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Let $X$ be a metric space with $p \in X$ a point, $C \subset X$ a subset. Show $C$ is closed iff $C \cap \overline{B_R(p)}$ is closed for any $R>0$.

Supposing $C$ is closed is pretty easy as intersecting it with closed ball is still closed.

So then assume $C \cap \overline{B_R(p)}$ is closed (so it equals its closure) and want to show $C$ is closed, i.e., $C = \overline{C}$.

Is this the way to go about it? Clearly $C \subset \overline{C}$, so we wish to show $\overline{C} \subset C$ but taking $x \in \overline{C}$ and showing $x \in C$?

Because then $x$ is a limit point of $C$ so any open ball (for any choice of $R>0$) centered at $p$ intersects $C$ nontrivially? Am I on the right track? Just a hint will suffice not an entire solution. Thanks!

MyMathYourMath
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    the only thing that worries me about what you are saying, is that it does not seem to involve the arbitrary radius $R$ – 311411 Jul 06 '20 at 21:59
  • Yes, I agree with 311411, you are on the right bath, but you can not drop the assumption $R$ is arbitrary. – Medo Jul 06 '20 at 22:02
  • @Hossien Sahebjame Is $C$ bounded ? If $C$ is bounded then the proof is easy: choose $R>0$ large enough so that $C\subset B_{R}(p)$ and consequently $C\cap B_{R}(p)=C$ is closed :) – Medo Jul 06 '20 at 22:14
  • @Medo no not bounded lol, ironically I thought the EXACT same thing initially, since $C$ is a subset of the ball, the intersection is closed but I don't have that $C$ is bounded unfortunately :( – MyMathYourMath Jul 06 '20 at 22:15
  • Thanks guys, that is the subtly I was worried most about Was arbitrariness of $R$, so this allows for any choice of $R$? – MyMathYourMath Jul 06 '20 at 22:16
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    maybe $d(p,x)$ is important. For consider the example X = real line, with C = $[0,2] \setminus 1$, while $p=5$. – 311411 Jul 06 '20 at 22:32
  • @Medo Im sorry that confused me more, are you swaying away from my initial question? So you're saying intersection can be closed when one component is not closed? – MyMathYourMath Jul 06 '20 at 22:56

1 Answers1

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suppose $x$ is a limit point for $C$, but maybe not an element of $C$. Let $R\,=\,d(p,x)\,+\,5.$ By hypothesis, the set

$$C\,\cap\,\overline{B_R(p)}\,\cap\,\overline{B_1(x)}$$

is closed. There is a sequence in $C$ that converges to $x$. This sequence is ultimately in the unit ball at $x$, and we have made $\overline{B_R(p)}$ big enough to include that unit ball. The triple intersection above is closed, so contains all of its limit points. Hence $x$ belongs to $C$.

311411
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  • omg wow thanks!!!!!! question, what guarantees the sequence to be inside of the unit ball at $x$? – MyMathYourMath Jul 06 '20 at 23:10
  • Also can you make the 5 a 1 and it would still work? – MyMathYourMath Jul 06 '20 at 23:19
  • actually i think the $5$ is important because we work in a very general abstract metric space. Remember that such spaces can be very anti-intuitive. For example $B(x,r) \subsetneq \overline{B(x,r)} \subsetneq \overline{B}(x,r) \ $ can occur. See https://math.stackexchange.com/questions/66020/proper-inclusion-between-open-ball-closure-of-open-ball-and-the-closed-ball-in?rq=1 – 311411 Jul 06 '20 at 23:21
  • Because of the fact that we may not have PROPER containment, correct? So you make it at least 2 just to guarantee proper containment – MyMathYourMath Jul 06 '20 at 23:24
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    on second thought, I think that replacing the 5 with a 1 will be ok! Because (check this) $B_1(x) \subset B_{1,+,d(x,p)}(p)$. From this $\overline{B_1(x)} \subset \overline{B_{1,+,d(x,p)}(p)}$ follows. – 311411 Jul 06 '20 at 23:38
  • Question, could we replace the number with any $\epsilon >0$ so you basically choose $R$ big enough so that $B_\epsilon(x) \subset B_R(p)$ then this forces $x \in C$. – MyMathYourMath Jul 08 '20 at 23:53
  • I think so. There is nothing magical about a radius of one. – 311411 Jul 09 '20 at 00:03
  • cool thanks I finally got it 100% down! thank you a million – MyMathYourMath Jul 09 '20 at 01:15