Sometimes people say that the concept of a continuous map express, intuitively, that "small changes in the domain must yield small changes in the respective images, in the codomain".

However, this is not the definition we actually use. We say instead: "If one wants small changes in the image, say, by $\epsilon$, one can find a $\delta$ (depending on $\epsilon$), such that, changes in the domain of at most $\delta$, reflects small changes in the images, by the $\epsilon$ previously given."

So, thinking of the intuition given above, consider the following definition.

A function $f:\mathbb{R}\to \mathbb{R}$ is said to be continuous at $x\in \mathbb{R}$ if for all $\delta>0$, there is a $\epsilon(\delta)$ such that $f((x-\delta,x+\delta))\subset (f(x)-\epsilon,f(x)+\epsilon)$.

Ok, so here I'm just switching the symbols of $\forall$ and $\exists$ in the usual definition of continuity. This definition is clearly bad, because, given a $\delta$, I can get a tremendously big $\epsilon$ and many functions that we don't want to be continuous would be it, with this definition.

But, what if we add to that definition:

A function $f:\mathbb{R}\to \mathbb{R}$ is said to be continuous at $x\in \mathbb{R}$ if there is a (choice) function $\epsilon_x:\mathbb{R}_+\to \mathbb{R}_+$ such that:

(1) $f((x-\delta,x+\delta))\subset (f(x)-\epsilon_x(\delta),f(x)+\epsilon_x(\delta))$, for all $\delta>0$;

(2) $\lim\limits_{\delta\to 0^+}\epsilon_x(\delta)=0$.

We say that $f$ is continuous if $\epsilon_x$ exists for all $x\in \mathbb{R}$.

Now, I'm saying that, when $\delta$ goes small, so does $\epsilon$. Is this definition still bad?

The next step would be to check if this definition implies the usual one. Well, let $\epsilon>0$ be given. By (2), there is a $\delta'>0$ such that, $0<\delta<\delta'\implies 0<\epsilon_x(\delta)<\epsilon$, and using $\delta'$ in the usual definition allows us to conclude that $f$ is continuous at $x$ in the usual sense.

So, are these definitions the same? I'm a little confused about the converse... (it smells like axiom of choice...)

  • $\begingroup$ I am not sure what you're asking. Are you asking about the axiom of choice, or about your definition? I don't see why set theory is related here... $\endgroup$ – Asaf Karagila Sep 14 '18 at 17:55
  • $\begingroup$ @AsafKaragila Set theory may be related if we need the axiom of choice. Not sure, though. What I'm asking is: Is my definition equivalent to the usual one? $\endgroup$ – Ders Sep 14 '18 at 17:57
  • $\begingroup$ The axiom of choice is needed to have the equivalence in general between $\varepsilon$-$\delta$ and preserving limits. I would first figure out if your definitions are equivalent in general, then I'll try to see what the axiom of choice has to do with the equivalence of whatever definitions. $\endgroup$ – Asaf Karagila Sep 14 '18 at 17:58
  • $\begingroup$ I don't see any connection to the axiom of chioce. But, towards the bottom line, - even if these are the same, what is the goal of making the definition of continuity that difficult-to-formulate? $\endgroup$ – metamorphy Sep 14 '18 at 17:59

As you showed, if $f$ satisfies your definition it is continuous at $x$. Conversely, suppose $f$ is continuous at $x$. We may take $$\epsilon_x(\delta) = \sup_{t \in [x-\delta, x+\delta]} |f(x)-f(t)| $$ (using the fact that a continuous function on a closed interval has a maximum there) and your condition will be satisfied. No axiom of choice needed here.

  • $\begingroup$ Nice. So the intuitive notion of continuity is indeed "correct" (agrees with the usual formal definition), if the intuitive definition is that given by me (there could be other possible formulations) $\endgroup$ – Ders Sep 14 '18 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.