Characterization of Continuity in terms of Nets

Theorem
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological spaces.

Let $f : \struct {X, \tau_X} \to \struct {Y, \tau_Y}$ be a function.

Let $x \in X$.

Then $f$ is continuous at $x$ :


 * for every directed set $\struct {\Lambda, \preceq}$ and Moore-Smith sequence $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, \tau_X}$, we have that:


 * the Moore-Smith sequence $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$ in $\struct {Y, \tau_Y}$.

Necessary Condition
Suppose that $f$ is continuous at $x$.

Let $\struct {\Lambda, \preceq}$ be a directed set.

Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a Moore-Smith sequence such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.

Let $U$ be an open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$.

Then there exists an open neighborhood $V$ of $x$ in $\struct {X, \tau_X}$ such that:


 * $f \sqbrk V \subseteq U$

Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_0 \in \Lambda$ such that:


 * $x_n \in V$ for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.

Then:


 * $\map f {x_n} \in U$

for $\lambda_0 \preceq \lambda$.

Sufficient Condition
Suppose that $f$ is not continuous at $x$.

Let $\Lambda$ be the set of open neighborhoods of $\map f x$.

Then $\struct {\Lambda, \supseteq}$ is a directed set from Open Neighborhoods of Point form Directed Ordering.

Since $f$ is not continuous at $x$, there exists an open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$ such that:


 * $f \sqbrk V \not \subseteq U$

for all open neighborhoods $V$ of $x$ in $\struct {X, \tau_X}$.

So for each open neighborhoods $V$ of $x$, we can pick $x_V \in V$ such that:


 * $\map f {x_V} \not \in U$

Since $U$ is an open neighborhood of $\map f x$, we certainly have:


 * $\family {\map f {x_U} }_{U \in \Lambda}$ does not converge to $\map f x$.

We show that:


 * $\family {x_U}_{U \in \Lambda}$ converges to $x$.

If $V$ is an open neighborhood of $x$ and $V \supseteq U$, we have:


 * $x_U \in U$

so that:


 * $x_U \in V$

So $\family {x_U}_{U \in \Lambda}$ converges to $x$.