# A compact Hausdorff space that is not metrizable

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.

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You need to find an example of a compact Hausdorff space which is not second-countable (as metrizability is equivalent to this) –  Mariano Suárez-Alvarez Oct 22 '11 at 22:17
I'd be surprised if that didn't come up many times before... anybody? –  t.b. Oct 22 '11 at 22:27
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. –  gary Oct 23 '11 at 2:01

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 $[0,\omega_1]$.

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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.

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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?)

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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, 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 so easier to disprove metrizability.

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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.
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? –  t.b. Oct 23 '11 at 3:17