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Prove that if $p_n \to p$ in a metric space then the set of points $\{p,p_1,p_2, ...,\}$ are closed.

A theorem in my book states that a set $S$ in a metric space is closed if and only if whenever $q_1,q_2,...$ is a sequence of points in $S$ that is convergent then the $\lim_{n\to \infty}q_n \in S$

What I did was try and prove it by contradiction. So suppose the set $S$ is not closed which will imply that $S^c$ is not open (where "not open" apparently doesn't mean "closed") then there exists a point in $S^c$; lets denote this point $p\in S^c$ such that there is an open ball $B(p,r)\in S^c$ ...

But I don't know how to finish it.

Tom
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4 Answers4

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To complete your proof : You want to show that $S^c$ is open. Suppose not, then there exists a point $x \in S^c$ such that, for any $\epsilon > 0$, $B(x,\epsilon)$ is not a subset of $S^c$. In other words, $$ B(x,\epsilon)\cap S \neq \emptyset \quad\forall \epsilon > 0 $$ For $\epsilon = 1/k$, there is a point $p_{n_k} \in S$ such that $$ p_{n_k} \in B(x,1/k) $$ Now the subsequence $\{p_{n_k}\}$ converges to $x$ (Why?)

However, $p_n \to p$, so $p_{n_k} \to p$ for any subsequence (Why?)

Hence, $x=p$ (Why?)

Thus, $p \in S^c$, which is a contradiction. Thus, $S^c$ must be open, and hence $S$ must be closed. Answering all the "why"s above should solve your problem :)

Prahlad Vaidyanathan
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  • When you said "$B(x,\epsilon)$ is not a subset of $S^c$" did you mean $S$ instead of $S^c$? – Tom Sep 24 '13 at 08:58
  • And since $x=p$ and $x\in S^c$ why would $p\in S^c$ be a contradiction? – Tom Sep 24 '13 at 09:03
  • @Tom : I did mean $B(x,\epsilon)$ is not a subset of $S^c$ (remember, you are assuming $S^c$ is not open). And $p\in S^c$ is a contradiction because $p\in S$! – Prahlad Vaidyanathan Sep 24 '13 at 09:41
  • Can you clarify some misconceptions I have. You claimed there is a point $x\in S^c$ such that there is ball $B(x,\epsilon)$ which is not a subset of $S^c$. Can you explain why it is not a subset? What I was thinking was since $x\in S^c$ then it is a subset of $S^c$. – Tom Sep 24 '13 at 20:02
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Let's prove that $p$ can be the only accumulation point of the set $S=\{p,p_1,\dotsc\}$

Suppose $q$ is an accumulation point of $S$ distinct from $p$, so $d(p,q)>0$. Set $\varepsilon=d(p,q)/2$; then there exists $k$ such that $p_n\in B(p,\varepsilon)$, for all $n>k$. This leaves only a finite number of elements of $S\cap B(q,\varepsilon)$, contradicting the fact that $q$ is an accumulation point. Indeed $$ S\cap B(q,\varepsilon)\subseteq\{p_1,p_2,\dots,p_k\} $$ because $B(q,\varepsilon)\cap B(p,\varepsilon)=\emptyset$.

Note also that, if $S$ has no accumulation points at all, it's closed.

Notation: $d$ is the metric on the space and $B(x,r)$ denotes the open ball with center $x$ and radius $r$.

Let's try it differently, by showing that the complementary set of $S$ is open, that is, it contains an open ball around any of its points.

Suppose $x\in X\setminus S$ (where $X$ is the whole space) and that every open ball $B(x,\varepsilon)$ intersects $S$.

If, for some $\varepsilon>0$, $p\notin B(x,\varepsilon)$, then $d(x,p)>\varepsilon$ and so $$ B(x,\varepsilon)\cap B(p,r)=\emptyset $$ where $r=d(x,p)-\varepsilon$. Since the sequence converges to $p$, there is an integer $k$ such that, for $n>k$, $p_n\in B(p,r)$. Therefore $$ B(x,\varepsilon)\cap S\subseteq\{p_1,p_2,\dots,p_k\} $$ and so, taking $$ \delta=\min\{d(x,p_1),d(x,p_2),\dots,d(x,p_k)\} $$ we have that $B(x,\delta)\cap S=\emptyset$, a contradiction.

Therefore, $p\in B(x,\varepsilon)$, for every $\epsilon>0$, which means that $x=p$, contradiction.

Thus, for any $x\in X\setminus S$, there is an open ball $B(x,\varepsilon)$ such that $B(x,\varepsilon)\cap S=\emptyset$, that is, $B(x,\varepsilon)\subseteq X\setminus S$, as we wanted to prove.

A different proof would be by noting that $S$ is compact, so closed in the space it lives in. But probably this is not a “simple” proof. ;-)

egreg
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  • This is very similar to pppqqq answer. But aren't we suppose to show that it is closed? – Tom Sep 24 '13 at 20:06
  • @Tom Since $p\in S$, the only possible accumulation point belongs to $S$, so $S$ is closed. I'm going to rewrite the answer, though. – egreg Sep 24 '13 at 20:10
  • Actually I see now, if it contains all its accumulation points then it is closed but we haven't been taught "accumulation points" yet so I doubt my professor will accept this as an answer. – Tom Sep 24 '13 at 20:10
  • @Tom Accumulation points are very basic tools in topology. – egreg Sep 24 '13 at 20:18
  • I sure in future lectures we are going to talk about them. But at this particular moment the definition we have of closed set is that it is not open. – Tom Sep 24 '13 at 20:36
  • @Tom That's a wrong definition of closed set; a set is closed if and only if its complementary is open. And a set may be both open and closed! – egreg Sep 24 '13 at 20:40
  • Yes that is what I meant. A set is closed if it's complementary is open. I tried proving it by contradiction which is the method I prefer. – Tom Sep 24 '13 at 20:46
  • Thanks for the new proof. Quick question, why will "every open ball $B(x,ε)$ intersects $S$? And by $d(x,p)>\epsilon$ you mean the distance between $x$ and $p$? – Tom Sep 25 '13 at 03:14
  • If one open ball around $x$ doesn't intersect $S$, you've found a neighborhood of $x$ contained in $X\setminus S$. So, if $X\setminus S$ is not open, there is a point $x$ such that all neighborhoods of $x$ intersect $S$. – egreg Sep 25 '13 at 13:19
  • I still am confused. I understood everything else except that part. Thanks for helping me though. – Tom Sep 25 '13 at 18:29
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... there is an open ball $B(p, r)$ that intersects $S$ for every value of $r$, from which you can construct a sequence in $S$ converging to a point in $S^c$...

But I don't know how to complete it!

Daniel Donnelly
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I'd rather go for a direct proof: let $S={p,p_1,p_2,...}$ and $p*$ be an accumulation point of $S$. Let $\varepsilon>0$ be given. Since $p_n\rightarrow p$, there exists $n$ such that $$d(p_m,p)<\varepsilon/2$$ for all $m>n$. On the other hand, since $p*$ is an accumulation point, there exists $m>n$ such that $$d(p_m,p*)<\varepsilon/2.$$ So we have:$$d(p,p*)\leq d(p,p_m)+d(p_m,p*)<\varepsilon.$$ That implies that $d(p,p*)=0$ and so $p=p*$. But $p\in S$, so $S$ contains all its accumulation points.

Actually this is also a proof of the uniqueness of the limit.

pppqqq
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  • Arent we suppose to show that it is closed? – Tom Sep 24 '13 at 20:08
  • Actually I see now, if it contains all its accumulation points then it is closed but we haven't been taught "accumulation points" yet so I doubt my professor will accept this as an answer. – Tom Sep 24 '13 at 20:09
  • If you haven't been taught about "accumulation points", then what's your definition of "closed set"? – pppqqq Sep 24 '13 at 20:32
  • That it is not open. lol – Tom Sep 24 '13 at 20:34
  • Well, so you have a wrong definition. A right one would be "a set which has an open complement". "Closed set" is not the opposite of "open set" and this is an important thing to grasp. – pppqqq Sep 24 '13 at 20:45
  • Yes forgive my lack of clarity, that is what I meant. I should be more precise next time. But the method I prefer is similar to Prahlad. – Tom Sep 24 '13 at 20:47