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The Baire category theorem implies that a nonempty complete metric space without isolated points must be uncountable. In many situations I have encountered, the "natural examples" of complete metric spaces without isolated points (of a certain type, or possibly with some additional structure) in fact have at least continuum cardinality. This is not so surprising, since if Cantor's continuum hypothesis holds, then uncountable is equivalent to at least continuum cardinality.

However, if we do not wish to assume CH -- and, ever since Godel and Cohen proved that (G)CH is independent of ZFC set theory, this seems to be the prevalent attitude -- what can be said about the existence of such spaces of uncountable cardinality less than the continuum?

I asked this question to someone before, and I seem to remember that it is known that one cannot unconditionally improve the conclusion of this application of Baire category to say "continuum cardinality". But could someone say a little bit about how this goes? Preferably in words that are comprehensible to a non-set theorist like myself?

Addendum: Thanks to Sergei Ivanov for a quick and convincing answer: evidently I was making things much more complicated than I needed to. Just to get myself reoriented properly, I would like to try to remember where the set-theoretic subtleties come in. Suppose I ask about the conclusion of BCT itself, rather than this particular corollary: not assuming CH, what can we say about the minimal cardinality of a covering family of nowhere dense subsets in a complete metric space?

Second Addendum: I was even more turned around than I had realized: I was (i) worrying needlessly about uncountable cardinals smaller than the continuum and (ii) not worrying enough about cardinals greater than the continuum! In particular, I was under the misimpression that for any cardinal $\kappa \geq 2^{\aleph_0}$, $\kappa^{\aleph_0} = \kappa$. This led me to incorrectly guess the strong form a classical theorem of F.K. Schmidt. I think I have it right now: if you are interested, see pp. 13-16 of


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Maybe the generalization you originally had in mind was that the union of fewer than $2^{\aleph_0}$ (instead of countably many) closed nowhere dense sets cannot cover the whole space? –  François G. Dorais Apr 28 '10 at 15:08

2 Answers 2

up vote 11 down vote accepted

A complete space without isolated points has at least continuum cardinality. At least if you agree to use (some form of) Axiom of Choice.

Choose two disjoint closed balls $B_1$ and $B_2$. Inside $B_1$, choose disjoint closed balls $B_{11}$ and $B_{12}$. Inside $B_2$, choose disjoint closed balls $B_{21}$ and $B_{22}$. And so on. At $n$th step, you have $2^n$ disjoint balls indexed by binary words of length $n$, and you choose two disjoint balls of level $n+1$ inside each ball of level $n$. This is possible because the balls are not single points. Make sure that radii go to zero. Now you have continuum of sequences of nested balls each having a common point.

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Hmm, rather embarrassingly this simple answer sure looks like it is correct. I must be a little confused about my previous question: I guess it applies to the Baire Category Theorem itself, not this special case. –  Pete L. Clark Apr 28 '10 at 10:08
Notice that this standard argument shows that a complete space without isolated points contains a subset homeomorphic to the Cantor set. I always prove this in a first course in real analysis because it is so useful for doing other things. It follows immediately that $[0,1]$ is the continuous image of every uncountable Polish space. –  Bill Johnson Apr 28 '10 at 18:51
The argument you give uses only DC, rather than AC. I'm not sure what might happen without DC; for example, if there are infinite Dedekind finite sets. Can they admit a complete metric? They cannot have size continuum, for this would violate Dedekind finiteness. –  Joel David Hamkins Apr 29 '10 at 2:00

The number of meager sets needed to cover the real line is a "cardinal invariant of the continuum". It is one of the invariants in Cichoń's diagram. In particular, it is Cov(K) in Cichoń's diagram on Wikipedia.

Looking at nowhere-dense sets instead of meager sets would not change this invariant, because of basic cardinal arithmetic.

I am not certain, off the top of my head, if the invariants are the same for every uncountable complete separable metric space.

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Yes, category invariants are always the same for every uncountable Polish space. This is because every perfect Polish space has a dense $G_\delta$ set which is homeomorphic to Baire space. –  François G. Dorais Apr 28 '10 at 12:03

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