We know that every ferfect set $E$ in a complete metric space $X$ is uncountable. My question is if there exists a metric space which is not complete, but every ferfect set in it is uncountable. The ferfect set here means a closed set which has no isolated points.
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1$\begingroup$ (I first thought about giving an example with no "ferfect" sets, but seemed to fail in doing so...) How about just a half-open interval $(0,1]$? Are you sure you have formulated your question correctly? $\endgroup$– Tapio RajalaCommented Mar 28, 2012 at 14:23
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1$\begingroup$ @Tapio: $(0,1]$ is not complete with the usual distance but completely metrisable. $\endgroup$– Jochen WengenrothCommented Mar 28, 2012 at 14:49
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1$\begingroup$ So leaving out a countable subset does not destroy the property that each perfect set is uncountable. So for example $[0,1]\setminus (\mathbb{Q}\cap[0,1])$ might be a more interesting example. $\endgroup$– HenrikRüpingCommented Mar 28, 2012 at 15:23
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2$\begingroup$ Unfortunately, $[0,1]\backslash(\mathbf{Q}\cap[0,1])$ is also completely metrisable: take the distance to be one over the first index at which the continued fraction expressions differ. $\endgroup$– Sean EberhardCommented Mar 28, 2012 at 15:56
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1 Answer
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It is not so easy to construct an example of this kind, I think, because of the Hurewicz theorem: if $X$ is a coanalytic separable metrizable separable space, then either it is Polish (in particular, its perfect subspaces are uncountable) or it contains a closed subset homeomorphic to the rationals (so, a countable perfect closed subset). This is corollary 21.21 in Kechris's book "Classical Descriptive Set Theory".
Allowing for the axiom of choice, a Bernstein set will provide a counterexample. This is a subspace $A$ of the real line built by transfinite recursion in such a way that both $A$ and its complement intersect any nonempty perfect subspace of the reals. (this construction is also explained by Kechris, 8.24).
EDIT: my earlier argument was incomplete, as pointed out by Andreas Blass, so I'm following his idea below to show that a Bernstein set is indeed a counterexample.
If $C$ were a countable closed perfect subset of $A$, then the closure of $C$ would be perfect and closed in $\mathbb R$, hence uncountable. So $\overline C \setminus C$ is an uncountable Borel set contained in the complement of $A$; since an uncountable Borel set contains a Cantor set, the complement of $A$ must then contain a Cantor set, which is impossible since $A$ is a Bernstein set.
Thus $A$ does not contain any countable closed perfect set, and $A$ is certainly not completely metrizable since $A$ is not even Borel.
It seems plausible to me that, if you want an example which is a subset of $\mathbb R$, then you need some form of the axiom of choice, though I have not thought it through carefully; maybe the residing set theorists will know just how much choice is needed to answer this question...
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1$\begingroup$ @Julien: A Bernstein set $A$ contains no perfect set of the ambient space ($\mathbb R$ in the situation you described), but the OP defined "perfect" (or "ferfect") in terms of closed subsets of the space $X$ itself. So I don't think the "there are none!" argument suffices. The example might work anyway, because if a countable set were perfect in $A$, its closure in $\mathbb R$ would contain an uncountable Borel subset of the complement of $A$, and there are none of those. (Caution: I haven't checked this argument carefully, and I probably won't have time to do so soon.) $\endgroup$ Commented Mar 28, 2012 at 17:37
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$\begingroup$ @Andreas: thank you, you are right, I did not think about this carefully enough. I think your suggestion works and I modified my answer accordingly. $\endgroup$ Commented Mar 28, 2012 at 18:27