# prove that every infinite subset of compact set H has a limit point (Explanation)

Let S be an infinite subset of H

Assume that no point of H were a limit point of S.

Since H is compact, there exist a collection of open sets V={Vq: q∈H }

Vq∩S is finite (Is this true? If yes why?)

Since S is assumed to be infinite

⋃Vq ⊃ S cannot cover S

Since S⊂H

⋃Vq ⊃ H (cannot cover H)

This contradict compactness of H.

Hence every infinite subset of H has a limit point (Why do we conclude this?)

## 2 Answers

Suppose $$S$$ has no limit point in $$H$$. So we are proving the statement by contradiction.

$$S$$ has no limit point in $$H$$ means "for all $$q \in H$$, $$q$$ is not a limit point of $$S$$".

This implies there is for each $$q \in H$$ some open set $$V_q$$ that contains $$q$$ and such that $$V_q \cap S$$ is finite.

This is the fact you're missing; it's what not being a limit point implies, because if $$q$$ is a limit point, then every open set $$V$$ containing $$q$$ has a point in $$S$$ unequal to $$q$$. The negation says that there exists some open set $$V$$ such that $$V \ni q$$ and $$V \cap S$$ can be at most $$\{q\}$$ (or empty). Or maybe you use the notion of a $$\omega$$-limit point: every open set containing $$q$$ intersects $$S$$ in an infinite set, and then the negation says that there exists an open set that only intersects $$S$$ in a finite set. Either way we can find such a $$V_q$$ for all $$q$$, if $$S$$ has no ($$\omega$$-)limit point.

Now as $$H$$ is compact, $$H = \bigcup\{V_{q_i}: i = 1,\ldots n\}$$, for some finitely many points of $$H$$. But then $$S = S \cap H = \bigcup \{V_{q_i} \cap S: i=1,\ldots, n\}$$ which is a finite union of finite sets by how we chose all $$V_q$$, but this contradicts that $$S$$ is infinite. So the initial assumption is false and $$S$$ must have some ($$\omega$$-)limit point.

You can prove this directly: since $$S$$ has no limit point in $$H$$ it is closed in $$H$$. Also, for each $$p\in S$$ there is an open neighborhood $$U_p$$ of $$p$$ such that $$U_p\cap S=\left \{ p \right \}.$$ But then, $$\left \{ U_p:p\in S, S^c \right \}$$ is an infinite open cover of $$H$$ with no finite subcover.