When I read and try to understand the uncountability of a perfect set, all of them was in complete metric spaces and $\mathbb{R}^{k}$ taking open intervals and getting contradiction, is there any generalization of the uncountability of perfect sets for any $X$ topological space? So doesnt it have to be uncountable for any space $X$?
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2$\Bbb Q$ is perfect as a subspace of itself – Alessandro Codenotti Apr 06 '21 at 09:43
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@AlessandroCodenotti thank you so much sir, I think for the completely metrizable spaces, the perfect subsets are uncountable – Woodx Apr 06 '21 at 10:01
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Here is a proof for completely metrizable spaces; it actually shows that the cardinality is $2^\omega=\mathfrak{c}$. – Brian M. Scott Apr 06 '21 at 21:25
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It's much more general: it holds for completely metrisable spaces, not just $\Bbb R^n$, e.g. And crowded Baire spaces of course, another large class.
Henno Brandsma
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I got it sir, thank you so much, so how can I start to show that a perfect space is uncountable in completely metrizable space? – Woodx Apr 06 '21 at 10:28
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@Woodx essentially the same proof as for $\Bbb R^n$. Assume $A$ is perfect and countable and construct a Cauchy sequence converging to a point not in it, showing it's not closed. – Henno Brandsma Apr 06 '21 at 10:28
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can we say that the uncountability of Cantor set from this result? so it is perfect and closed so completely metrizable, that is why it is uncountable? @HennoBrandsma – Woodx Apr 06 '21 at 11:00
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It's even equinumerous with ${0,1}^{\Bbb N}$ so size continuum. @Woodx it's not based on that theorem. – Henno Brandsma Apr 06 '21 at 11:01