1
$\begingroup$

OK, My book has a proof that a continious function defined on $[0,1]$ attains all values between $f(0)$ and $f(1)$ using some ultra case bashy stuff, but I have two different proofs, is those correct ?

(a) Let the desired value be $m$. We prove that there exists a sequence of reals $\{a_i\}_{i=0}^{\infty}$ such that $lim \; a$ exists, and $0 \leq a_i \leq 1$ for all $i$, satsifying $lim\; f(a_i) = m$. So, by the defination of cont

Construction: From the defination of continiuty, for $x \in [0,1]$ given any $\epsilon$, there exists a nonzero $h(\epsilon, x):= \delta$ such that for all $y$ with $|y - x| < \delta$, $|f(y) - f(x) | < \epsilon$ Let $g(\epsilon) := min\{h(\epsilon, x) | x \in [0,1]\}$.

Now, set $\epsilon_0 := 10^{-10000}$. split $[0,1]$ into $ \lfloor \frac{1}{g(\epsilon_0)} \rfloor $ equal intervals $I_1, I_2, \cdots, I_{\text{A big number}}$ , and let $a_0$ be the lower bound of the interval $I_i$, of which $\max{f(x) | x \in I_i} \geq m \geq \min{f(x) | x \in I_i}$.

Now, set $\epsilon_{1} = \epsilon_0^{100000}$, and divide the $I_i$ into $ \lfloor \frac{1}{g(\epsilon_1)} \rfloor $ equal interval, and choose $a_1$ to be the lower bound of the interval $I_{i_{j}}$ for which $\max{f(x) | x \in I_{i_{j}}} \geq m \geq \min{f(x) | x \in I_{i_{j}}}$

Repeat the process.

(b) Another proof: Assume WLOG $f(0) < f(1)$. Let the desired number be $m$. If $f(0) = m$ or $f(1) = m$, then we're done. Otherwise, divide the reals in $[0,1]$ into sets $L$ and $R$ such that:

  1. $y \in L$ if and only if $max\{ f(x) | 0 \leq x \leq y \} \leq a$
  2. Otherwise, put $y$ in $R$.

Now, $L$ exists because as $m \neq f(0)$, we can pick very small epsilon $\epsilon$ such that for $0 \leq x \leq \epsilon$, $f(x) < m$, and $R$ exists by the analogous arguement on $f(1)$.

Now it's well known that a number $y$ exists such that for all member of $L$ is smaller or equal to it, and all members of $R$ is larger or equal to it. As $y$ must be inside $[0,1]$, we're done.

$\endgroup$
  • 1
    $\begingroup$ In your title you should change "mean" to "intermediate". $\endgroup$ – amsmath Oct 24 '17 at 17:45
  • $\begingroup$ lol $\endgroup$ – cdt Oct 24 '17 at 17:54
  • $\begingroup$ Your second proof is fine in its overall shape, though you mean $m$ instead of $a$, and depending on the context of the proof (first course in analysis?) you might want to be clear why you can WLOG that. You should also clarify that the set in your step 1 exists because the continuous image of a compact set is compact. And the word "exists" in the following paragraph should be replaced by "is non-empty", and you should probably spell out exactly why the number $y$ has $f(y) = m$. $\endgroup$ – Patrick Stevens Oct 24 '17 at 17:59
  • $\begingroup$ @PatrickStevens Thanks for your reply ! Unfortunately, I don't know what compactness means still now :( (BTW, The notaion in first and second proofs are seperate..., so there is no $a$ in the second proof. ) $\endgroup$ – cdt Oct 24 '17 at 18:06
  • $\begingroup$ @AlexKChen: "compact" is simply a short way of saying "both closed and bounded". If you have a compact set S and a function f which is continuous everywhere on S then the set of f(x) for all x in S is also compact. $\endgroup$ – Eric Lippert Oct 24 '17 at 18:17
1
$\begingroup$

There are fundamental issues with both approaches. You assume that things like $\min, \max$ exist. They do exist if the function under consideration is continuous but that's another deep theorem (extreme value theorem, EVT) which is at the same level of complexity as the intermediate value theorem (IVT) which you are trying to prove. Also the fact that $g(\epsilon) $ exists and is positive is a property which goes by the name uniform continuity. This seems to suggest that IVT depends on EVT or uniform continuity. This is not true.

The proof strategy works in both cases (I do have a few reservations about the choice of values of $\epsilon$ in first proof, you need to fix that somehow) but it is undeniably complicated and uses EVT unnecessarily. Moreover you have to establish that $f(a) =m$ in each of the proofs.

Much easier and simpler to understand proofs exist for IVT and all of them are based on different notions of completeness. I have presented a few proofs in this blog post.

$\endgroup$
  • $\begingroup$ To me, the intermediate value theorem is much more about connectedness than completeness (even though completeness is used as a tool in proving $[a,b]$ is connected). $\endgroup$ – Daniel Schepler Oct 25 '17 at 0:02
  • $\begingroup$ @DanielSchepler: Fully agree! My point was to highlight that these theorems are all based on the distinguishing property of real numbers which normally people call by the name "completeness". The idea of connectedness and compactness are at a higher level to give different perspectives, but on a lower level all this is the result of the fine structure of real numbers. $\endgroup$ – Paramanand Singh Oct 25 '17 at 0:09
  • $\begingroup$ I am not sure do I fully understand you two, at a first glance, I do not see big problems if we were to construct functions with intermediate-value property on sets that are not connected. @DanielSchepler $\endgroup$ – user480281 Oct 25 '17 at 0:17
  • $\begingroup$ @AntoinePalAdeen: I meant that IVT expresses the idea that continous functions map connected spaces to connected spaces. $\endgroup$ – Paramanand Singh Oct 25 '17 at 0:24
  • $\begingroup$ @ParamanandSingh Do you think that non-continuous cannot do that? $\endgroup$ – user480281 Oct 25 '17 at 0:25
0
$\begingroup$

In your first proof, you define $g(\epsilon):=\min\{h(\epsilon,x)|x\in[0,1]\}$ and then go on to consider $1/g(\epsilon)$. However, to do so you would need to prove that $g(\epsilon)>0$ (which is not always true in your definition).

The correct way of proving (defining $g(\epsilon)$) is through the usual compactness argument: $\{B_{h(\epsilon,x)}(x)\}_{x\in[0,1]}$ (balls) form an open cover of $[0,1]$, from which you can extract a finite subcover indexed by $x_i$. Then redefine $g(\epsilon)$ as the minimum over the finite set of $h(\epsilon,x)$, so then $g(\epsilon)$ is positive.

I also don't understand your choice of $\epsilon_0$. Presumably you meant it to be small, but small is relative to $f(x)$, so I find it hard to believe your $\epsilon_0$ can be independent of $f(x)$.

$\endgroup$
  • $\begingroup$ I can let $g(\epsilon)\to 0$ since $h$ is not properly defined. $\endgroup$ – amsmath Oct 24 '17 at 18:04
  • $\begingroup$ @amsmath: To be clear I've redefined $g$ over a finite set of $x_i$, so how could it go to 0? $\endgroup$ – Alex R. Oct 24 '17 at 18:06
  • $\begingroup$ Can you give me some example where $g(\epsilon)$ is equal to $0$ for nonnegative and nonzero $\epsilon$ ? $\endgroup$ – cdt Oct 24 '17 at 18:10
  • $\begingroup$ Sorry, what I wanted to say is that $h$ is not properly defined, so $g(\epsilon)$ might be zero. $\endgroup$ – amsmath Oct 24 '17 at 18:11
  • $\begingroup$ Why $h$ is not properly defined ? $\endgroup$ – cdt Oct 24 '17 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.