# Constant Function is Continuous/Metric Space/Proof 1

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## Theorem

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:

- $\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$

That is, every point in $A_1$ maps to the same point $c$ in $A_2$.

Then $f_c$ is continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

## Proof

Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.

From Constant Function is Uniformly Continuous, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

The result follows from Uniformly Continuous Function is Continuous.

$\blacksquare$