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Definition : Let $S$ be a subset of a metric space $(X,d)$. A point $x \in S$ is called an isolated point if $\exists \varepsilon >0 $ s.t $B(x,\varepsilon) \cap S \setminus \{x\}= \emptyset. $

Baire's Theorem : Let $ (X,d)$ be a complete metric space. Then, the interior of any countable union of closed subsets is nonempty implies the interior of one of the closed subsets is nonempty as well.

With the theorem and the definiton in mind, I am supposed to prove the following; Propositon : If $X$ is a complete metric space which has $\textbf{no}$ isolated points, then $X$ is uncountable.

I wrote down as follows : By having no isolated points , we have $\forall x \in X, \quad \forall \varepsilon> 0 \quad B(x,\varepsilon) \cap X\ \{x\} \ne \emptyset.$ And I assume the assertion of the proposition is false. In other words, $X $ is countable and it can be rewritten as $X= (S^c) \cup S$ for some $S^c $ which is countable and and $S = \bigcup_{n=1}^\infty S_n$ is uncountable and each $S_n$ is closed. I somehow must try to get a contradiction with the Baire's Theorem. But I could not. Help me if my starting point is wrong and if it is guide me to another direction.

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  • $\begingroup$ Nitpicking: You should also say that $ X$ is not empty. $\endgroup$ – DanielWainfleet Dec 24 '16 at 12:02
  • $\begingroup$ We can show by very elementary means that a non-empty complete metric space with no isolated points has a subspace homeomorphic to the Cantor set. But it's much longer than Brian M. Scott's answer to your Q. $\endgroup$ – DanielWainfleet Dec 24 '16 at 12:30
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HINT: Turn it around: show that if $X$ is a countable metric space without isolated points, then $X$ is not complete. Suppose, then, that $X$ is a countable metric space without isolated points.

  • Show first that $X$ must be countably infinite.

Thus, we can enumerate $X=\{x_n:n\in\Bbb N\}$.

  • Show that each singleton set $\{x_n\}$ is closed and has empty interior.
  • Use the Baire category theorem and the fact that $X=\bigcup_{n\in\Bbb N}\{x_n\}$ to show that $X$ is not complete.
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  • $\begingroup$ If the space is not complete how does baire theorem in my question help?(i do not know about cathegory theorem). You re suggesting me that i use the other direction of the baire theorem? $\endgroup$ – Quantes Dec 24 '16 at 3:12
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    $\begingroup$ @Quantes: Baire category theorem is the more common name of what you called simply the Baire theorem: it’s the same theorem. You use it to show that if $X$ were complete, $X$ would be a counterexample to it. $\endgroup$ – Brian M. Scott Dec 24 '16 at 3:14
  • $\begingroup$ Ok. I will try it. Thank you so much $\endgroup$ – Quantes Dec 24 '16 at 3:17
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    $\begingroup$ @David: For me the term Baire category theorem covers all of the variants, including the ones for locally compact Hausdorff spaces. $\endgroup$ – Brian M. Scott Dec 24 '16 at 3:35
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    $\begingroup$ @Quantes: If $p$ is an isolated point then $\{p\}$ is still closed but it has nonempty interior (specifically, the interior of $\{p\}$ is $\{p\}$). So then $\{p\}$ is not nowhere-dense and the proof falls apart. Indeed, it is entirely possible that a complete metric space with isolated points can be countable (consider $\mathbb{Z}$). $\endgroup$ – Nate Eldredge Dec 24 '16 at 16:32
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If $Y$ is a countable subset of $X$ (where $X$ is not empty and $X$ has no isolated points) then $F=\{\{p\}:p\in Y\}$ is a countable family of closed nowhere-dense sets, so (Baire) $\cup F$ is co-dense in $X.$ Since $X$ is not empty, therefore $\emptyset \ne X$ \ $\cup F=X$ \ $Y.$ That is, $X\ne Y.$

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