(False?) Proof that Monotone Darboux Functions are Continuous

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$$\tag{Wikipedia} I am somewhat familiar with a number of proofs of this statement, and have been trying to find new ways to prove it. I have come up with the following proof, with which I have some serious qualms: Assume f: [a,b] \to \mathbb{R} is an increasing Darboux Function. Let \epsilon>0 be given, and suppose that x<y with x,y\in[a,b]. Choose any L\in \left(f(x),f(y)\right), which is equivalent to f(x)<L<f(y) since f is increasing. If |f(x)-L|<\epsilon\, let y'=y and x'=x. If |f(x)-L|\ge\epsilon\, choose x' and y' such that L-\epsilon < f(x') < L < f(y') < L+\epsilon by Intermediate Value Property and note |f(x')-L|<\epsilon Let \delta = y' - x' Now, by Intermediate Value Theorem there exists c\in(x',y') such that f(x)=L. This implies that 0 < c-x' < y'-x' so that |x'-c|< y'-x'=\delta while |f(x')-f(c)| = |f(x')-L| < \epsilon Now, I am not confident that this proves continuity of f. Instead of following the standard process, namely 1. Starting with an \epsilon 2. Choosing some \delta and assuming |x-c|<\delta 3. Showing this implies |f(x)-f(c)|<\epsilon I feel that I have done everything out of order and thus my "proof" is invalid. Q: Assuming I am correct and that this proof is fundamentally flawed, is there any simply way to correct it without completely altering the proof? • A monotone increasing, Darboux function on [a, b] will necessarily have \alpha = f(a) \le f(x) \le f(b) = \beta, so it will map [a, b] \to [\alpha, \beta] surjectively because of the Darboux property. Then the answers given here apply immediately. – Chris Jul 28 '17 at 6:47 • @Chris completely true, and a very valid proof strategy. Nevertheless, I am already familiar with such a proof, and my post is concerned with validating the validity (or more likely, lack of validity) in my methods – Brevan Ellefsen Jul 28 '17 at 6:59 2 Answers I think that what you are doing is equivalent to : Let y_2<x<y_1. Given \epsilon >0, take y'_1\in (x,y_1) such that$$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. • 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? – Brevan Ellefsen Jul 31 '17 at 4:28 • The idea of letting$\delta= \min $(et cetera) is a common technique. – DanielWainfleet Jul 31 '17 at 4:32 • Common, yes. I've used it many times before. I just meant to say that I completely neglected that I could use it here! – Brevan Ellefsen Jul 31 '17 at 4:35 • I like your idea for a proof of continuity by this method. It's elegant and brief and does not require case-by-case analysis (As compared to showing that$\lim_{y\to x^-}f(y)=f(x)=\lim_{y\to x^+}f(y).$– DanielWainfleet Aug 1 '17 at 1:50 In the end, you still have show that the definition of continuity applies at any fixed point$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.

• 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! :) – Brevan Ellefsen Jul 31 '17 at 4:40
• @BrevanEllefsen: You're very welcome. Glad to make this active and help you crystallize your thinking. – RRL Jul 31 '17 at 5:38
• The last paragraph is the crux of the whole argument. +1 for the same. – Paramanand Singh Jul 31 '17 at 6:39