Metric Space Continuity by Neighborhood Basis

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Theorem

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.

Let $a \in A_1$ be a point in $A_1$.

Let $\BB_{\map f a}$ be a basis for the neighborhood system at $\map f a$.


$f$ is continuous at $a$ with respect to the metrics $d_1$ and $d_2$ if and only if:

for each neighborhood $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.


Proof

By definition, $\BB_{\map f a}$ be a basis for the neighborhood system at $\map f a$ if and only if:

$\forall N_a \subseteq M_2: \exists N \in \BB_{\map f a}: N \subseteq N_a$

where $N_a$ denotes a neighborhood of $\map f a$ in $M_2$.


Necessary Condition

Let $f$ be continuous at $a$ with respect to the metrics $d_1$ and $d_2$.

Then by Metric Space Continuity by Inverse of Mapping between Neighborhoods:

for each neighborhood $N$ of $\map f a$ in $M_2$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.

In particular, let $N \in \BB_{\map f a}$.

Then $f^{-1} \sqbrk N$ is a neighborhood of $a$.

$\Box$


Sufficient Condition

Let $f$ be such that:

for each neighborhood $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a neighborhood of $a$.

Let $N'$ be any neighborhood of $\map f a$.

Then by definition:

$\exists B \in \BB_{\map f a}: B \subseteq N'$

From Preimage of Subset is Subset of Preimage:

$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {N'}$

By hypothesis $f^{-1} \sqbrk B$ is a neighborhood of $a$.

From Superset of Neighborhood in Metric Space is Neighborhood it follows that $f^{-1} \sqbrk {N'}$ is a neighborhood of $a$.

As $N'$ is arbitrary, the result follows.

$\blacksquare$


Sources

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