# Isn't the intermediate value theorem self-evident for continuous functions?

The way I understand the intermediate value theorem is this: if you have a function f that is continuous over a domain $[a,b]$ then there is a value $f(c)$, where $f(a)≤f(c)≤f(b)$, such that $a≤c≤b$.

This seems self-evident. If $f$ is continuous, then there exists an $f(c)$ such that $a≤c≤b$. But isn't this just a restatement of the fact that $f$ is continuous?

Isn't the intermediate value theorem self-evident for continuous functions?

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No you need the fact that the real line as a topological space is connected (plus the fact that the image of a connected set under a continuous function is connected). – BenjaLim Feb 22 '12 at 0:53

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"$f$ is continuous on $[a,b]$" means "for every $x$ in $[a,b]$, for every $\epsilon >0$, there exists $\delta >0$ such that, for every $y$ in $[a,b]$, if $|x-y|<\delta$, then $|f(x)-f(y)|<\epsilon$". Can you please explain how the intermediate value theorem is an obvious restatement of this condition? – Chris Eagle Feb 22 '12 at 0:54

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The intermediate value theorem is intimately connected with the fact that the reals are complete (every nonempty set bounded above has a supremum; equivalently, every Cauchy sequence coverges). In fact, it is equivalent to the supremum property/every Cauchy sequence converges, so it is not entirely "obvious" (in that the proof relies on these rather deep properties of the real numbers that are not easy to prove from first principles). – Arturo Magidin Feb 22 '12 at 0:55

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You have a misunderstanding of the intermediate value theorem. The intermediate value theorem says that $f(x)$ must take every possible value between $f(a)$ and $f(b)$. It might "seem" self-evident, but it isn't true for example, if you look at continuous functions on the rational numbers, for example. – Thomas Andrews Feb 22 '12 at 0:55

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My favorite example of a continuous function from rationals to rationals that fails the intermediate value test is simply $f(x)=1/(2-x^2)$: rational function, gotta be continuous, but it blows up around the “missing” point $\sqrt{2}$. – Lubin Feb 22 '12 at 3:36
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## 7 Answers

It's not self-evident; rather, it's a justification for the epsilon-delta definition of "continuous"! It's the bridge between the epsilon-delta view of the world that we learn in university, and the "point and line" view of the world which we know from high school.

This is the content of two 1-hour lectures, but let me try to summarize it in a nutshell:

The fundamental problem in calculus can be thought of as:

Estimate $f(a)$ using information about $f(b)$ where $b$ is some number near $a$. (You can directly measure only $f(b)$ but not $f(a)$).

So at the most primitive level, before we discuss derivatives or anything, we simply want to characterize functions for which $f(b)$ gives some estimate of $f(a)$ for $b$ close enough to $a$. These are the functions for which it makes some kind of sense to even try to make such an estimate. So what are these functions?

1. Function $f$ doesn't a jump or snap at $a$. For example, if $f$ is "tension in a string", and the string breaks at $a$, measuring tension just before $a$ won't tell you the tension at $a$.
2. There isn't some kind of crazy $f(x)=sin\left(\frac{1}{x}\right)$ at $x\neq 0$ and $f(0)=0$ "resonance" "shaking itself to pieces" phenomenon at $a$.

In other words, $f$ is continuous at $a$ ($f(a)$ can be estimated by looking at values of $f$ for inputs close enough to $a$) if no terrible crisis occurs for function $f$ at $a$.

Good. How do you formalize this? Along comes Arbogast at the end of the 18th century, and suggests that the property you need to demand is the intermediate value property. This is a good first-approximation to a formalization of "continuity", because is takes care of "breakage" ("issue 1"). But it does not take care of not "shaking" ("issue 2")- the topologist's sine curve satisfies the intermediate value property at $0$, but isn't "continuous". Even worse, it's satisfied by crazy functions like the Conway base 13 function which shake so badly that it makes no sense to try to estimate in anywhere.

Then, along comes Bernard Bolzano, works out epsilon-delta, and gives the modern definition of continuity. And, because he's a Catholic Priest and not a great and famous mathematician like Arbogast, he has to probe that his definition implies the Intermediate Value Property in order for anyone to take him seriously. But, personal reputation of Bolzano aside, why in fact is this property so central?

The epsilon-delta view of the world, put forward by Bolzano, posits that a line is more than just an infinite connection of points- rather, it is made up of interlocking epsilon-delta fuzz (using more technical terms, you're imposing equiping the real line with the metric topology). One way of thinking about this is that, in reality, you never know exactly what a real number is- you can write down the first zillion digits of pi, but never all of it. So a real number is an inherently fuzzy concept, and the real line (as a topological space) is made up of interlocking fuzz rather than being made up of points.

Writing out the epsilon-delta construction of a continuous function (or any epsilon-delta construction) is painful, because you have all of these nested quantifiers to deal with fuzziness (for any epsilon there exists delta such that for any x between bla bla bla). It's not human language- it looks more like some kind of awful computer code. On the other hand, the intermediate value property is talking about points. It tells you that there exists a point c between $a$ and $b$ such that bla bla bla.

Punchline: The intermediate value theorem is the bridge between fuzz world and point world.

Every other bridge between fuzz-world and point-world goes through it (all the ones you learn in undergrad calculus do anyway). So it's not at all obvious... it's the statement that the horribly convoluted non-human-language epsilon-delta language captures all the properties of a continuous function that you want; and then some. It's the statement that the convoluted non-intuitive epsilon-delta formalism happens to be the one which captures your geometric intuition. How surprising is that!

Formally, the intermediate value theorem is a consequence of completeness of the reals, which is roughly the statement that the real line doesn't have microscopic holes in it like the line of rational numbers has. So again- the passage from fuzz to points and back again has to make use completeness of the real numbers, and the intermediate value theorem is where that happens.

The intermediate value theorem isn't important because it's surprising. It's important because it gives you the bridge which you need in order to cross from epsilon-delta world to point-world and back.

Anyway, I've written far too much by now... Sorry for this tl;dr answer!

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You can define continuity on the rational numbers, $\mathbb Q$, using the same definition that you use for real functions.

But if you define $f(x)=0$ if $x^2<2$ and $f(x)=1$ if $x^2>2$, then you can show that this function is continuous on all the rationals. This function is continuous on the rational numbers precisely because $\sqrt 2$ is not a rational number.

The "intermediate value theorem," then, is not true for the rationals.

Basically, there is a property, completeness, of the real numbers which makes this sort of thing impossible for continuous functions defined on them. There are essentially no "gaps" in the real numbers. "Obvious?" I don't think so.

One way of stating this property is that any non-empty subset of the real numbers which is bounded above has a "least upper bound," also known as a supremum.

To prove the intermediate value theorem, let $a<b$ and $f$ a continuous function with $f(a)<f(b)$. Let $C\in(f(a),f(b))$. Let $U=\{x:f(x)<C\}$. Then $U$ is a non-empty subset of the real numbers and all the elements of $U$ are less than $b$, so $U$ is bounded above. It turns out you can show the supremum of $U$ is a value such that $f(x)=C$.

[There is a sense in which the irrational reals can be seen as the set of non-descreasing onto functions from $\mathbb Q \rightarrow \{0,1\}$ that are continuous as functions on the rationals.]

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To adopt a view slightly different to that of the other answerers:

In terms of an intuitive picture of the real numbers, as a continuum, and an intuitive picture of a continuous function, as one whose graph is drawn without having to lift the pencil from the paper, the intermediate value theorem is intuitively clear.

However, if one wants to reason carefully about the notion of real number, or function, and to prove statements and analyze phenomena that go beyond the intermediate value theorem, then it becomes important to have precise technical definitions of all the concepts involved: namely of real numbers, and of a continuous function. The key technical point in the characterization of real numbers is their completeness (as many people here have mentioned), while the technical definition of continuity is somewhat involved. In my view, the point of the rigorous proof of the intermediate value theorem is not primarily to confirm an intuitively obvious statement (although there is a certain pleasure in doing this), but rather to (a) confirm that the technical formalizations of real numbers and continuous functions capture the intuition they were intended to; and (b) to provide a model of how these formalizations can be harmonized with our intuitions. Point (b) is very important, since formalizations are only as powerful as our ability to wield them, and intuition (for most of us) provides the basic guide as to how to construct our mathematical arguments; thus it is important to learn how to meld intuition with formal arguments, and constructing formal proofs of intuitively clear statements is one of the best ways to do this.

Note that the formal proof also suggest the development of concepts (connectedness of topological spaces, and its preservation under formation of continuous images) which go well beyond the intermediate value theorem, and extend to contexts in which there was no a priori intuition at all. (This is one of the fantastic features of formal mathematics: the formalism itself, when applied in a new context, can provide intuition! In other words, we can draw analogies between what are apparently very disparate situations by virtue of certain structural similarities that they share in common.)

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First of all, the statement you've given is not the Intermediate Value Theorem. In fact, what you've written is true for any function: one can take $c = a$ or $c = b$. So as a first step in understanding these "big theorems" in calculus, you need to be excruciatingly careful to make sure you get the statements correct.

(Often when I teach freshman calculus I ask for statements of theorems like IVT on midterm exams. It never fails to surprise me how often students get these questions wrong, even though I've long since come around to telling them which three or four theorems they should have memorized for any given exam. I think part of the problem is that -- given that they don't understand the statement of the theorem, at least not directly in terms of parsing the sentence; rather they may have a mental picture that they think the sentence corresponds to -- it is very hard for them to know whether they are reproducing the statement correctly or not. I still find it surprising that their rote memorization skills are not stronger, given that these skills are being tested at a fairly high level in certain other university classes. It's all a bit mysterious...)

But, no, the correct statement is not close at all to the definition of continuity. It is however quite close to the rough intuitive idea that you may have of continuity...but that doesn't make IVT obvious! Rather, it makes it a very important theorem, one which ensures that a formal definition captures certain elements of the intuition that led to it.

Anyway, in September 2011 a short article by S. Walk was published on exactly this: the title is "The Intermediate Value Theorem is Not Obvious -- and I Am Going to Prove It to You". I recommend it.

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Dear Pete, I just finished writing an answer I began an hour or so ago, and it makes a similar point to your third paragraph, which I think is a (maybe the) key point. Best wishes, – Matt E Feb 22 '12 at 3:17
+1.Leave it to Pete to make the important point everyone except Brian made and to make it more forcefully then any of us.We don't make hay of this point because most of us think it goes without saying. It DOESN'T and it's important to point this out to a beginner.The important point isn't that there's a arbitrary point c on the compact interval where the function is defined between the endpoints,but rather that there necessarily is a point DIFFERENT FROM AND BETWEEN THE ENDPOINTS where the value of the function exists and is located between the values of the endpoints in the range. – Mathemagician1234 Feb 22 '12 at 5:56

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@Mathemagician1234: No, it's not that there is such a point, but that the function takes every value between the values at the endpoints. – joriki Feb 22 '12 at 15:15
@joriki: Right! There are two mistakes in the OP's statement of IVT. The quantification error is certainly the more serious: I just wanted to point out that, in conjunction with the non-restriction to interior points, it gives a statement which holds trivially for all functions. – Pete L. Clark Feb 22 '12 at 17:27
@jorki,Pete: See,I made a different mistake when stating what it meant.That's what I intended since the point between the endpoints was arbitrary-but it wasn't clear in how I stated it.In other words, the image of whatever curve has endpoints a and b is a subset of the range of f if the conditions of the IVT are met. THAT'S what I should have said. – Mathemagician1234 Feb 22 '12 at 22:39

The construction of the real numbers as a geometric line is a deep and significant result. This forms the headwaters from which the important theorems of analysis on the line flow: completeness of the real numbers, the Heine-Borel Theorem and the Weierstrass Intermediate Value Theorem. The elegance of the Dedekind cut construction underlies this all. It is hardly self-evident or trivial.

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No, the intermediate value theorem is not a restatement of continuity, especially when it’s stated correctly: if $f$ is continuous on $[a,b]$, and $c$ is any real number between $f(a)$ and $f(b)$, then there is some $x_0$ between $a$ and $b$ such that $c=f(x_0)$. Try to see how this differs from what you wrote.

To see why this isn’t just a restatement of continuity, consider the function $$f:\mathbb{Q}\to\mathbb{R}:x\mapsto x\;,$$ where $\mathbb{Q}$ is the set of rational numbers. This function, considered only as a function on the rationals, is continuous, $f(0)=0$, and $f(2)=2$, but there is no number $p$ anywhere in the domain of $f$, let alone between $0$ and $2$, such that $f(p)=\sqrt2$.

For another example, consider the function $$f:\mathbb{R}\setminus\{0\}\to\mathbb{R}\setminus\{0\}:x\mapsto \frac1x\;;$$ here $\mathbb{R}\setminus\{0\}$ means all of the real numbers except $0$. This function $g$ is continuous on its domain $\mathbb{R}\setminus\{0\}$, $g(-1)=-1$, $g(1)=1$, and $-1<\frac12<1$, but there is no $x\in\mathbb{R}\setminus\{0\}$ such that $-1\le x\le 1$ and $g(x)=\frac12$.

The problem with $f$ and $g$ is that their domains are not connected: intuitively speaking, they have ‘holes’ in them. $\mathbb{Q}$ is full of holes, with one at every irrational number; $\mathbb{R}\setminus\{0\}$ has only one hole, at $0$.

The intermediate value theorem is actually a special case of a more general result, namely, that if $f$ is a continuous function on a connected set $A$ (what connected means in general isn’t really important right now), then $f[A]$ is also a connected set. The connected sets in the real line are precisely the intervals, open, closed, or half-open, and bounded or unbounded. Thus, if $f$ is a continuous function on an interval $[a,b]$, the image of that interval must also be an interval of come kind; call it $J$. Since $f(a)$ and $f(b)$ are in $J$, and $J$ is an interval, everything between $f(a)$ and $f(b)$ must also be in $J$. Thus, if I pick any $c$ between $f(a)$ and $f(b)$, there must be some $x_0$ between $a$ and $b$ such that $c=f(x_0)$ $-$ which is exactly what the intermediate value theorem says.

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+1 for the wonderful explanation, would +2 if I could. – Alex Becker Feb 22 '12 at 1:10

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Your first $f$ can't possibly take the value $\sqrt2$, because it is a function from $\mathbb Q$ to $\mathbb Q$. Perhaps a better example would be that $f : \mathbb Q \to \mathbb Q : x \mapsto x^2$ has no $p$ between $1$ and $2$ such that $f(p) = 2$. – Rahul Narain Feb 22 '12 at 1:23
@Rahul: That was actually a typo: I wanted the codomain to be $\mathbb{R}$. Thanks for catching it. Your example (and Thomas Andrews’) is also nice. – Brian M. Scott Feb 22 '12 at 1:33
+1 for the terrific explanation and I doubt any of us will be able to give a better one.It's interesting since I was taught the intermediate value theorem is really a trivial result from the standpoint of topology-the really deep and difficult result the IVT relies on is the fact that the real line with the usual topology is a connected space. The best discussion and proof of this fact I've ever seen is in a most unlikely source: Elliott Mendelson's THE NUMBER SYSTEMS AND THE FOUNDATIONS OF ANALYSIS,now in a nice cheap Dover paperback.Elliott once told me he was rather proud of it. – Mathemagician1234 Feb 22 '12 at 5:45

Absolutely not! The definition of continuous function is quite technical (for every $\epsilon$ there exists a $\delta$, I forget the rest). There is no strong reason to believe that the graphs of continuous functions, as continuous function is ordinarily defined, will behave like the smooth curves of our imagination. For instance, there is a continuous nowhere differentiable function, indeed most continuous functions are nowhere differentiable.

The Intermediate Value Theorem for $f$ is not a restatement of the fact that $f$ is continuous. For example, derivatives satisfy the IVT, and are not necessarily continuous.

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I know that nowhere-differentiable functions are dense in $C(\mathbb R,\mathbb R)$ (by the Baire category theorem), but in what sense are most continuous functions nowhere differentiable? Cardinality? Measure? If so, under what measure? – Alex Becker Feb 22 '12 at 1:07
Perhaps one can look at homepages.math.uic.edu/~marker/math414/fs.pdf. There are many other sources. And it is in the sense of category. – André Nicolas Feb 22 '12 at 1:32

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@Alex: If I am remembering correctly, then indeed the set of differentiable functions is meager in $C(\mathbb{R},\mathbb{R})$ in an appropriate topology, so the "most" is in the sense of Baire category. (Certainly the cardinalities of the two sets are the same.) – Pete L. Clark Feb 22 '12 at 1:51
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