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I'm trying to solve the following exercise from Munkres book. Can you please check it?

Let $X$ be a second countable space and let $A$ be an uncountable subset of $X$. Show that an uncountable number of points of $A$ are limit points of $A$.


Let $C = \{x \in A: \textrm{x is a limit point of A} \}$. We want to show that $C$ is uncountable, so suppose not, then C is countable.

Let $B$ be a countable basis for $X$. Then for each $a \in A \setminus C$ we have that a is not a limit point of $A$ so we can find a basis element $U_{a}$ such that $U_{a} \cap A = \{a\}$.

The above defines an injection from the set $A \setminus C$ to $B$ by just mapping each $a \in A \setminus C$ to $U_{a}$. Since $B$ is countable (because $X$ is 2nd countable) then $A \setminus C$ is countable.

But then $(A \setminus C) \cup C = A$ is countable, contradiction because $A$ is uncountable.

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This looks perfectly fine. Just one minor point: I'd prefer to say that the set $I$ of isolated points is countable. Since $A$ is uncountable, so is $A \setminus I (=C)$ – t.b. Jan 11 '11 at 1:17
Can I just say the cardinality of points in $A$ being not a limit point of $A$ is at most countably many due to the second countable nature of the space and thus the conclusion? – Xiaochuan Jan 11 '11 at 2:01
@Xiaochuan: sure! – t.b. Jan 11 '11 at 2:13
There is absolutely no need to write this as a proof by contradiction, because you never used the fact that $C$ is countable. Your argument shows that $A\setminus C$ is countable, so from $A=C\cup(A\setminus C)$ you immediately conclude that $C$ is uncountable; direct proof, no contradiction involved at all. – Arturo Magidin Jan 11 '11 at 2:50

Arturo Magidin made a good point: if the assumption "to the contrary" is never used in a proof by contradiction, it is not really a proof by contradiction. This is a nice direct proof: let $F$ be the set of points of $A$ that are not limit points of $A$. For each $a\in F$ there is a basis element containing $a$ and nothing else from $A$. All such $U_a$ are distinct by construction. Hence, $F$ is at most countable, and therefore $A\setminus F$ must be uncountable.

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