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I am working on Mendelson's Introduction to topology. This question arose after completing question $8$ from the exercise on page $87.$

A subset $A$ of a topological space $(X,\mathfrak{I})$ is said to be dense in $X$ if $\overline{A}=X.$ Prove that if for each non-empty open set $O$ we have $A\cap O \neq \emptyset,$ then $A$ is dense in $X.$

I proved the contrapositive:

Suppose $A$ is not dense in $X,$ that is, $\overline{A}\neq X=\overline{A}\cup \operatorname{Int}(C(A)).$ Then $\operatorname{Int}(C(A))\neq\ \emptyset.$ Thus $\ \exists\ $ a non-empty open set $O:=\operatorname{Int}(C(A))$ such that $A\cap O = \emptyset.$

I am wondering if there is also a direct proof.

Adam Rubinson
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4 Answers4

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Using Mendelson's definition of closure, we have that if $x\in \overline{A}$, then for every neighborhood $N$ of x, we have that $N \cap A \not=\emptyset$. Take $x\in X$ and note that every neighborhood $N$ of $x$ has $x \in O \subset N$ for some open $O$. Thus $N\cap A \supset O\cap A$ is nonempty and $x\in \overline{A}$, showing $X \subset \overline{A}$. Now to show $\overline{A} \subset X$, use Lemma 4.3 and note that $X$ is a closed set that contains $A$.

Vuong Nguyen Hoang
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We must show $\overline{A} = A \cup \partial A = X$. Let $O'$ be open and nonempty in $X$. Either $\DeclareMathOperator{Int}{\mathrm{Int}}$

  • $O' \subset A$: we need not consider this case further; or
  • $O' \not\subset A$: so there is an $x \in O' \smallsetminus A \subset X \smallsetminus A$. We wish to show $x \in \partial A$.

Let $O$ be any open set containing $x$. By construction, $O$ contains $x \not\in A$ and $A \cap O \neq \varnothing$. By definition , $x \in \partial A$. Therefore, every point of $X$ is either in $A$ or in $\partial A$, and so $\overline{A} = X$.

Eric Towers
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You could use the following definition(or equivalence if you have other definition) of closure:

$x \in \bar{A} \Leftrightarrow \forall U \text{ neighborhood of } x, U \cap A \neq \emptyset$

This a definition use in some text. From this, the exercise is almost trivial.

Carlos Licea
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Let $U=(\overline A)'$ be the complement of $\overline A$ in $X$. It is an open set and $U\cap A=\emptyset.$ So, $U=\emptyset$ and thus $\overline A=X.$

Bob Dobbs
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