# Nonmetrizable compact totally disconnected spaces without isolated points

Brouwer proved that a topological space is homeomorphic to the Cantor set if and only if

1) it's non-empty,

2) it's compact,

3) it's totally disconnected,

4) it has no isolated points, and

5) it's metrizable.

What are some nice examples of spaces that meet conditions 1)-4) but not 5)?

Perhaps it's worth noting that a compact Hausdorff space is metrizable if and only if it's second-countable, by a spinoff of the Urysohn Metrization Theorem. I suppose I'd prefer my examples to be Hausdorff. If so, they must be non-second-countable. Such spaces might thus be seen as 'like the Cantor set, but bigger'.

[1] L. E. J. Brouwer, Over de structuur der perfekte puntverzamelingen, Amst. Ak. Versl. 18 (1910) 833-842.

• Uncountable direct product of copies of the Cantor set. Nov 13, 2013 at 7:15
• $\beta N \setminus N$. Nov 13, 2013 at 7:28
• @marguax: Or in other words, the product topology on $2^\kappa$ for uncountable $\kappa$. Nov 13, 2013 at 7:30
• The dual (in the sense of Stone duality) of an uncountable atomless Boolean algebra. Nov 13, 2013 at 10:40
• To keep going, one can list the maximal ideal space of the commutative Banach algebra $L_\infty(\mu)$ for some atomless measure (this is a special case of shane's comment). Nov 13, 2013 at 11:52

## 1 Answer

The boolean algebra of clopen subsets of a space satistying conditions 1,2 and 3 is always a base for the topology (i.e. the space is zero-dimensional). So the space is homeomorphic to the Stone space of that boolean algebra. The space will satisfy condition 4) if and only if the algebra is atomless. Non-metrizability corresponds exactly to the algebra being uncountable. So Shane's comment actually gives a characterization of such spaces.

There is only one countable atomless boolean algebra and that is why 1)-5) characterizes a single space which happens to be Cantor's space.

The space $2^\kappa$ in Nates's comment correspond to the free boolean algebra in $\kappa$ generators. A space satisfies 1)-4) if and only if it is homeomorphic to a perfect subspace of $2^\kappa$ for some $\kappa$ (the smallest $\kappa$ for which this happens coincides with the weight of the space, and such space will be metrizable if and only if it has countable weight).