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I have been attempting to show that strictly increasing Darboux Functions are continuous, where $f$ is a Darboux Function if it has the so called "Intermediate Value Theorem", i.e.

for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$$\tag{Wikipedia}$

I am somewhat familiar with a number of proofs of this statement, and have been trying to find new ways to prove it. I have come up with the following proof, with which I have some serious qualms:

Assume $f: [a,b] \to \mathbb{R}$ is an increasing Darboux Function.
Let $\epsilon>0$ be given, and suppose that $x<y$ with $x,y\in[a,b]$. Choose any $L\in \left(f(x),f(y)\right)$, which is equivalent to $f(x)<L<f(y)$ since $f$ is increasing.

If $|f(x)-L|<\epsilon\,$ let $y'=y$ and $x'=x$.
If $|f(x)-L|\ge\epsilon\,$ choose $x'$ and $y'$ such that $L-\epsilon < f(x') < L < f(y') < L+\epsilon$ by Intermediate Value Property and note $|f(x')-L|<\epsilon$
Let $\delta = y' - x'$

Now, by Intermediate Value Theorem there exists $c\in(x',y')$ such that $f(x)=L$. This implies that $0 < c-x' < y'-x'$ so that $|x'-c|< y'-x'=\delta$ while $|f(x')-f(c)| = |f(x')-L| < \epsilon$

Now, I am not confident that this proves continuity of $f$. Instead of following the standard process, namely

  1. Starting with an $\epsilon$
  2. Choosing some $\delta$ and assuming $|x-c|<\delta$
  3. Showing this implies $|f(x)-f(c)|<\epsilon$

I feel that I have done everything out of order and thus my "proof" is invalid.

Q: Assuming I am correct and that this proof is fundamentally flawed, is there any simply way to correct it without completely altering the proof?

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  • $\begingroup$ A monotone increasing, Darboux function on $[a, b]$ will necessarily have $\alpha = f(a) \le f(x) \le f(b) = \beta$, so it will map $[a, b] \to [\alpha, \beta]$ surjectively because of the Darboux property. Then the answers given here apply immediately. $\endgroup$ – Chris Jul 28 '17 at 6:47
  • $\begingroup$ @Chris completely true, and a very valid proof strategy. Nevertheless, I am already familiar with such a proof, and my post is concerned with validating the validity (or more likely, lack of validity) in my methods $\endgroup$ – Brevan Ellefsen Jul 28 '17 at 6:59
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I think that what you are doing is equivalent to : Let $y_2<x<y_1.$ Given $\epsilon >0,$ take $y'_1\in (x,y_1)$ such that $$f(y'_1)\in [f(x),f(y_1)]\cap [f(x),f(x)+\epsilon)$$ and take $y'_2 \in (y_2,x)$ such that $$f(y'_2)\in [f(y_2),f(x)]\cap (f(x)-\epsilon, f(x)].$$ Let $\delta=\min (y'_1-x,x-y'_2).$ Then $\delta >0$ and $$\forall x'\in (-\delta+x,\delta+x)\; (\;-\epsilon +f(x)<f(y'_2)\leq f(x')\leq f(y'_1)< \epsilon +f(x)\;).$$

Remark: It is common usage that "$f$ is increasing" means $x<y\implies f(x)\leq f(y),$ and that "$f$ is strictly increasing" means $x<y\implies f(x)<f(y).$ Note that what I have written above does not require $f$ to be strictly increasing.

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  • $\begingroup$ wow. This is really fantastic way to rework my argument! I've diagrammed each step carefully to make sure this lines up with the geometric argument I had in my head. Great idea letting $\delta = \operatorname{min}(y'_1 - x, x - y'_2)$... I hadn't thought at all to try this! As near as I can tell, your argument here is exactly what I was thinking of but couldn't sort out enough to put into a logical flow. With your (serious) rewording and reorganization, is this now a valid proof of continuity? $\endgroup$ – Brevan Ellefsen Jul 31 '17 at 4:28
  • $\begingroup$ The idea of letting $\delta= \min $(et cetera) is a common technique. $\endgroup$ – DanielWainfleet Jul 31 '17 at 4:32
  • $\begingroup$ Common, yes. I've used it many times before. I just meant to say that I completely neglected that I could use it here! $\endgroup$ – Brevan Ellefsen Jul 31 '17 at 4:35
  • $\begingroup$ I like your idea for a proof of continuity by this method. It's elegant and brief and does not require case-by-case analysis (As compared to showing that $\lim_{y\to x^-}f(y)=f(x)=\lim_{y\to x^+}f(y).$ $\endgroup$ – DanielWainfleet Aug 1 '17 at 1:50
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In the end, you still have show that the definition of continuity applies at any fixed point $c$, finding an appropriate $\delta$ for any choice of $\epsilon > 0$.

You are correct that this proof is flawed and it breaks down even in the easier case where you assume $|f(x) - L| < \epsilon$.

Choosing an arbitrary $\epsilon > 0$ is a good first step. Next you choose arbitrary $x < y$ in $[a,b]$. By placing $L$ between $f(x)$ and $f(y)$, you have done nothing more than to fix a point $c$ between $x$ and $y$ where $f(c) = L$. As you say the existence of this point is guaranteed by the IVT. However, this is also fine in that you now have a point specified where you are trying to establish continuity.

For the first case you assume that $|f(x) - f(c)| = |f(x) - L| < \epsilon$. Of course this condition is only possible through the continuity you are trying to prove, but, nevertheless, you can proceed under this assumption and see where it leads. You set $\delta = y'-x' = y-x$ and assert that now we have both $|x' - c| < \delta $ and $|f(x') - f(c)| < \epsilon$ where the latter is true only by hypothesis. As this is an independent case, it must demonstrate that continuity at $c$ holds.

Now that $\delta$ has been fixed, if you choose any other point $x''$ satisfying $|x'' - c| < \delta$ it must be shown that $|f(x'') - f(c)| < \epsilon$. Since $x' < c < y'$ and $\delta = y' - x'$, if you choose an $x''$ between $x'$ and $c$, then $|f(x'') - f(c)| < \epsilon$, by the intermediate value property. Unfortunately, you can choose $x''$ such that $|x'' -c| < \delta$ and $x'' > y'$ and nothing is known about $f(x'')$ other than $f(x'') > f(y)$.

The only way to fix this is to replicate the standard argument where it is shown that a monotone function has a left and right limit at every point -- a consequence of the completeness of the real numbers.

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  • $\begingroup$ Thank you very much RRL. Your answer goes perfectly with Daniel's, as your answer has helped me find the flaws in my proof that I need to fix and Daniel's answer has shown me what patching those holes and rewriting the proof looks like. If I could accept both I would! :) $\endgroup$ – Brevan Ellefsen Jul 31 '17 at 4:40
  • $\begingroup$ @BrevanEllefsen: You're very welcome. Glad to make this active and help you crystallize your thinking. $\endgroup$ – RRL Jul 31 '17 at 5:38
  • $\begingroup$ The last paragraph is the crux of the whole argument. +1 for the same. $\endgroup$ – Paramanand Singh Jul 31 '17 at 6:39

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