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Text I am reading states:

Let $S$ be a compact set. Suppose $p \notin S$. For each point $q \in S$, consider an open neighborhood $W_q$ of radius less than half $d(p, q)$. Such a system of open neighborhoods covers $S$ because $q \in W_q$ for each $q \in S$. Since $S$ is compact, there are finitely many points $q_1 , q_2 , . . . , q_n$ in S such that $$S\subseteq \cup^n_{i=1} W_{q_i} = W.$$

So far so good, I understand everything clearly.

Now let $q^∗$ be the point in $\{q_1 , . . . , q_n\}$ closest to $p$ and let $V_{q^*}$ be a neighborhood of $p$ with radius less than half $d(p, q^∗ )$. Then $V_{q^I}$ is a neighborhood of $p$ that does not intersect $W$.

First problem: This is the first and only mentioning of $V_{q^I}$ in the text to follow. Perhaps it is a mistake and the author meant $V_{q^*}$. We continue

Hence $V_{q^*}\subset \bar S$, so $p$ is an interior point of $\bar S$. Hence, if there is a sequence $s_n$ of points in $S$ converging to $p$, we must have $p \in S$. So $S$ is closed.

Question

Why would there be a sequence of points converging to $p$. This statement seems pulled out of thin air... I understand that if a sequence points in $S$ that converges to a point $p$, than the set is closed. But why would there be such a sequence at all?

Edit

On second though, I need a clarification of the entire last quoteblock. I thank you in advance. The text starts with $p\notin S$. How can it conclude with $p\in S$ without it being a contradiction? This makes no sense to me...

Asaf Karagila
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LeastSquaresWonderer
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  • Are there any hypotheses about $p$--is $p \in \bar{S}$? – eepperly16 Dec 13 '17 at 06:29
  • No, I copied the text word for word... I added an edit to the question. The text seems to state both, $p\notin S$ and $p\in S$, which is what confuses me... – LeastSquaresWonderer Dec 13 '17 at 06:32
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    This looks to me like a proof by contradiction that compact implies closed. Under this interpretation, the author presumably means to assume $p\in\bar S$ by hypothesis – eepperly16 Dec 13 '17 at 06:42
  • Could you please expand on this interpretation in an answer and help me understand? I will accept your answer. Thank you! – LeastSquaresWonderer Dec 13 '17 at 06:43
  • " if there is a sequence" and "But why would there be such a sequence at all" Who said there was? It said IF there was a sequence. Not there IS a sequence. – fleablood Dec 13 '17 at 06:43
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    $\bar{S}$ should be replaced with the set-theoretic complement $S^{c}$ of $S$. then the proof shows that $S^{c}$ is open, so that $S$ is closed. – user363464 Dec 13 '17 at 06:44
  • @user363464 can you please explain how this is done? I mean, how should the proof begin, for it to be more clear to me, that this is what we are trying to prove (that $S^c$ is open). – LeastSquaresWonderer Dec 13 '17 at 06:45
  • @fleablood is stil dont understand what is its place in the proof. What if there is no such sequence. What than, how do I conclude the set is closed, if there is no such sequence? – LeastSquaresWonderer Dec 13 '17 at 06:47
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    Note that $V_{q^{*}}$ is a neighborhood of $p$ disjoint from $S$. So $p$ is interior to $S^{c}$. But $p$ is chosen arbitrarily among points in $S^{c}$, thus $S^{c}$ is open. And complements of open sets are closed, hence $S$ is closed. – user363464 Dec 13 '17 at 06:47
  • @user363464 Thank you.Now I understand. If you please expand your comment in an answer, I will gladly accept! – LeastSquaresWonderer Dec 13 '17 at 06:50
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    If there is no sequence then it is not a limit point and we don't have to worry about whether it is in the set or not. The set must be closed because there are not any limit points that are not in the set. – fleablood Dec 13 '17 at 07:01

1 Answers1

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The author probably intends to show that $S$ is closed in these sentences, and you are right that $V_{q'}$ should mean $V_{q^{*}}$. I've never seen the notation $\bar{S}$ used to denote the set-theoretic complement of $S$, but if the intent is to prove that $S$ is closed, $\bar{S}$ must either mean the set-theoretic complement, or it should be replaced by $S^{c}$, the notation I am familiar with in denoting such.

In any case, note that $V_{q_{*}}$ is a neighborhood of $p$ disjoint from $S$, so $p$ is interior to $S^{c}$. Since $p$ was chosen arbitrarily among points in $S^{c}$, it follows that $S^{c}$ is open. Hence $S$ is closed.

user363464
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