Show two notions of dense are equivalent

Question

This question follows from another one Topology proof: dense sets and no trivial intersection

Show that given a topological space $(X, \mathcal{T}), D \subseteq X$

Then $D$ is dense iff $\forall U \in \mathcal{T}, U \neq \varnothing, D \cap U \neq \varnothing$

Here my try:

$(\Rightarrow)$ $D$ is dense if $\overline D = X$. By definition, $\overline{D} = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap D \neq \varnothing\}$ so $\overline{\overline D} = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap \overline D \neq \varnothing\}$

Since $\overline D = X$, and $U \subseteq X$, therefore $\forall x \in X, \forall U \in \mathcal{T}, x \in U \implies U \cap \overline D \neq \varnothing$

But $\overline D = \overline{\overline D}$, therefore $\forall x \in X, \forall U \in \mathcal{T}, x \in U \implies U \cap D \neq \varnothing \Leftrightarrow \forall U \in \mathcal{T}, U \neq \varnothing, D \cap U \neq \varnothing$

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($\Leftarrow$) Given $D \subseteq X$, $\forall U \in \mathcal{T}, U \subseteq X, D \cap U \neq \varnothing$, we want to show that $\overline D = X$. I think this proof is immediate but am I not sure how to show this...

Can someone check my attempt and show me how to continue with $\Leftarrow$

Answer 1

Your first proof is correct but unnecessarily complicated. Suppose that $\operatorname{cl}D=X$, and let $U$ be any non-empty open set in $X$. $U\ne\varnothing$, so there is an $x\in U$; and $x\in X=\operatorname{cl}D$, so $U\cap D\ne\varnothing$.

For the other direction assume that each non-empty open subset of $X$ meets $D$, and let $x\in X$ be arbitrary. If $U$ is any open nbhd of $x$, then $U\cap D\ne\varnothing$, so $x\in\operatorname{cl}D$. Thus, $X\subseteq\operatorname{cl}D\subseteq X$, so $\operatorname{cl}D=X$.

By the way, in general proofs are easier to read if you don’t go overboard on the symbols and do include some plain English ‘connective tissue’ to make the flow of logic clearer; I said a little more about this in this answer.