Metric Space Continuity by Inverse of Mapping between Neighborhoods

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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 $A_1$.


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

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


Proof

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

for each neighborhood $N$ of $\map f a$ in $M_2$ there exists a corresponding neighborhood $N'$ of $a$ in $M_1$ such that $f \sqbrk {N'} \subseteq N$.


For a mapping $f: X \to Y$ we have:

$f \sqbrk U \subseteq V \iff U \subseteq f^{-1} \sqbrk V$

where $U \subseteq X$ and $V \subseteq Y$.

Hence the result.

$\blacksquare$


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