Doubt about definition of continuity involving open ball.

Question

I've seen a definition of continuity on the resolution of an excersice that left me a little bit sceptical.

The definition stated that a function $f$ is continuous at a point $x$ if $\forall \epsilon>0$ there's a $\delta>0$ such that $|f(z)-f(y)|<\epsilon$ whenever $z, y \in B(x,\delta)$.

I may be ignoring something, but it seems that this definition is not ok. What about removable discontinuities? Any thoughts?

Answer 1

This is not the $\varepsilon-\delta$ definition of continuity in the context of metric spaces. However, it is equivalent to the usual definition, which is$$(\forall\varepsilon>0)(\exists\delta>0):d(x,y)<\delta\implies d\bigl(f(x),f(y)\bigr)<\varepsilon.\tag1$$In fact, if your condition holds, then just take $z=x$ abd $(1)$ will hold. And if $(1)$ holds, then, given $\varepsilon>0$, take $\delta>0$ such that $d(x,y)<\delta\implies d\bigl(f(x),f(y)\bigr)<\frac\varepsilon2$. Then the condition that you stated will hold.

Answer 2

The given condition implies $|f(x)-f(y)| <\epsilon$ whenever $y \in B(x,\delta)$ (Just take $z-x$). Conversely if we have $|f(x)-f(y)| <\epsilon$ whenever $y \in B(x,\delta)$ we can use the fact that $|f(z)-f(y)| \leq |f(z)-f(x)|+|f(x)-f(y)|$ to see that the condition holds when $f$ is continuous.