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I was wondering

if a set $A$ has no accumulation point, is this set $A$ closed?

I think this is true, but I'm not quite sure.

Here's my thinking:

By closed set definition: A set $A$ is closed if every accumulation point of $A$ is a point of $A$.

Since $A$ has no accumulation points, it is closed. Am I saying this right?

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    $\begingroup$ You have $\text{cl}(A) = A \cup A' = A \cup \emptyset = A.$ $\endgroup$
    – IAmNoOne
    Commented Oct 26, 2014 at 3:09
  • $\begingroup$ Whenever you say "open" or "closed", you must say it in relation to some specific topology. Each A is always open and closed in relation to A. $\endgroup$
    – Ivan Kuckir
    Commented Oct 26, 2014 at 9:57
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    $\begingroup$ @IvanKuckir Actually, one should also talk about accumulation points with respect to a specific topology. Given this, the only reasonable reading of the submitter's statement is "If a set A has no accumulation points with respect to a topology T, is the set A closed with respect to the same topology T?" The answer is of course "yes". $\endgroup$
    – cefstat
    Commented Oct 26, 2014 at 11:37

3 Answers 3

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Well, we need to be a little careful with our wording.

Consider the set $S = \{\frac{1}{n} : n \in \mathbb{N} \}$. Certainly $0$ is an accumulation point, but $0 \not\in S$. Therefore, $S$ has no accumulation point within $S$, but $S$ is certainly not closed relative to the larger metric space $\mathbb{R}$. However, as cesfat was saying below, if we look at $S$ in its own right with the subspace topology induced from $\mathbb{R}$, then $S$ is closed relative to itself (since it contains no accumulation points within itself).

Bottom line: when we talk about closure, we do so relative to a given metric space.

So to answer your question, if a set $A$ is embedded in a larger metric space $X$ and $A$ has no accumulation point anywhere in $X$, then it is vacuously true that $A$ is closed in $X$.

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  • $\begingroup$ $S$ is certainly closed in itself. If you restrict your attention to $S$ with the subspace topology induced from $\mathbb{R}$ then $S$ has no accumulation points (in itself) and it is closed. If now you see $S$ as a subset of $\mathbb{R}$ and you work with the standard topology in $\mathbb{R}$ then $S$ has the accumulation point $0 \not\in S$ and $S$ is not closed. One should be careful to discuss accumulation points and closedness with respect to the same topology. $\endgroup$
    – cefstat
    Commented Oct 26, 2014 at 11:47
  • $\begingroup$ Good point @cefstat. That was kind of the idea I was trying to convey. I'll edit my post shortly to better reflect what you're saying. $\endgroup$
    – Kaj Hansen
    Commented Oct 26, 2014 at 19:45
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The set of all accumulation points of S is the empty set and every set has the empty set as a subset. Therefore, S contains all its accumulation pointz and thus S is closed. Very simple.

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Yes it is just a consequence of logic and has nothing to do with the properties of closed sets, which may be why you find it strange. Anyway for instance an empty set is closed.

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