Let $(X,\tau)$ be a topological space and $V\subseteq X$. For $x\in X$ are equivalent:
I) $x\in\overline{V}$
II) For every neighborhood $U$ of $x$ is $U\cap V\neq\emptyset$.
III) It exists a net $(x_\lambda)_{\lambda\in\Lambda}$ in $V$ with $x_\lambda\to x$
I want to show, that I) $\Rightarrow$ II) $\Rightarrow$ III) $\Rightarrow$ I)
But I do not really get started... :(
ad I)$\Rightarrow$ II):
Let $U$ be a neighborhood of $x$. Since $x\in\overline{V}$ it is $x$ element of every closed set $A$ which contains $V$.
But I do not see, how I now proof, that $U\cap V\neq\emptyset$. Can you give me a hint?
ad II)$\Rightarrow$ III)
A net $(x_\lambda)_{\lambda\in\Lambda}$ converges to $x\in V$ iff for all open $O\subseteq V$ with $x\in O$ exists $\lambda_0\in\Lambda$ such that for all $\lambda\geq\lambda_0$ is $x_\lambda\in O$.
ad III) $\Rightarrow$ I)
My problem here is, that I do not see how I get from the open sets to closed sets, so I can conclude $x\in\overline{V}$
I would like to try this again, but at the moment I need some hints. Thanks in advance
Well, closed sets are complements of open sets and vice versa. :)
I) $\Rightarrow$ II)
Let $x\in\overline{V}$ and assume that $U$ is a neighbourhood of $x$ such that $U\cap V=\emptyset$. Then $V\subseteq X\backslash U$ and since $X\backslash U$ is closed (being a complement of an open subset) then $\overline{V}\subseteq X\backslash U$ and therefore $U\cap\overline{V}=\emptyset$. Since $x\in U$ then $x\not\in\overline{V}$. Contradiction.
II) $\Rightarrow$ III)
Let $$\Lambda=\{U\subseteq X\ |\ U\mbox{ is open and }x\in U\}$$ be ordered by reversed inclusion (i.e. $U< W$ iff $W\subsetneq U$). Obviously it's a directed set. For any $U\in\Lambda$ define $x_U$ to be any element of $U\cap V$ (by assumption this subset is nonempty). Almost by definition $(x_U)_{U\in\Lambda}$ is convergent to $x$.
III) $\Rightarrow$ I)
Let $x\in X$ and $(x_\lambda)_{\lambda\in\Lambda}$ be a net convergent to $x$ with $x_\lambda\in V$ for any $\lambda$.
It is enough to show that for any closed subset $C$ such that $V\subseteq C$ we have that $x\in C$. So assume this is not true, i.e. we have a closed subset $C$ with $V\subseteq C$ and $x\not\in C$. Let $U=X\backslash C$. By definition $U\cap V=\emptyset$. But since $(x_{\lambda})$ converges to $x$ then $x_\theta\in U$ for some $\theta\in\Lambda$. By assumption $x_\theta\in V$ and thus $U\cap V\neq\emptyset$. Contradiction.