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Say that we have a metric space $X$ and that $Y= \{y_1, y_2,\ldots\}$ is a countable collection of points in $X$ such that for any two points in $Y$, we have $d(y_n, y_m) \geq1$, i.e. the distance between those points is $\geq1$.

We want to show that $Y$ is closed in $X$ and that when equipped with the subspace topology, $Y$ is not compact. Couldn't we go about using contradiction to show the first claim? I'm thinking we could assume $Y$ is not closed in $X$, which would mean that there exists an accumulation point, say $x$ of $Y$, such that $x \notin Y$. How would we go from there? Not exactly sure how to tackle the second part of the question. I guess once the second part is shown, we will in fact have that $X$ is not compact, since every closed subspace of a compact space is compact.

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The current answers seem sufficient to me - I would just add, to really get an idea of what would be useful in the proof, come up with an example using a real-ly easy metric space ... –  Neal Oct 20 '12 at 20:25
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2 Answers 2

up vote 3 down vote accepted

For each $x\in X$ let $B(x)=\left\{z\in X:d(x,z)<\frac12\right\}$. Show that for each $x\in X$, $B(x)\cap Y$ contains at most one point. Conclude that $Y$ is closed in $X$. Does $\{B(y)\cap Y:y\in Y\}$ have a finite subcover?

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For $Y$ not to be compact in the subspace topology, an open cover must not be able to exhibit a finite subcover. So in answer to your question, I would say no. Just need to see how. –  Libertron Oct 20 '12 at 20:22
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@Sachin: It follows from the first thing that I suggested you show: that for each $x\in X$ (and therefore certainly for each $x\in Y$) $B(x)\cap Y$ contains at most one point. –  Brian M. Scott Oct 20 '12 at 20:28
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Hints:

  1. Use sequences and their limits.
  2. Cover $Y$ by balls of radius $1/2$.
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Using sequences and limits seems more complicated than what is needed. If you cover $Y$ with balls of radius $1/2$, then no point of $Y$ except the center of the ball is within the ball, and that's enough to show that the center of the ball is not a limit point of $Y$, without thinking about sequences. –  Michael Hardy Oct 20 '12 at 21:51
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