Constant Function is Uniformly Continuous

Topological Space
Let $$T_1 = \left({A_1, \vartheta_1}\right)$$ and $$T_2 = \left({A_2, \vartheta_2}\right)$$ be topological spaces.

Let $$f_C: A_1 \to A_2$$ be the constant mapping from $$A_1$$ to $$A_2$$:
 * $$\exists C \in \vartheta_2: \forall U \in \vartheta_1: f_C \left({U}\right) = C$$

That is, every open set in $$\vartheta_1$$ maps to the same open set $$C$$ in $$\vartheta_2$$.

Then $$f_C$$ is continuous with respect to $\vartheta_1$ and $\vartheta_2$.

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

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 $d_2$.

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

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