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 $x$ if and only if:

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

the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$ in $\struct {Y, \tau_Y}$.

Proof

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 net 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 $\lambda_0 \preceq \lambda$.

Since $U$ was an arbitrary open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$, we have that $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.

$\Box$

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 neighborhood $V$ 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:

$\family {\map f {x_V} }_{V \in \Lambda}$ 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 $x$.

So:

if $f$ is not continuous at $x$, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, \tau_X}$ but $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ does not converge to $\map f x$ in $\struct {Y, \tau_Y}$.

By contraposition, we have the demand.

$\blacksquare$