Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product

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 continuous mapping.

Let $\AA = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
 * $\ds \map {d_\infty} {x, y} = \max \set {\map {d_1} {x_1, y_1}, \map {d_2} {x_2, y_2} }$

where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.

Let $\Gamma_f$ be the graph of $f$.

Then $\Gamma_f$ is a closed set of $\struct {\AA, d_\infty}$.

Proof
Let $f: A_1 \to A_2$ be continuous.

Then:
 * $\forall a \in A_1: \ds \lim_{n \mathop \to \infty} x_n = a \implies \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$

This can be extended to ordered pairs and ordered tuples, because:


 * $\ds \lim_{n \mathop \to \infty} x_n = a \iff \exists N \in \N: n > N \implies x_n \in V$

where $V$ is some neighborhood of $a$.

That is:
 * $n > N \implies x_n \in \map {B_\delta} a$

for some $N \in \N$ and all $\delta \in \R_{>0}$.

So:

That is, each sequence of points $\sequence {a_n, b_n}$ such that:
 * $\map f {a_n} = b_n$

converges to a point $\tuple {a, b}$ such that:
 * $\map f a = b$

Hence:
 * $\ds \lim_{n \mathop \to \infty} \tuple {a_n, b_n} = \tuple {a, \map f a} = \Gamma_f$

That is, $\Gamma_f$ contains all its limit points.

Hence the result by definition of closed set.

Also see

 * Mapping whose Graph is Closed in Chebyshev Product is not necessarily Continuous