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 limit 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 you could help me with this problem.
Best Answer
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.