Let $f:X \rightarrow Y$ (and let $A \subset X$ where $f$ is continuous.$)^{(1)}$

If $f$ is continuous on $A \equiv $ { $x\in X$: $\forall ε>0 , \exists δ>0$ such as that $d_Y(f(z),f(y))<ε ,\forall \ z,y \in B(x,δ)$}$^{(2)}$ (where $d_Y$ metric on Y)

My thoughts are that, the deffinition describes a uniformly continuous function, as for my doubts, they are created by the fact that the handscript I was reading claims that this definition is equal to continuous function. Also I am aware of the fact that a uniformly continuous function is a continuous one and this suggests that $A \subseteq (2)$ whereas the sentence above claims that $A \equiv (2) $

  • $\begingroup$ The notation is a little unclear to me: if in your definition 2 the delta is independent of x then it is describing uniform continuity. Otherwise it is the definition of just continuity. To me it looks like the delta is independent of x, as it is written. $\endgroup$ – Brahadeesh Dec 4 '17 at 13:20
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    $\begingroup$ I read it as, choose those $x \in X$ such that for all $\epsilon$ there exists $\delta$. Which sounds as continuity (not uniform) to me. @Brahadeesh $\endgroup$ – Olba12 Dec 4 '17 at 13:22
  • $\begingroup$ That's exactly my problem I can't figure out if it's independent or not I've read the sentence dozens of times and I started wondering if I miss something. Although what I stated above is exactly what the handscript says. $\endgroup$ – Pookaros Dec 4 '17 at 13:25
  • $\begingroup$ @Olba12 right, this makes more sense. I take back what I said earlier. It looks like continuity. $\endgroup$ – Brahadeesh Dec 4 '17 at 13:31
  • $\begingroup$ The relation between $ε,δ$ if one exists, is possible to be ascertained in the sentence as a deduction and not be given when $ε,δ$ are defined (for example if I change $f(z)$ with $f(x)$). But i guess u are right after all@Olba12 $\endgroup$ – Pookaros Dec 4 '17 at 13:31

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