If $f$ and $g$ are continuous in an arbitrary metric space $X$, show that $f + g$ is also continuous
I'm trying to prove this, but a problem arises, which I'll show below :
We'll let $X$ and $Y$ be metric spaces, with $E_1 \cup E_2 \subset E \subset X$, and $H_1 \subset Y$ and $H_2 \subset Y$.
Put $f : E_1 \to H_1$ and $g: E_2 \to H_2$
Since $(f+g)(x) := f(x) + g(x)$. I would think to satisfy the definition of continuity we'd need to show that for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$d_Y(f(x) + g(x), f(p) + g(p)) < \epsilon \ \ \ \ \ \ (1)$$ for all points $x \in E$ for which $d_E(x, p) < \delta $.
The Problem
Put the problem is that $a + b$, is not defined in a arbitrary metric space. Binary operations need not be defined in arbitrary metric spaces, and addition is a binary operation. Thus $(1)$ is meaningless.
So how would I go about proving this considering the fact that addition isn't even defined in arbitrary metric spaces?
It says "$f$ and $g$ are continuous in an arbitrary metric space $X$". Implicit in that are: $(1)$ that the domains of $f$ and $g$ are both equal to the entire space $X$ (so you need not consider any other supspaces of $X$ than $X$ itself, in the role of domains) and $(2)$ the codomain is some metric space such as $\mathbb R$ or $\mathbb R^n$ in which addition is defined. In effect those are hypotheses of the theorem, presupposed by the way the theorem is stated.