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I am interested in proving a weak version of the intermediate value theorem, specifically:

Suppose that $f$ is continuous on $[a,b]$, and $f(a)<0<f(b)$. Then, there exists a number $x\in(a,b)$ such that $f(x)=0$.

The usual proof of this uses contradiction. I am curious as to whether there is a proof that doesn't use contradiction, and also doesn't require too much machinery.

For reference, here is the argument presented in Michael Spivak's Calculus. We need the following lemma, which can be easily proven using the epsilon-delta definition of the limit:

Suppose that $\lim_{x \to \lambda}f(x)=f(\lambda)$, and $f(\lambda)<0$. Then, there is a $\delta>0$ such that if $|x-\lambda|<\delta$, then $f(x)<0$. Similarly, if $\lim_{x\to\lambda}f(x)=f(\lambda)$ and $f(\lambda)>0$, then there is an open interval containing $\lambda$ for which $f$ is positive. This lemma also applies to "one-sided continuity": if $\lim_{x \to a^+}f(x)=f(a)$ and $f(a)<0$, then there is a $\delta>0$ such that if $0\le x-a<\delta$, then $f(x)<0$; likewise, if $\lim_{x \to b^-}f(x)=f(b)$ and $f(b)>0$, then there is a $\delta>0$ such that if $0\le b-x<\delta$, then $f(x)>0$.

Consider the set $$E=\{x\in[a,b]:\text{$f$ is negative on $[a,x]$}\} \, .$$From the above lemma we know that $E$ contains values greater than $a$, and that all points sufficiently close to $b$ are not in $E$. Since the real numbers satisfy the least upper bound property, we know that $\sup E$ exists, and we will denote it as $\alpha$. Note that $\alpha$ satisfies $a<\alpha<b$. We can prove that $f(\alpha)=0$ by contradiction. If $f(\alpha)$ were smaller than $0$, then because $f$ is continuous at $\alpha$, there would be a $\delta>0$ such that if $|x-\alpha|<\delta$ then $f(x)<0$. This means that there would be numbers $x>\alpha$ such that $x\in A$, contradicting the fact that $\alpha$ is an upper bound of $E$. Similarly, if $f(\alpha)$ were greater than $0$, then there would be a $\delta>0$ such that if $|x-\alpha|<\delta$, then $f(x)>0$. This means that there would be numbers $x<\alpha$ that are not in $E$, contradicting the fact that $\alpha$ is the least upper bound of $E$. So $f(\alpha)$ must be equal to $0$, and since $a<\alpha<b$, the theorem is proven.

As you can see, this proof functions by ruling out the possibilities $f(\alpha)>0$ and $f(\alpha)<0$ using contradiction. Is there a way of directly showing that $f(\alpha)=0$ instead?

Joe
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    Once you defined $x = \sup \lbrace y \in [a,b], f(y) < 0\rbrace$, there is no need to use contradictions : by continuity of $f$, you have $f(x) \leq 0$, and by definition of $x$ as a $\sup$, you have $f(x) \geq 0$, so you get finally $f(x)=0$. – TheSilverDoe Jul 14 '21 at 12:30
  • @TheSilverDoe: Can you elaborate on "by continuity of $f$, you have $f(x)\le0$"? I don't immediately see why this is the case. – Joe Jul 14 '21 at 12:50
  • @TheSilverDoe: Also, the set ${y\in[a,b]: f(y)<0}$ is not the same as $E$. – Joe Jul 14 '21 at 13:03
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    @Joe : using ${y\in [a,b]: f(y)<0}$, you will find the largest zero of $f$ in $[a,b]$, while using ${y\in [a,b]: f|_{[a,y]}<0}$ you will find the smallest – SolubleFish Jul 14 '21 at 13:15
  • The Least Upper Bound gives us the existence of the point $\alpha$. The lemma shows us that $f(\alpha) \geq 0$ and also $f(\alpha) \leq 0$. I don't think this is quite a full proof by contradiction, a la irrationality of $\sqrt{2}$. Proof by contradiction would begin with something like, "suppose $f(x) \neq 0$ for all $x$ in $[a,b]$," and would show how this leads to some absurdity. – Ben Jul 14 '21 at 13:34
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    @Ben: I just realised that, if the proof is worded more carefully, then it is a proof by contrapositive, not contradiction. I'm writing up an answer now. – Joe Jul 14 '21 at 13:38
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    One can prove it using the bisection method. – Arctic Char Jul 14 '21 at 13:56
  • @SolubleFish minor quibble: $\alpha = \sup{y\in[a,b]: f(y)<0}$ will give a zero but not the largest zero in the case where $f$ is constant zero over some interval past $\alpha$. (${y\in[a,b]: f(y)\leq0}$ would work though.) – Ben Jul 14 '21 at 14:27
  • @Ben: Is it possible if you check that my answer is correct, please? – Joe Jul 14 '21 at 14:30
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    @Joe I should say, the Least Upper Bound gives us the existence of the point $\alpha$. The trichotomy law tells us $f(\alpha)$ is either zero, positive, or negative, and the lemma lets us eliminate the latter 2 (through contradiction). – Ben Jul 14 '21 at 14:32
  • @Ben: Yes. The alternative proof I came up with also used the trichotomy law, but it proved that $f(\alpha)\le0$ and $f(\alpha)\ge0$. This (I think) allows the intermediate value theorem to be proven without contradiction. – Joe Jul 14 '21 at 14:33
  • @Joe it looks ok to me but I'm not really skilled enough to say. I don't have much training in formal logic, I'm afraid. – Ben Jul 14 '21 at 14:33
  • @Ben: your approach also works. Assume that $f$ is non-zero and then we can apply Heine Borel to get a partition of $[a, b] $ into finite number of intervals where $f$ has same sign. – Paramanand Singh Jul 15 '21 at 15:40
  • @Paramanand Singh: have you confused me with the OP? – Ben Jul 15 '21 at 15:48
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    @Ben: no, I was just mentioning about a proof by contradiction. You had mentioned that a proof via contradiction would begin with "suppose $f(x) \neq 0$...". – Paramanand Singh Jul 15 '21 at 18:57
  • @Paramanand Singh ah, understood. I mentioned that to contrast a "full" proof by contradiction with Spivak's proof (which OP originally mistook for a proof by contradiction.) – Ben Jul 15 '21 at 19:03
  • @ParamanandSingh: In my answer below, I modify Spivak's proof a little to turn it into a proof by contraposition, not contradiction. The basic idea is the same. – Joe Jul 15 '21 at 19:05

2 Answers2

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Why not Cantor's Intersection Theorem (=CIT)? Let $\;f:[a,b]\to\Bbb R\;$ be continuous such that $\;f(a)f(b)<0\;$ Let $\;z:=\cfrac{a+b}2\;$ be the middle point of the interval $\;[a,b]\;$. If $\;f(z)=0\;$ we're done, otherwise either $\;f(a)f(z)<0\;$ or else $\;f(z)f(b)<0\;$ . Suppose WLOG that this happens on $\;[a,z]\;$ and denote this new subinterval as $\;[a_1,b_1]\;$ , meaning: $\;a=a_1\,,\,\,b_1=z\;$ , so that $\;f(a_1)f(b_1)<0\;$ .

Once again subdivide as above this interval and...etc., just as before. At the end you get all the conditions of CIT: a (decreasing) sequence of non-empty nested closed and bounded intervals whose sequence of lengths tend to zero and thus their interesection is one single point, say $\;c\;$:

$$\bigcap_{n\in\Bbb N}[a_n,b_n]=\{c\}$$

Well, now...what is $\;f(c)\;$ ? Exactly, it must be $\;f(c)=0\;$ since:

$$f(a_n)\nearrow f(c)\;,\;\;\;f(b_n)\searrow f(c)\implies (\text{remember!}\;0>)\; f(a_n)f(b_n)\xrightarrow[n\to\infty]{}f(c)^2\le0$$

But of course also trivially $\;f(c)^2\ge 0\;\implies f(c)^2=0\iff f(c)=0\;$ and we're done, directly and without contradiction.

Just to make sure we understand: where did we use continuity of $\;f\;$ in the above ?

DonAntonio
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  • Do you think Cantor's Intersection Theorem would be understandable to me if I've only read Spivak's Calculus? – Joe Jul 19 '21 at 15:50
  • @Joe Yes. It is exercise 14 in page 142 (4th Edition), at the end of chapter II "Foundations". It is called there "Nested intervals Theorem", and then there come some uses of it in the following exericses. – DonAntonio Jul 19 '21 at 21:51
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I just realised that if the proof is worded more carefully, then it is a proof by contraposition, not contradiction. I thought that this was worth mentioning because many mathematicians have a preference for contraposition over contradiction (see here). Let $$E=\{x\in[a,b]:\text{$f$ is negative on $[a,x]$}\} \, .$$It is not difficult to prove that (a) if $\alpha=\sup E$, then $f(\alpha)\le0$, and (b) if $\alpha=\sup E$, then $f(\alpha)\ge0$. This method is still not as "direct" as I would like, but I thought it would be worth mentioning anyway.

The contrapositive of (a) is "if $f(\alpha)>0$, then $\alpha\ne\sup E$". Since $f$ is continuous at $\alpha$, there is an open interval $I$ containing $\alpha$ such that for all $x\in I$, we have $f(x)>0$. Let $x_0$ be a member of $I$ satisfying $x_0<\alpha$. Since $x_0$ is an upper bound of $E$, we find that $\alpha\ne\sup E$.

The contrapositive of (b) is "if $f(\alpha)<0$, then $\alpha\ne\sup E$". Since $f$ is continuous at $\alpha$, there is an open interval $I$ containing $\alpha$ such that for all $x\in I$, we have $f(x)<0$. Let $x_0$ be a member of $I$ satisfying $x_0>\alpha$. Since $x_0\in E$, $\alpha$ is not even an upper bound of $E$, and so $\alpha\ne\sup E$.

Combining (a) and (b) gives us $f(\alpha)=0$, as desired.

Joe
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