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.

  • $\begingroup$ You need to find an example of a compact Hausdorff space which is not second-countable (as metrizability is equivalent to this) $\endgroup$ – Mariano Suárez-Álvarez Oct 22 '11 at 22:17
  • $\begingroup$ I'd be surprised if that didn't come up many times before... anybody? $\endgroup$ – t.b. Oct 22 '11 at 22:27
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    $\begingroup$ I think this should work:The uncountable product of non-trivial metric spaces is not metrizable. Take, then, e.g., uncountably-many copies of [0,1] with the subspace metric. The product is compact, by Tychonoff,and Hausdorff, but not metrizable. $\endgroup$ – gary Oct 23 '11 at 2:01

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.


A compact metric space is separable. A metric space is first countable. Fairly simple examples of compact Hausdorff spaces which are neither include the one-point compactification of an uncountable discrete space and the ordinal space $[0,\omega_1]$.


The linearly ordered space $\omega_1+1$ is an easy example. Another is $\beta\omega\setminus\omega$, the Čech-Stone compactification of $\omega$. $\{0,1\}^\kappa$ for $\kappa>2^\omega$ will also work, since it isn’t separable.

  • $\begingroup$ $\{0,1\}^\kappa$ for $\kappa>\omega$ works as well as the density of $C(K)$ is the same as weight of $K$ and $\{0,1\}^\kappa$ has weight $\kappa$. We have $K$ metrisable if and only if $C(K)$ is separable (has density $\omega$). $\endgroup$ – Tomasz Kania Sep 19 '14 at 15:35

You are missing hypothesis on $X$ and $Y$. And I see no reason to restricting yourself to continuous functions. Just take an uncountable $X$ and some $Y$ with more then one element with the topology for point-wise convergence. Then you will basically get the following case.

The space $\{0,1\}^{\mathbb{R}}$ with the product topology is compact but is not metrizable. If this topology was metrizable, then there would be a family of neighborhoods $V_1 \supset V_2 \supset \dotsc$ for $(0,0,0,\dotsc)$ such that $$\{(0,0,0,\dotsc)\} = \bigcap V_n.$$ But this cannot happen! (why?)


The uncountable product of non-trivial metric spaces is not metrizable. Take, then, e.g., uncountably-many copies of [0,1] with the subspace metric. The product is compact, by Tychonoff, and it is Hausdorff, but it is not metrizable.

Edit As t.b pointed out, my answer does not add much that was not previously said, so hopefully this comment will add something: notice that the metric $\Sigma^{\infty}_{i=1} \frac{d_i(x_i,y_i)}{2^i}$ (which is one of the metrics that) generates the product metric for a countably-infinite product will not work for an uncountably-infinite product since, among other things, the sum will diverge for pairs of points that have more than countably-many different entries from each other. Of course, this is not a disproof.

  • $\begingroup$ gary: no offense, but in view of your meta complaint why do you post an answer that is an $\varepsilon$-variation of what was already given by Brian and especially André? $\endgroup$ – t.b. Oct 23 '11 at 3:17
  • $\begingroup$ I don't see how this relates to my complaint, which was about anonymous downvoting.Still, I did not read their posts before posting my own answer. $\endgroup$ – gary Oct 23 '11 at 3:24
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    $\begingroup$ anonymous downvoting and downvoting in general, I think. Anyway, look: Chris gives 2 examples. Brian gives 3 examples including a variation of yours. André gives the metrizability argument and adds other information - then what do you really add? Not reading the other answers is not an excuse, especially if you're more than half an hour late to the party. I'm just saying: if you don't like downvotes, you shouldn't do that. $\endgroup$ – t.b. Oct 23 '11 at 3:30
  • $\begingroup$ O.K, thanks for the comment; I will consider it. What upset me was the lack of comments associated with the downvote. I was actually thinking of including an argument about the product not being 1st countable, in case the OP was interested. Still, weren't mine and Andre's post almost simultaneous? I saw the 1hr ago marker on both, or is this a discrete marker after 1hr? $\endgroup$ – gary Oct 23 '11 at 3:44
  • $\begingroup$ If you hover your mouse over the hour marker you can see when exactly the answers were posted. André posted at 1:43, you at 2:08. André outlines the first countability argument already. $\endgroup$ – t.b. Oct 23 '11 at 3:49

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