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Note: There is another question with the same title, but it is essentially different.

The usual definition of continuity of a real-valued function of a single real variable:

Definition 1: $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous at $a$ if for every $\epsilon>0$, there is $\delta>0$ such that $|x-a|<\delta\implies |f(x)-f(a)|<\epsilon$

An alternative definition of continuity:

Definition 2: $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous at $a$ if there is a function $b:\mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}\cup\{\infty\}$ such that:

  1. If $|x-a|<\delta$, then $|f(x)-f(a)|<b(\delta)$
  2. $\lim_{\delta\rightarrow 0}b(\delta)=0$

If $f$ is continuous at $a$ in the second sense, then it is continuous in the first sense; this can be proven without using the Axiom of Choice.

But if $f$ is continuous at $a$ in the first sense, and we want to construct a function $b$ to show that it is also continuous in the second sense, we have to use the Axiom of Choice...right?

Question: Is there a model of the real numbers where there is a function which is continuous in the first sense but not the second?

semisimpleton
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  • You might want to allow your function $b$ to possibly take value $\infty$, as otherwise the function $f$ can be continuous at $a$ but behave wildly outside a small interval around $a$, in which case if $\delta$ is larger than the radius of that small interval, you cannot guarantee the existence of a non-infinite $b(\delta)$. – David Gao Nov 19 '24 at 02:52
  • Oh right, thanks! – semisimpleton Nov 19 '24 at 02:55

1 Answers1

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No. The axiom of choice is not needed to go from the first definition of continuity to the second. Indeed, assume $f$ is continuous at $a$ in the first sense. Then we may define,

$$b(\delta) = \delta + \sup_{x: |x-a| < \delta} |f(x)-f(a)|$$

It is easy to check that $b(\delta)$ satisfies the conditions in Definition $2$, so $f$ is continuous in the second sense as well.

David Gao
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    I don't know why I didn't think of that! Also, doesn't $b(\delta)=\sup |f(x)-f(a)|$ suffice? Why the extra $\delta$ term? – semisimpleton Nov 19 '24 at 03:39
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    @semisimpleton If $b(\delta)$ is just the supremum, then you only get $|f(x)-f(a)|\leq b(\delta)$ instead of the strict inequality you required in your Definition $2$. Also, the supremum could be $0$, which is not in $\mathbb{R}_{>0}\cup{\infty}$. So, adding $\delta$ is just a technicality, to satisfy those two requirements. – David Gao Nov 19 '24 at 04:12