Constant Function is Uniformly Continuous

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
Let $$f_c: \R \to \R$$ be the constant mapping:
 * $$\exists c \in \R: \forall a \in \R: f_c \left({a}\right) = c$$

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

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 $$M_2 = \left({A_2, d_2}\right)$$.

Let $$\epsilon \in \R: \epsilon > 0$$.

Pick any $$\delta \in R$$ such that $$\delta > 0$$.

Let $$x, y \in A_1$$ such that $$d_1 \left({x, y}\right) < \delta$$.

Now we have:
 * $$f_c \left({x}\right) = c = f_c \left({y}\right)$$

Hence:
 * $$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$$

By definition of a metric:
 * $$d_2 \left({c, c}\right) = 0 < \epsilon$$.

Thus we have that:
 * $$\forall \epsilon > 0: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$$

Hence by definition of uniform continuity on a metric space, $$f_c$$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

The result for real numbers follows directly from the fact that the Real Number Line is Metric Space.