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Prove that if we take away supremum axiom of real numbers, that is to say, if there exists a subset of real numbers that is bounded from above but does not have a supremum, then there exists a function defined on a closed interval that is differentiable but does not satisfy the intermediate value theorem nor the Lagrange mean value theorem, that is bounded but does not have maximum or minimum.

My try: if we delete some elements from the set of real numbers, then what is left does not satisfy the supremum axiom. For example if $\pi$ is removed, then $\sin(x)$ would not satisfy the intermediate value theorem. But if an arbitrary real number is removed, how do I construct such a function?

Any help is appreciated!

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  • $\begingroup$ IVT doesnt only apply at 0’s. Perhaps look at the more general version of the theorem. $\endgroup$ – T. Fo Jan 8 at 5:32
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So there is some subset bounded above with no supremum. Let's call that set $A$.

Let $f(x) = 0$ if $x$ is not any upper bound of $A$ and let $f(x) =1$ if $x$ is an upper bound of $A$..

Trick is to show that this actually satisfies the definition of differentiable.

The thing is normally continuity (and differentiability) will fail at $x = \sup A$ and this is the only point it does fail. But if $x = \sup A$ simply does not exist.....

... try to work it out on your own. If you need more, read on.....

If $x$ is not an upper bound of $A$ then there is $a \in A; a > x$ and so all $y < a$ are also not upper bounds of $A$. So for any $h$ so that $h \ne 0$ and $|h| < a-x$, then $x+h$ is not an upper bound of $A$ so $f(x+h) = f(x) = 0$.

And so $\lim\limits_{h\to 0} \frac {f(x+h) - f(x)}h = \lim\limits_{h\to 0}\frac {0-0}h = 0$. So $f$ is differentiable at $x$ if $x$ is not an upper bound of $A$.

And if $x$ is an upper bound of $A$, $x$ is not the supremum or least upper bound of $A$ (as that doesnt exist!) so there is a $b< x$ that is also an upperbound of $A$. And all $y > b$ are also upper bounds of $A$. So for any $h$ so that $h \ne 0$ and $|h| < x-b$, then $x+h$ is an upper bound of $A$ so $f(x+h) = f(x) = 1$.

And so $\lim\limits_{h\to 0} \frac {f(x+h) - f(x)}h = \lim\limits_{h\to 0}\frac {1-1}h = 0$. So $f$ is differentiable at $x$ if $x$ is an upper bound of $A$.

So $f$ is differentiable at all points in $\mathbb R$.

But it clearly fails the Intermediate Value Theorem. Let $c$ not be an upper bound of $A$ and let $d$ be an upper bound then $f(c) = 0$ and $f(d) = 1$ but there is no $e \in [c,d]$ so that $0 < f(e) < 1$.

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This fails if $\mathbb R$ has the supremum axiom.

If we allow the supremum axiom. then there is some $\alpha = \sup A$.

And $f$ is neither differentiable nor continuous at $x = \alpha$.

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  • $\begingroup$ Thank! That's very clever! $\endgroup$ – Jiu Jan 8 at 9:01
  • $\begingroup$ If you frame the question "what is it about the axiom of suprema that allows IVT" it's that differentiability implies continuity and you cant jump from a value to another without passing between the points in between. But if there is no supremum you can jump from points from points in a set to points above a set. $\endgroup$ – fleablood Jan 8 at 18:10
  • $\begingroup$ yes I realized that we lose connectedness when we remove the axiom of suprema. $\endgroup$ – Jiu Jan 9 at 3:44

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