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.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.7$