I'm having trouble with an example from Munkres dealing with limit point compactness. The example is as follows:

Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$

So what I don't understand is how is $X$ limit point compact. The way I picture this is by taking the real line, picking the positive integers, and having two point above it. Now if we give each point an epsilon neighbourhood, then it would either contain both points or none. How would this imply that it is limit has a limit point?


1 Answer 1


Say $Y=\{a,b\}$. If $S$ is a subset of $\Bbb Z_+\times Y$ and $(n,a)\in S$, then $(n,b)$ is a limit point of $S.$ Conversely, if $(n,b)\in S$, then $(n,a)$ is a limit point of $S.$

There is also the notion of countable compactness: A space is called countably compact if every countable open cover has finite subcover.

Every countably compact space is limit point compact. The converse can fail, as your example demonstrates. This is mainly due to the absence of the $T_1$ property, as every limit point compact $T_1$ space is also countably compact.
Another example of such a space is $X=\Bbb N$ with the topology generated by the sets $A_n=\{k\mid k< n\}$ for all natural $n$. It is limit point compact but not countably compact.

  • $\begingroup$ would this change if the topology was the discrete topology as opposed to indiscrete? $\endgroup$ Oct 2, 2013 at 21:58
  • 1
    $\begingroup$ @nonameswereavailable: Yes, absolutely. $X$ wouldn't be limit point compact anymore. $\endgroup$ Oct 2, 2013 at 22:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.