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In the proof which says Compactness implies limit point compactness, Munkres writes ‘$A$ has no limit point, hence $A$ contains all its limit points and is closed. Is it not contradictory?

danny
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    If $A$ has no limit point then the set of limit points of $A$ is empty set which is contained in $A$. – Infinity_hunter Oct 12 '22 at 17:51
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    Ask yourself also this question then. Would you not want the finite sets of the real line to be compact? . If no, then how can you claim that a finite set is closed? . You can precisely do it by saying that the set has no limit point and hence the set of limit points which is a null set is contained in $A$.(an empty set is always a subset of any set) . – Mr.Gandalf Sauron Oct 12 '22 at 17:57
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    It may help to think as follows. Because $A$ has no limit points, you cannot produce a counterexample to the assertion that $A$ contains all of its limit points -- in other words, there does not exist a limit point $a$ of $A$ such that $A$ fails to contain $a$. – MPW Oct 12 '22 at 18:03

2 Answers2

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This is a vacuous logic argument. (Some further discussion here, and elsewhere on MSE.)

You can think of it as "$A$ contains all zero of its limit points."

Alternatively, if the claim "$A$ does not contain all its limit points" is untrue, then $A$ has a limit point $x$ which is not in $A$, i.e. $\exists x \in A'$ where $x \not \in A$. But we see that $A' = \varnothing$ since $A$ has no limit points, and hence a contradiction is given.

PrincessEev
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Yes, as @PrincessEev wrote it is a vacuous truth of logic:

$(\forall x)(x\in A \land x\notin A')$

The negation of which is:

$(\exists x)(x \notin A \lor x\notin A')$

So from the truth table of $\lor$ there exists the case, $x\notin A \land x\in A'$.

But this is the negation of the vacuous formula.

ryaron
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