Understanding the definition of continuity between metric spaces. As we know that $\epsilon-\delta$ definition of continuity between metric spaces $X$ and $Y$ can be stated as follows:
A map f:$(X, d_X)\rightarrow (Y, d_Y)$ is said to be continuous at a
point $p\in$ X if for a given $\epsilon >0$, $\exists$  $\delta >0$ such
that $d_X(p,x)< \delta\Rightarrow d_Y(f(p),f(x))< \epsilon
$
I need help to understand this definition. Can we interpret it geometrically?
How this definition is related with the definition of continuity of real variables?
Sincerely thanks
 A: With real variables, we were saying that given an open interval around $f(p)$ of radius $\epsilon$--that is, $(f(p)-\epsilon,f(p)+\epsilon)$--we could find an open interval around $p$ of sufficiently small radius $\delta$--$(p-\delta,p+\delta)$--such that whenever $x$ was in the interval around $p$, then $f(x)$ would be in the corresponding interval around $f(p)$.
It is the same idea in metric spaces. Instead of open intervals of given radius, we talk of open "balls" of given radius--so the ball of radius $r$ about a point $y$ in the space $Y$, for instance, would be $\{z\in Y:d_Y(y,z)<r\}$.
Intuitively, it just means we can tighten our focus in the domain sufficiently so that the corresponding range values remain within any given ball around a given range value, no matter how small the ball.
A: Let's look at the $\varepsilon$-$\delta$-definition in $\mathbb R$:
We say that $f$ is continuous at a point $p$ if for $\varepsilon > 0$ we can find a $\delta > 0$ such that if $|x-p| < \delta$ then we have $|f(p) - f(x)| < \varepsilon$.
What does the set of $x$ with $|x-p| < \delta$ look like? It looks like this: $(p-\delta, p + \delta)$.
Now the definition of continuity of $f$ tells you that $f((p-\delta, p + \delta)) \subset (f(p) - \varepsilon, f(p) + \varepsilon)$. Geometrically, this says that $f$ maps points that are $\delta$-close to $p$, to points $\epsilon$-close to $f(p)$.
This is the same as saying that given $\varepsilon >0$ there exists  $\delta >0$ such that $d_X(p,x)< \delta \Rightarrow d_Y(f(p),f(x))< \varepsilon$.
