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

• Which inequality can be strict or not? There are two inequalities in that definition of continuity. Commented Dec 1, 2022 at 20:59
• How does strict or non-strict inequality make a difference here? Commented Dec 1, 2022 at 21:02
• What JCS is getting at is the fact that it is no big deal if we have $\overline{B}_\delta(x_0) \subset f^{-1}\left(B_\epsilon(f(x_0))\right)$ because we still have $B_\delta(x_0) \subset \overline{B}_\delta(x_0)$. Commented Dec 1, 2022 at 21:03
• The one in metric space is always strict from what I have seen so far, but in $\mathbb{R}$ I have always see a non-strict inequality. The difference is that if $\epsilon=5$ and $\delta=3$ for exemple, I did not end with the same sets $B_{\delta}(x_0)$ and $B_{\epsilon}(f(x_0))$, is this not a problem ? Or is there an $\epsilon-\delta$ argument that I ignores ? Commented Dec 1, 2022 at 21:16
• @coboy Can you state precisely for which quantity you would like to have a large inequality? Is it on $\varepsilon$, on $\delta$, on $|x-x_0|$, on $|f(x)-f(x_0)|$? Commented Dec 1, 2022 at 21:43

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 $$a, then $$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.

• Thank you a lot Didier ! Your answer really makes sense !! Commented Dec 1, 2022 at 22:29