I am trying to prove that a metric space in which every infinite subset has a limit point is compact. I am trying to prove it with Heine-Borel theorem for general metric spaces, since I have not studied countable bases and separability.

My proof is as follows:

  • Suppose you have a metric space where every infinite subset has a limit point.
  • If the metric space is unbounded (thus infinite), one can form an infinite subset with points at distance between any two points greater than a certain minimum, which will be without a limit point since the space is unbounded.
  • If the metric space is bounded and finite then it won't be a problem as there will be no infinite set.
  • But if the metric space is bounded and infinite, then it has to be complete because if not, then there is a point not belonging to the metric space, to which a Cauchy sequence converges (a limit point not in the metric space), but in the given metric space every infinite subset has a limit point. So the metric space has to be bounded and complete, meaning complete, by the Heine-Borel theorem for general metric spaces.

If I missed something or did something wrong, please help me complete it.

  • $\begingroup$ Looks like a good proof. $\endgroup$
    – user416426
    Feb 26, 2017 at 10:15
  • $\begingroup$ If you know the definition of sequential compactness, then every sequence will converge by assumption, which is equivalent to compactness in a metric space. $\endgroup$ Feb 26, 2017 at 10:28
  • $\begingroup$ i need to prove boundedness though right? $\endgroup$
    – jnyan
    Feb 26, 2017 at 10:34
  • 1
    $\begingroup$ A metric space $X$ which is non-compact has an infinite closed discrete subspace $ Y.$ So if $Z=\{y_n:n\in \mathbb N\} \subset Y$ with $(m\ne n\implies y_m\ne y_n)$ then $ Z$ is an infinite subset with no limit points. $\endgroup$ Mar 10, 2017 at 0:44

1 Answer 1


Your proof is nearly correct, except that your second bullet needs a little more work before we can conclude that such a set is necessarily bounded. You've asserted that there is such a sequence, but you need to prove such a sequence exists. One way is as follows:

We construct the sequence iteratively. First, pick any point to be $a_0$. Draw a ball of radius $1$ around $a_0$ and pick any point not in that ball to be $a_1$. We continue similarly. At each step, we draw a ball of radius $1$ around each point and pick some point not in the union of the balls to be the next element of the sequence. We know that we can do this, because otherwise the set would be bounded by the max distance between two points plus $2$. This sequence contradicts the assumption, so the metric space is bounded.

  • $\begingroup$ you may want to make sure that the radius sufficiently small, since the metric space may have diameter less than one $\endgroup$
    – C Squared
    Mar 18 at 3:20
  • $\begingroup$ The metric space is unbounded by assumption, making this a non-issue. $\endgroup$ Mar 26 at 13:40

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