Constant Function is Uniformly Continuous

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Metric Space

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 uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Real Function

This result can be directly applied to the real numbers considered as a metric space:

Let $f_c: \R \to \R$ be the constant mapping:

$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$

Then $f_c$ is uniformly continuous on $\R$.