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If A is a nowhere dense set, it means that "A closure" has an empty interior. Canwe also say that "A" too has an empty interior?

I believe Yes, we can say so.

Because "A closure" is just the union of A and it's limit points. The set of limit points are always closed(since they are not open). Now when we talk about "A Closure's interior", we refer to A's interior. And so if we say A closure has an empty interior, it means "A Closure" sans the closed subset of A (which is the set of it's limit points) has an empty interior. In other words, A has an empty interior? Am I Right?

Krishan
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  • Judging by one of your claims it would seem that you believe a subset (of a given topological space) to be closed if it isn't open. Beware of that, for it is not correct: there may very well exist subsets which are neither open nor closed. As to the claim itself that the derivative of any subset is always closed, that is indeed true in $\mathrm{T}_1$-spaces but not in general for arbitrary spaces. – ΑΘΩ Nov 18 '20 at 12:08

2 Answers2

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Definitely closure of a set having empty interior implies that set also has empty interior.

As far as I understand from your argument, you accept the definition a set is nowhere dense if its closure has empty interior.

You said

A Closure's interior", we refer to A's interior

If you are trying to prove if $A$ has an empty interior then $A$ is a nowhere dense, here is a counter example- Take $A$ be the set of rational numbers in real metric space with respect to Euclidean metric then the interior of rational numbers is empty but the interior of limit points of $A$ is $\mathbb{R}$.

David C. Ullrich
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Infinity_hunter
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  • Dear @Infinity_hunter,I was trying to prove that if A is nowhere dense then not only does A closure have an empty interior, but also A has an empty interior. – Krishan Nov 18 '20 at 13:34
  • For a nowhere dense set A, if we say Int(A closure)=phi is same as Int(A)=phi, isn't A assumed to be closed (since we assume, A=A closure )? – Krishan Nov 18 '20 at 15:12
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    @Krishan if $Int(A) \ne \phi$ then definitely $Int(Cl(A)) \ne \phi$. So by contrapositive statement of this it would imply that Interior of nowhere dense set is empty – Infinity_hunter Nov 18 '20 at 15:47
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If $A$ is nowhere dense, then, as $\operatorname{int}(A) \subseteq \operatorname{int}(\overline{A}) = \emptyset$, $A$ has empty interior.

But e.g. $([0,1] \cap \Bbb Q)$ has empty interior in $\Bbb R$, but is not nowhere dense, so it's quite a bit stronger than having empty interior.

Henno Brandsma
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