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Every uncountable subset $A$ of $\mathbb{R^n}$ has cardinality of the continuum?

Proof: We know that an uncountable subset has uncountable many points that are limit points ( there is a proof in thie link: Uncountably many points of an uncountable set in a second countable are limit points )

Let $B=\{ x \in A : x \text{is not limit point of A}\}$. If $B$ is uncountable then there is a point $y\in B$ which is a limit point of $B$ this implies that $y$ is a limit point of $A$, but this is a contradiction because if $y\in B$ then $y$ is a limt point of $A$

Then $B$ is at most countable. Let $P=A\setminus B$ then $P$ is perfect so its cardinality is at least the continuum $c$

So $c \le \text{card}(P) \le \text{card}(A) \le \text{card}(\mathbb{R^n} ) = c$ therefore $A$ has cardinality of the continuum.

But I think this proof is incorrect due to tha fact that I´ve seen this "theorem" but with an extra hypothesis: $A$ must be closed... so I don´t know where is my mistake.

I would really appreciate if ypu can help me with this problem.

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    $\begingroup$ Why is $P$ perfect? You may test your argument with $A=\mathbb{R}\setminus\mathbb{Q}$. $\endgroup$ – Sangchul Lee Jul 4 '18 at 4:01
  • $\begingroup$ And interestingly enough, every Borel subset of $\mathbb{R}^n$ is either (at most) countable or has cardinality of the continuum. In particular, this claim is true for any open/closed subsets. $\endgroup$ – Sangchul Lee Jul 4 '18 at 4:08
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No, this is clearly false, as it would imply the Continuum Hypothesis, which cannot be proved from ZFC (as Cohen showed first).

You claim $P$ is perfect and this is in general false. The fact that $B$ is at most countable shows that $|A \setminus B| = |A|$. Perfect sets are closed. $B \setminus A$ is merely a crowded subset, not a perfect subset.

Recall $P$ is perfect when $P=P'$ where $P'$ is the set of limit points, and whenever $A' \subseteq A$, $A$ is closed, for any $A$. So perfect implies closed. And there is no reason $A \setminus B$ would be closed.

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  • $\begingroup$ So if $A$ is closed then $A \setminus B$ would be perfect right? every point of $A \setminus B $ would be a limit point of $A \setminus B$? $\endgroup$ – user128422 Jul 4 '18 at 4:35
  • $\begingroup$ @user128422 yes as $B$ is open in $A$. But then it’s ok. An uncountable closed set has size continuum $\endgroup$ – Henno Brandsma Jul 4 '18 at 4:40
  • $\begingroup$ @user128422 It not not true in general that that set would be perfect. $ A - B $ is the set of all limit points of $ A $, which is normally denoted $ A' $. This is not necessarily perfect, even if $ A $ is closed. For example, take $ A = \{0\} \cup \{\frac{1}{n} : n \in \mathbb{N} \} $. Then $ A' = \{0\} $, which is not perfect. $\endgroup$ – user571438 Jul 4 '18 at 5:07
  • $\begingroup$ So how can we use the fact that $A$ is also uncountable to prove this? $\endgroup$ – user128422 Jul 4 '18 at 5:43
  • $\begingroup$ You can use the Cantor Bendixson Theorem, which states that every closed subset of $\mathbb R$ can be written uniquely as a disjoint union of a perfect set and a countable set. From here, the proof you had above will work. The proof of this is nontrivial. One method I know is to use transfinite recursion to essentially iterate the process of $A \mapsto A' $. Alternatively, we say $ x \in A $ is a condensation point of $ A $ if every open neighborhood of $ x $ intersects with $ A $ in an uncountable set. Then let $ \text{kernel}(A) $ be the condensation points of $ A $. That's your perfect set. $\endgroup$ – user571438 Jul 4 '18 at 6:14

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