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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 $A$.

"Proof".

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.

  • 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 01:17
  • 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 02:01
  • @Xiaochuan: sure! – t.b. Jan 11 '11 at 02:13
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    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 02:50

2 Answers2

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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.

ˈjuː.zɚ79365
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Just an observation: for the real line (or any Euclidean space, in fact), the limit points of an uncountable subset (all limit points, not only those which are elements of the uncountable set) should have cardinality of the continuum, since we can apply Cantor-Bendixson theorem to the uncountable closed set $\overline{A} = A \cup A' = A \cup (A' \setminus A)$ in order to get a perfect set (thus, of size continuum) included in $\overline{A}$, so either $A$ or $A' \setminus A$ should include continuum many limit points of $A$.

Samuel G. Silva
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