can be seen as the subspace $(0,1) \times \{1\}$ ). So even very nice compact spaces need not be metrizable. Proofs can be found here, e.g.
The lexicographically ordered unit square, also discussed in the previous link, is another example, which is less nice (not separable), but for which it is easier to disprove metrizability, as it's compact and not separable.
Another classical example from the same Aleksandrov paper IIRC, is the double of $[0,1]$, which is also $X = [0,1] \times \{0,1\}$, and where a basic neighbourhood of $(x,1)$ is just $\{(x,1)\}$ (these are isolated points), but a basic neighbourhood of $(x,0)$ is of the form $O = (I \times \{0,1\}) \setminus \{(x,1)\}$ where $I$ is any standard open set of $[0,1]$ containing $x$. This is compact as $[0,1]$ is, but has an uncountable discrete open subspace $[0,1] \times \{1\}$, making it not separable and not second countable, so not metrisable.