I wonder if there is an example of a set $A$ that is closed and that does not contain any nonempty open set.
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2Take a single point for example. – M. Winter May 31 '17 at 16:19
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since single point have no limit point , it is closed set?? – wlaudsla123 May 31 '17 at 16:21
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A single point is its own and only limit point. A set with a single point does only have the constant sequence which converges to the only point present. So it contains all its limit points, hence is closed. – M. Winter May 31 '17 at 16:23
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You should specify if you're talking about $A$ as a subset of $\mathbb{R}^n,$ a metric space, a topological space etc... – Riccardo Ceccon May 31 '17 at 16:25
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A as a subset of a metric space – wlaudsla123 May 31 '17 at 16:30
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I will just point out that not containing a non-empty open set is equivalent to having empty interior. Such sets are also called co-dense (see, for example, this answer) and less frequently boundary set or border set (probably influenced by Polish authors - it's called Zbiór brzegowy in Polish). – Martin Sleziak Jul 27 '17 at 06:30
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Here is an example of such a set with infinitely many elements:
The set of integers in $\mathbb R$ is closed, and any nonempty subset of integers is closed as well.
Alex
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In any Hausdorff space, any finite set is closed.
So in any connected Hausdorff space a singleton does contain any non-empty open set.
(I mentioned connectedness as then the singleton cannot be open since it is already closed.)
Arpan1729
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The empty set is closed and does not contain any nonempty open set. Probably you meant to ask for a nonempty closed set which does not contain any nonempty open set. The following theorem answers this question in a general topological space X.
Theorem. If in the space $X$ there is a closed set which is not open, then there is a nonempty closed set which does not contain any nonempty open set.
Proof. If there is a closed set which is not open, then its complement, call it $U,$ is an open set which is not closed. Of course $U\ne\emptyset,$ since $\emptyset$ is closed. Assuming the axiom of choice, we can extend $\{U\}$ to a maximal collection $\mathcal U$ of pairwise disjoint nonempty open sets. Then the set $A=X\setminus\bigcup\mathcal U$ is a closed set which does not contain any nonempty open set. I claim that $A\ne\emptyset.$
If we had $A=\emptyset,$ it would follow that $\bigcup\mathcal U=X$ and that $\bigcup(\mathcal U\setminus\{U\})=X\setminus U,$ so that $X\setminus U$ would be an open set, contradicting the fact that $U$ is not closed.
bof
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