Definition of continuous function with $\delta-\epsilon$

The definition says: Let $(X,d_{1})$ and $(Y,d_{2})$ be a metric space. A map $f:X\rightarrow Y$ is called continuous if for every $x\in X$ and $\epsilon>0$ there exists a $\delta>0$ such that:

$d_{1}(x,y)<\delta\Rightarrow d_{2}(f(x),f(y))<\epsilon$

does it mean that $x,y\in X$ and $f(x),f(y)\in Y$?

• Yes, that's exactly what that means. – DMcMor May 4 '17 at 20:12
• @DMcMor ok, so the use of elements $x,y$ was confusing, thanks! – gbox May 4 '17 at 20:13

Your wording is a bit unclear but here's a rephrasing of the definition that may be a bit more clear:

Fix $x\in X$. We say that $f:X\to Y$ is continuous at $x$ if for every $\epsilon>0$, there exists $\delta>0$ such that the following condition holds: if $y\in X$ such that $d_1(x,y)<\delta$, then $d_2(f(x),f(y))<\epsilon$.

We say $f$ is continuous if it is continuous at every $x\in X$.

If I've interpreted your question wrong, let me know and I can try to update my answer.

$\;\;\;\;f$ is continuous at $X$ $$\iff$$

$(\forall x\in X )\;\;\;f$ is continuous at $x$ $$\iff$$ $$(\forall x\in X) \;\;(\forall \epsilon>0)\;\;(\exists \delta>0 )\;:\;\color {green}{(\forall y\in X )}$$ $$(d_1 (x,y)<\delta\implies d_2 (f (x),f (y))<\epsilon).$$

of course $f (x)$ and $f (y)$ are in $Y$ since $f :X\to Y$.