Open ball and continuity in metric space I'm studying a bit of topology in metric space, and my text defines an open ball as follows : $(X,d)$ is a metric space, then an open ball centered at $x_0\in X$ is $B_{\epsilon}(x_0)=\{x\in X : d(x,x_0)<\epsilon\}\quad\epsilon>0$
Then we define continuity in metric space as follows : Let $(X,d_{X})$ and $(Y,d_{Y})$ be two metric spaces and $f: X\to Y$. Then $f$ is said to be continuous at $x_0$ if
$\forall\epsilon>0,\exists\delta: d_{X}(x,x_0)<\delta\implies d_{Y}(f(x),f(x_0))<\epsilon $
Seeing this definition, I can't help but make a connection with the definition of continuity in $\mathbb{R}$, but in $\mathbb{R}$ whether the equality is strict or not doesn't change anything, yet in this definition it does. So if I want to take $X =Y= R$ and $d_{X} = d_{Y} = \lvert .\rvert$ I would have to be satisfied with a strict inequality and then I have the impression that something is missing.
Could someone help me better understand what's going on here please?
Thank you a lot
 A: Fix a function $f\colon \Bbb R \to \Bbb R$, a point $x_0$, and consider the following properties:

Property 1:
$$
\forall \varepsilon > 0,\exists \delta > 0,\quad |x-x_0|\leqslant \delta \implies |f(x)-f(x_0)|\leqslant \varepsilon.
$$
Property 2:
$$
\forall \varepsilon > 0,\exists \delta > 0,\quad |x-x_0|< \delta \implies |f(x)-f(x_0)|< \varepsilon.
$$

Let us show that these two properties are equivalent.
Assume that property 1 holds.
Fix $\varepsilon >0$.
By assumption applied on $\frac{\varepsilon}{2}$, there exists $\delta >0$, such that
$$
|x-x_0|\leqslant \delta \implies |f(x)-f(x_0)|\leqslant \frac{\varepsilon}{2}.
$$
In particular,
$$
|x-x_0| < \delta \implies |x-x_0|\leqslant \delta \implies |f(x)-f(x_0)| \leqslant \frac{\varepsilon}{2} \implies |f(x)-f(x_0)|< \varepsilon,
$$
the last implication resulting on the fact that $0<\frac{\varepsilon}{2}<\varepsilon$.
Hence, we have proven property 2.
Assume that property 2 holds.
Fix $\varepsilon >0$.
By assumption, there exists $\delta > 0$ such that
$$
|x-x_0| < 2\delta \implies |f(x)-f(x_0)| <\varepsilon.
$$
Since $0<\delta <2 \delta$, it follows that
$$
|x-x_0| \leqslant \delta \implies |x-x_0| < 2\delta \implies |f(x)-f(x_0)| < \varepsilon \implies |f(x)-f(x_0)|\leqslant \varepsilon.
$$
The last inequality coming from the fact that if $a<b$, then $a\leqslant b$.
Hence, we have proven property 1.

The same ideas work for general metric spaces, since for $\varepsilon,\delta >0$, we have
$$
\renewcommand{\bar}{\overline}
B_{\delta}(x_0) \subset \bar{B}_{\delta}(x_0)\subset B_{2\delta}(x_0) \quad \text{and} \quad \bar{B}_{\frac{\varepsilon}{2}}(f(x_0)) \subset B_{\varepsilon}(f(x_0)) \subset \bar{B}_{\varepsilon}(f(x_0)),
$$
and
$$
A\subset B \implies f^{-1}(A)\subset f^{-1}(B).
$$
The reason most authors choose to define continuity in terms of a strict inequality rather than a large inequality is because in a more general topological space, continuity is defined in terms of open subsets.
