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My textbook gave a proof by contradiction of the following theorem:

Theorem: If $f(x)$ is a real-valued function that is continuous on $[a,b]$ then it is uniformly continuous on $[a,b]$

The proof by contradiction seemed overly complex and indirect. Was wondering if the following direct proof is valid.

Direct proof of Theorem

Since $f(x)$ is continuous on $[a,b]$ then for any $\epsilon > 0$ and $y,x \in [a,b],\; \exists \delta_x(\epsilon)>0 $ such that $|y-x|<\delta_x(\epsilon) \implies |f(y)-f(x)|< \epsilon$.

Letting $\delta(\epsilon) := \min_{x \in [a,b]} \delta_x(\epsilon)$ Then for each $\epsilon > 0$ and $x,y \in [a,b]$, we have $|x-y|<\delta(\epsilon) \leq \delta_x(\epsilon) \implies |f(x)-f(y)|< \epsilon\;\square$

This seems pretty direct to me and was wondering if I am missing something in this proof.

EDIT: As the great answers and comments have shown, my mistake/challenge is due to the ill-defined $\delta_x(\epsilon)$ function. I was assuming there was some (undefined) process that would select a finite positive number for each $\delta_x(\epsilon)$, but I did not specify it. I also rely on the continuity of $\delta_x(\epsilon)$ and that is not assured or proved.

  • Your $\delta(\epsilon)$ may be $0$. – Kavi Rama Murthy Oct 11 '22 at 05:37
  • @geetha290krm thanks! yes, what I had in my head and what I wrote were not the same -- I've updated it to better show my actual intent. –  Oct 11 '22 at 20:59
  • See https://math.stackexchange.com/questions/110573/continuous-mapping-on-a-compact-metric-space-is-uniformly-continuous#:~:text=The%20answer%20is%20yes%2C%20if,%2Cf(x)). If you are not familiar with metric spaces, replace $d(x,y)$ for $|x-y|$ – nicoyanovsky Oct 11 '22 at 21:14

2 Answers2

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The problem is in the argument that $\delta(\epsilon)$ exists because the minimum exists. Here, you are using a theorem that says that a minimum of a continuous function over a compact set always exists.

However, you don't have any reason to believe that the function $x\mapsto \delta_x(\epsilon)$ is continuous, so the argument is incorrect.

Magdiragdag
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5xum
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  • From continuity of $f(x)$ on $[a,b]$ why doesn’t the minimum $\delta$ exist ? The minimum is in the delta of a delta-epsilon definition of continuity. –  Oct 11 '22 at 06:18
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    @User5678 But there is more than one viable $\delta$ for each $\epsilon$. And for each $\epsilon$, the infimum of all $\delta$ values that can satisfy the inequality is actually $0$, because whenever $\delta$ is viable, then $\frac\delta2$ is also viable. – 5xum Oct 11 '22 at 06:25
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    @User5678 Also, "The minimum is in the delta of a delta-epsilon definition of continuity" is a pretty hand-wavy sentence, now that I think about it. Consider writing it in a more precise manner, currently it's so vague I don't really understand what you wanted to say with it. – 5xum Oct 11 '22 at 06:34
  • What I meant to say is that since the function $f$ is continuous on a closed interval, it then for every $x,y \in [a,b]$ and $\epsilon > 0$ we can find a $\delta_x(\epsilon) > 0$ such that $|x-y| < \delta_x(\epsilon) \implies |f(x)-f(y)| < \epsilon$. I did correct my earlier statement to use $\inf$ vs $\min$ as you correctly pointed out there is no reason that the $\delta_x$ needs to be continuous in $x$. I also corrected for your above comment that we can take arbitrarily small $\delta$ (I was picturing the largest such $\delta$ for each $x$ and $\epsilon$, so I made that explicit). –  Oct 11 '22 at 20:58
  • This was a great answer as well! +1 thank you :) –  Oct 11 '22 at 21:48
  • @User5678 The problem with taking the largest $\delta$ for each $x$ is that in that case, the function might not be infinite, or even well defined. For example, take any constant function $f$. Then, for any $\epsilon$, you can literally take any $\delta$, and the statement $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$ will be true (because $|f(x)-f(y)|$ will always be $0$. – 5xum Oct 12 '22 at 06:53
  • makes sense to me –  Oct 12 '22 at 14:33
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There are numerous problems with this approach:

  1. You are defining a function $\delta$ which depends on $x$ and $\varepsilon$—in your notation, $\delta_x(\varepsilon)$. However, it is not clear that this function is well-defined. For example, if $f(x) = C$ is a constant function on $[a,b]$, then it is possible to take $\delta_x(\varepsilon)$ to be any positive value you like for any $x \in [a,b]$. You are going to have similar problems for any assignment of $\delta_x(\varepsilon)$: if $\delta$ gets the job done for some particular choice of $x$ and $\varepsilon$, then $\delta/2$ will also get the job done.

    How do you choose $\delta_x(\varepsilon)$?

  2. You are attempting to invoke a version of the Extreme Value Theorem, which states

    If $g : [a,b] \to \mathbb{R}$ is continuous, then there are $c, d \in [a,b]$ such that $$ g(c) \le g(x) \le g(d)$$ for all $x \in [a,b]$.

    That is, continuous functions on closed, bounded (i.e. compact) domains attain their extreme values. But, as noted above, you have no control over the choice $\delta_x(\varepsilon)$, which means that you have no control on whether or not $\delta_x(\varepsilon)$ is a continuous function of $x$.

    For a rather extreme example, suppose that $f : [a,b] \to \mathbb{R} : x\mapsto C$ is constant function. Then for any $\varepsilon > 0$, any $\delta > 0$, and any $x\in [a,b]$, it will follow that if $y\in [a,b]$, then $$ |x-y| < \delta \implies |f(x) - f(y)| = 0 < \varepsilon. $$

    So, for example, I could define $$ \delta_x(\varepsilon) = \begin{cases} \dfrac{1}{p} & \text{if $x = \frac{p}{q} \in \mathbb{Q} \cap [a,b]$ and $\gcd(p,q)=1$, and} \\[1ex] 1 & \text{if $x \not\in\mathbb{Q} \cap [a,b]$.} \end{cases} $$ (If $0 \in [a,b]$, adopt some convention regarding the assignment of $\delta_0(\varepsilon)$). This function does not attain a minimum on the interval $[a,b]$.

    How do you know that it is possible to choose $\delta_x(\varepsilon)$ so that it is a continuous function of $x$?

Xander Henderson
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  • Thanks for your thoughtful critique -- I have noted that the use of "min" was problematic and moved to inf. However, that was still problematic b/c, as you and @5xum pointed out, you can chose arbitrarily small values of $\delta$ so it is not well defined. I've attempted to fix this by looking at the largest such delta for each $x,\epsilon$, which I think surely exists due to the assumed continuity of $f$, right? Still not sure its right and I've learned a ton from your and 5xum's responses :) +1 –  Oct 11 '22 at 21:16
  • @User5678 I genuinely think that there is no way of patching up your approach. You simply don't have enough control over the value of $\delta_x$. – Xander Henderson Oct 11 '22 at 21:23
  • i see -- so the issue is that even though we know we can chose a finite $\delta_x$ for each $x,\epsilon$ that doesn't guarantee that $\delta_x$ is bounded for any given $\epsilon$? maybe I was being too complex and all I need to use is that the continuity of $f$ implies that I can assign a finite $\delta$ to each $x$ for every choice of $\epsilon$, hence the supremum is bounded over $[a,b]$ and then uniform continuity follows? –  Oct 11 '22 at 21:32
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    Again, how do you get any control over $\delta_x$, as a function of $x$? What theorem or result can you invoke which guarantees that you will get a positive lower bound? If you want a direct proof, I would suggest that you focus not on the Extreme Value Theorem, but on compactness: The open intervals ${ (x-\delta_x, x+\delta_x) : x \in [a,b] }$ form an open cover of $[a,b]$. The interval $[a,b]$ is compact, so you can extract a finite subcover ${ (x_n - \delta_{x_n}, x_n + \delta_{x_n} }$. Take $\delta = \min_{n} \delta_{x_n}$. – Xander Henderson Oct 11 '22 at 21:42
  • so what you did is use compactness to force the (arbitrary) choice of $\delta_x$ down to a finite set and hence one that has a minimum. That is a great suggestion! Thank you. I see my mistake(s) clearly now -- I was trying to get $\delta_x$ to be selected by an undefined choice function over an uncountable set, whereas you were able to sidestep that by constructing a finite set from compactness. –  Oct 11 '22 at 21:48
  • just for clarity -- does the supremum exist for each set of valid values $\delta_x(\epsilon)$? –  Oct 11 '22 at 21:55
  • Xander henderson, tanking just min doesnt work. You need $\delta = \min {\dfrac{\delta_{xi}}{2}}$ to guarantee than whenever you take $x,y$ such that $|x-y| < \delta$ then both $x$ and $y$ lie on some $(x-\delta_{xi}, x+\delta{xi})$ – nicoyanovsky Oct 11 '22 at 22:16
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    @nicoyanovsky Indeed. I threw that out that as a quicky, without thinking about it too hard. Even taking half the minimal $\delta$ doesn't seem to quite work. Rather, the best option might be to take something akin to $\delta = \frac{1}{2} \min |x_n - x_m|$, taken over all $n,m$. In any event, I don't intended for my comments to be a complete answer---rather, a direct proof should probably make use of compactness in this way, with the details "left to the reader". – Xander Henderson Oct 11 '22 at 22:30