Characterization of Openness in terms of Nets

Theorem

Let $\struct {X, \tau}$ be a topological space.

Let $U \subseteq X$.

Then $U \in \tau$ if and only if for each:

$x \in U$

directed set $\tuple {\Lambda, \preceq}$

net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$

there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.

Suppose that $U \in \tau$.

Take $x \in U$, a directed set $\tuple {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$.

By the definition of convergence for net, we have:

there exists $\lambda_0 \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$ we have $x_\lambda \in U$.

In particular, $x_{\lambda_0} \in U$.

$\Box$

Suppose that for each:

$x \in U$

directed set $\tuple {\Lambda, \preceq}$

net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$

there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.

Aiming for a contradiction, suppose $U \not \in \tau$.

Then $X \setminus U$ is not closed.

From Set is Closed iff Equals Topological Closure, we have $\map \cl {X \setminus U} \ne X \setminus U$.

Let $x \in \map \cl {X \setminus U} \setminus \paren {X \setminus U}$.

That is, $x \in \map \cl {X \setminus U} \cap U$.

there exists directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $X \setminus U$ converging to $x$.

By hypothesis, since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda \in \Lambda$ such that $x_\lambda \in U$.

This contradicts that $x_\lambda \in X \setminus U$ for each $\lambda \in \Lambda$.

$\blacksquare$