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It's one of my real analysis professor's favourite sayings that "being obvious does not imply that it's true".

Now, I know a fair few examples of things that are obviously true and that can be proved to be true (like the Jordan curve theorem).

But what are some theorems (preferably short ones) which, when put into layman's terms, the average person would claim to be true, but, which, actually, are false (i.e. counter-intuitively-false theorems)?

The only ones that spring to my mind are the Monty Hall problem and the divergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n}$ (counter-intuitive for me, at least, since $\frac{1}{n} \to 0$ ).

I suppose, also, that $$\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e=\sum\limits_{n=0}^{\infty}\frac{1}{n!}$$ is not obvious, since one 'expects' that $\left(1+\frac{1}{n}\right)^n \to (1+0)^n=1$.

I'm looking just for theorems and not their (dis)proof -- I'm happy to research that myself.

Thanks!

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The Banach-Tarski paradox is a good one. –  mjqxxxx Jun 4 at 18:08
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related: mathoverflow.net/questions/23478/… –  Will Jagy Jun 4 at 18:29
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This question is hard to answer, because an essential skill for a mathematician is being able to rapidly retune your intuition to match the truth. So once you know that a fact is false, very soon it no longer seems obvious. –  Nate Eldredge Jun 4 at 19:53
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I strongly disagree that the Jordan Curve theorem is obviously true! I would agree that the Jordan Curve theorem for piecewise smooth curves is obviously true; but it's also pretty easy to prove, at least compared to the topological version. (Is it really obvious that the Koch snowflake doesn't have some pathological path from the inside to the outside?) –  Mike Miller Jun 6 at 7:04
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"Every true statement can be proved" –  sinelaw Jun 9 at 17:27

62 Answers 62

up vote 120 down vote accepted

Theorem (false):

One can rearrange the terms in a convergent series.

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The series $\sum \frac{(-1)^i}{i}$ is a counterexample; see this Wikipedia article for a discussion. –  MJD Jun 4 at 18:39
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@alexqwx Roughly, that some terms can be pushed 'infinitely far away', so that any two finite partial sums differ by arbitrarily many terms. Imagine e.g. taking the positive numbers and rearranging them as 1, 2, 3, 5, 4, 7, 9, 11, 6, 13, 15, etc. with a growing number of odd terms between each two consecutive even terms; then every number still shows up in the sequence eventually, but that doesn't mean that any concept like a 'ratio' of odd-to-even (e.g. natural density) is preserved. –  Steven Stadnicki Jun 4 at 18:55
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Huh? One can rearrange the terms in any series, convergent or otherwise. Did you leave something out of the statement of your theorem? –  bof Jun 4 at 21:25
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Actually, you can rearrange terms in a convergent series when computing its value (or otherwise). You can even rearrange terms without changing the value, with a bit of care. What you cannot, is assume that any rearrangement will preserve convergence, or when it does preserve the value converged to. –  Marc van Leeuwen Jun 4 at 22:42
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@kutschkem You can always make finitely many swaps of terms (using commutativity) without changing the value, but you can't make infinitely many swaps; continuity isn't preserved across an infinite number of operations. –  Steven Stadnicki Jun 5 at 21:36

A shape with finite volume must have finite surface area.

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This is evidently false. Take a clay cylinder of diameter 1. Roll it in your hands so that its diameter becomes $\frac 12$. The volume is the same, but it is now four times as long as it was before, so it has twice the surface area of the original cylinder. (Half the circumference, but four times the length.) Roll it some more, and the surface area increases again. By making the snake very long and thin, you can increase the surface area to infinity while the volume remains constant. This is not only obvious, it's commonplace. –  MJD Jun 4 at 20:31
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Those are good examples. I know this wouldn't confuse anyone formally trained like most of the posters here, but I was trying to suggest an example for "layman's terms, the average person" as OP requested. It seemed most of the examples so far would not even be interpretable to an untrained individual. –  DanielV Jun 4 at 20:41
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@MJD: your cylinder example is nice, but it only shows that the surface area can be arbitrarily large. The OP is pointing out that the surface area can literally be infinite. –  Greg Martin Jun 5 at 0:00
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I dunno either, I'm a physics ignoramus. Are you sure clay is made of atoms? –  MJD Jun 5 at 1:11
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Mathematically this may be true, but physically impossible, as matter is discrete. The limit would be a row of atoms. –  Alex Jun 5 at 4:54

I keep harping on this, because I think it's a spectacular example of something that can be demonstrated to be completely obvious (not only because it seems so, but because it was so widely believed for so long) and yet is completely wrong:

Suppose $\Phi$ is a property that might or might not hold of some object. Then there is a collection $S_\Phi$ of all objects with property $\Phi$.

Many serious and even famous mathematicians went ahead with this intuitively obvious but utterly false principle, whose demolition shook mathematics to its foundations and marks the beginning of modern logic and set theory.

(There are many counterexamples, of which the most well known is $\Phi(x) = $ “$x$ is not a member of collection $x$”.)

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+1 for a great example, but I do think the use of "properties", "objects", and "collections" takes away from the impact by being imprecise. There are definitely consistent set theories where there are collections which are not objects and the statement is true in some sense. Stating it using the word sets and in the context of traditional naive set theory and modern widely-used formalizations of set theory would make this answer better in my opinion. –  R.. Jun 4 at 22:39
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Why is this wrong? –  mattecapu Jun 5 at 6:45
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I disagree, I think this works quite fine with undefined terms as long as you have a notion of containment, you don't need anything else. –  Joshua Biderman Jun 5 at 6:52
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I really think that is precisely not the problem. The problem here has nothing to do with ZF, with axioms, or with how we exactly define "objects"; it is a fundamental problem with the intuitive notion of what it means for things to have properties. How do you feel about this formulation: “For any property $\Psi$, one can construct a catalog that lists all the books with property $\Psi$?” And the answer is, that for some properties $\Psi$, you simply cannot. Is that sufficiently concrete and not-set-theoretic? –  MJD Jun 5 at 15:22
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This is Russell's Paradox, discovered in 1901, which destroyed Frege's Grundgesetze der Arithmetik (1903) and prior work. The immediate responses to the paradox include Whitehead and Russell's Principia Mathematica (1910) and Zermelo's work on axiomatic set theory starting in 1905, which eventually became ZFC, which dominates set theory to this day. –  MJD Jun 8 at 15:11

I wish I'd thought of this yesterday, when the question was fresh, because it's astounding. Suppose $A$ and $B$ are playing the following game: $A$ chooses two different numbers, via some method not known to $B$, writes them on slips of paper, and puts the slips in a hat.

$B$ draws one of the slips at random and examines its number. She then predicts whether it is the larger of the two numbers.

If $B$ simply flips a coin to decide her prediction, she will be correct half the time.

Obviously, there is no method that can do better than the coin flip.

But there is such a method, described in Thomas M. CoverPick the largest numberOpen Problems in Communication and Computation Springer-Verlag, 1987, p152.

which I described briefly here, and in detail here.

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Who said anything about a uniform distribution on $\Bbb N$? Certainly not I. –  MJD Jun 5 at 13:17
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Yes. That was what I intended by saying that the method was unknown to $B$, which is all that is important here. How could I make this clearer? –  MJD Jun 5 at 13:18
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Yes, absolutely. It is an actual strategy that one can actually execute, and actually predict correctly with probability greater than $\frac 12$. I was at some pains to explain this in my original post, including for example “methods for doing this are well-studied” and such in hope of persuading the doubtful. –  MJD Jun 5 at 20:47
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I don't think it's obvious that no such strategy exist, in fact I think most laypeople would intuitively apply it (in a heuristic form, i.e. "does that number sound big?" in the sense of "would I have been likely to choose two numbers and find this to be the larger one"). The thing that's obvious, and indeed correct, is that the strategy can easily be subverted by person A, by choosing both numbers very big and close to each other. –  leftaroundabout Jun 6 at 10:59
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This can be explained very simply. B guesses A's strategy. If she's wrong, her odds are 1/2. If she's right, her odds are better than 1/2. As long as her odds of guessing A's strategy are better than 0, her odds of winning are better than 1/2. –  Aleksandr Dubinsky Jun 10 at 19:54

This is elementary compared to most of the other examples, but how about

There are more rational numbers than there are integers (natural numbers).

?

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The beauty lies in it's simplicity. The falseness of this one of those very elementary observations the part of my brain that is driven by common sense still refuses to accept. –  aRestless Jun 5 at 23:11
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And there are half as many odd integers than (odd or even) integers. –  gnasher729 Jun 6 at 7:30
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@Jori: I haven’t studied the proofs and terminology in years, but I think a somewhat simple (simplistic?) way of putting it is that the existence of an injection mapping (not one-to-one) does not preclude the existence of a one-to-one mapping. P.S. Some people might consider 2∞>∞ to be obviously true. –  Scott Jun 7 at 20:08
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@Jori: (2) No! As Cornelius implied in his comment, Cantor’s diagonal argument (correct link) shows that there are infinite sets (such as the set of real numbers) that cannot be put into one-to-one correspondence with the (infinite) set of natural numbers. There are (infinitely many!) different transfinite cardinals. Crudely put, not all infinities are equal. –  Scott Jun 9 at 14:43

In a related mathOverflow thread, Gowers pointed out the following obvious but false claim:

Let $I_1, I_2, \ldots$ be subintervals of $[0,1]$ whose total length is strictly less than 1. Then the union of the $I_i$ cannot contain $\Bbb Q\cap [0,1]$.

(Note that if $\Bbb Q$ is replaced with $\Bbb R$, the claim is true.)

I find the fact that all of $\Bbb Q$ can be covered by an arbitrarily small family of intervals to be one of the most bizarrely counterintuitive in all of mathematics.

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You order the rationals $q_1, q_2, \ldots, q_n, \ldots$ and place an interval of $(\frac \epsilon 2)^n$ around each one, choosing $\epsilon$ to be less than $1$. –  dfan Jun 4 at 19:37
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Of course the theorem is still true if we have a finite number of intervals; one needs the infinite construction to make this work, which is likely where the gap in intuition lies. –  Steven Stadnicki Jun 4 at 19:48

The falseness of

Let $S$ be an infinite family of strictly positive numbers. Then $\sum S = \infty$

has been boggling people for thousands of years. It is the basis for Zeno's paradox, but if you think Zeno's paradox is old and tired, consider that it is also the basis for the Gabriel's Horn paradox (also mentioned in this thread), which still puzzles people.

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@jpmc26 It does always get bigger as you keep adding numbers; it just doesn't necessarily do so without limit. –  Mike Scott Jun 5 at 7:26
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On the other hand if the family is uncountable, it holds. –  user87690 Jun 5 at 13:27
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I just read this one and the comments... and it's really freaky. (I have a master's in physics, so I'm not entirely unfamiliar with these sorts of things.) –  Almo Jun 5 at 17:31
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I feel like this isn't that crazy, though maybe that's because I have experience, but I mean $1.1111\cdots=1+0.1+0.01+\cdots$. –  JLA Jun 5 at 21:10
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Ooooh! I read it as "strictly positive integers" which I believe would diverge. Right? (Too much programming) –  Almo Jun 5 at 22:30

If a function $f(x)$ has an horizontal asymptote, then $\lim f'(x) = 0$

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+1 Love this one! A counterexample is $\frac{\sin{x^2}}{x}$. –  Gamma Function Jun 5 at 10:30
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@A.P. or even just that $\lim f'(x)$ exists. –  Ant Jun 5 at 17:50
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As an undergraduate I attempted to write an $\varepsilon$-$\delta$ proof that if $f'(a)>0$ then there is some open interval containing $a$ on which $f$ is increasing. And I thought my ability to write $\varepsilon$-$\delta$ proofs was missing something because I couldn't figure out how to do it. –  Michael Hardy Jun 5 at 20:41

$ 0.\overline{9} < 1 $

Probably the most famous of the "obvious" but false.

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The question did ask what the average person would consider true. –  MichaelRushton Jun 9 at 7:56
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This is by far the best one on the page imo. Some of the others are so complex to the average person that they wouldn't be "obviously" true to almost anyone. –  Sam Creamer Jun 10 at 20:20

This part is true (Jordan-Brouwer separation theorem):

(a) Any imbedding of the $2$-sphere into $3$-dimensional Euclidean space separates the space into two disjoint regions.

But this part, which would seem to be a natural generalization of the Jordan-Schönflies Curve Theorem, is not true:

(b) The regions are homeomorphic to the inside and outside of the unit sphere.

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I was thinking of this example (the horned sphere) myself. My answer can be viewed as a sort of 2d analog in some ways... –  Steven Stadnicki Jun 4 at 18:37

Keller's conjecture is obviously true:

Let $\Bbb R^n$ be completely covered with identical, non-overlapping $n$-cubes. There must be two cubes that share a face.

(For example, when $n=2$ we cover the plane with little square tiles, and the conjecture states that there must be two tiles that share an edge. This is true.)

However, the conjecture is false for all $n>7$.

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I'm not sure I buy this example myself, because I'm not sure anything concerning 7-dimensional anythings can be considered intuitively obvious. –  MJD Jun 4 at 18:55
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Wouldn't the fact that it's true for $n = 1, 2, 3$ (and $n = 4, 5, 6$) lead you to suspect that it might be true for all $n$? –  Jesse Madnick Jun 5 at 4:47
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What does "face" even mean in 7 dimensions? Is it a 6-dimensional thing or a 2-dimensional thing? –  Mehrdad Jun 5 at 7:04
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What precisely do we mean by "share a face" in this context? –  goblin Jun 5 at 10:20
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what do you mean by "tiling"? –  electronp Jun 5 at 12:04

$i^i$ is imaginary.$\ \ \ \ \ \ $

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Believing $i^i$ is well-defined is illusory. –  Marc van Leeuwen Jun 6 at 13:26
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I'm certainly not saying anything profound. One simply should not write down things that look like constant expressions but aren't. –  Marc van Leeuwen Jun 6 at 13:40
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$i^i\approx 0.207879$ –  Anonymous Pi Jun 6 at 14:24
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@bof: Yes. And to say that "$i^i$ is imaginary" is false even when taking the multivalued nature into account. That's what I meant. –  mike4ty4 Jun 7 at 3:04
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@anorton Depends on how you define "imaginary". The usual definition is "having real part 0". and 0 certainly has real part 0. –  Snowbody Jun 9 at 16:50

Every chain of subsets of $\mathbb N$ is countable.

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So far I find this to be the most true for me. For the life of me I still cannot give myself a good metaphor that make it obviously false, even though I know mathematical counterexample. Other answers, either it never found them obvious, or if I did, I can quickly fix the misconception by an alternative way to look at it. You probably should add in counterexample for this one in the answer. –  Gina Jun 9 at 2:30
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By "chain", do we mean a collection of subsets of N that is totally ordered by the subset relation? You can represent the interval $[0, 1]$ as such a chain. For each number $x$ in that interval, the subset contains (rounding down) the first $9x$ one-digit numbers, the first $90x$ two-digit numbers, the first $900x$ three-digit numbers, and so on. –  Tanner Swett Jun 9 at 4:15
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@TannerSwett More straightforwardly, consider Dedekind cuts of rational numbers. –  Dustan Levenstein Jun 12 at 3:56
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@DustanLevenstein Sure. Or, let $\mathbb Z[i]$ be the set of all Gaussian integers, and let $S_\alpha=\{z\in\mathbb Z[i]:0\lt arg(z)\lt\alpha\}$. –  bof Jun 12 at 4:12

The real numbers/Cantor set are countable.

There are several false "obvious" proofs:

  1. "Proof". Consider the tree $\{0,\ldots,9\}^\Bbb N$, then every real number corresponds to a node in the tree. Since there are only countably many levels and each is finite, it follows that the real numbers are finite.

    Why does it fail? This set is actually not a tree. You can order it so it looks like a tree, but in fact the tree would be composed of initial segments of each functions ordered by continuation. This tree, then, would have a last level (namely a level that no point there has a successor), and it would be exactly the level of the functions themselves (the previous levels would be proper initial segments of the functions).

    If we remove that last level, then the tree is indeed countable, but now each real number corresponds to a branch in the tree rather than a node. (It's the unique branch whose limit equals to the function, which previously appeared on that final level.)

  2. "Proof". The rational numbers are countable, and between every two real numbers there is a rational number. Therefore this defines a bijection between pairs of real numbers and the rational numbers.

    Why does it fail? Because there are many, many, many pairs being mapped to the same rational number, this is not actually a bijection.

  3. "Proof". The Cantor set is closed, its complement is open, so it is a countable union of intervals, so the Cantor set is countable.

    Why does it fail? Because not every point in the Cantor set is an endpoint of such interval. For example $\frac14$. It is true that the endpoints of these intervals form a countable dense subset, though.

  4. BONUS!, $\mathcal P(\Bbb N)$ is countable.

    "Proof". For every finite $n$, $\mathcal P(n)$ is finite, and $\mathcal P(\Bbb N)=\mathcal P(\bigcup n)=\bigcup\mathcal P(n)$, is a countable union of finite sets, which is countable.

    Why does it fail? Because the union only includes finite subsets of $\Bbb N$, but none of its infinite subsets.

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I think the non-existence of irrational numbers may itself be in this category (obviously true, but false), but was discovered to be false so long ago that we no longer remember how bizarre it must have seemed. –  MJD Jun 4 at 19:29
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@Christopher: No, the tree has $\omega+1$ levels. The last is the infinite sequence $0.333\ldots$. –  Asaf Karagila Jun 5 at 6:27
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@Cory: Or $\frac13$, yes. There are many examples. But this doesn't quite explain why the proof fails. It just shows that it does. I was trying to actually explain the reason for the failure. –  Asaf Karagila yesterday

Theorems that are intuitively true, but actually flawed:

  • There is no continuous, nowhere-differentiable real function.

  • There is no real function that is differentiable and not monotonic on any non-trivial interval.

  • If a real function satisfies $\forall x, y, f(x+y) =f(x) +f(y) $, it is of the form $x\to ax$.

  • Infinite sums and integrals can be swapped anytime.

  • A connected metric space is path-connected.

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Can you show a counterexample for the third claim? (The one about linear functions) –  Alessandro Jun 4 at 19:28
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Almost any statement where the counterexample begins with “Let $H$ be a Hamel basis of $\Bbb R$…” is going to be one of those seems-true-but-isn't things. –  MJD Jun 4 at 19:50
  • If $U$ is an open subset of $\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$, one might think it "obvious" that it's in fact diffeomorphic to $\mathbb{R}^n$ (perhaps thinking something like "topologically it looks like $\mathbb{R}^n$, and differentiably it's locally trivial"). In fact, this is true (but by no means obvious!) for $n\neq 4$. But for $n=4$ it is false: there exist exotic $\mathbb{R}^4$'s (differentiable manifolds that are homeomorphic, but not diffeomorphic, to $\mathbb{R}^4$), including "small" ones which are diffeomorphic to an open subset of $\mathbb{R}^4$.

  • Much less profound, but still fun: it's "obvious" that the sum of two convex open sets in the plane whose border is $C^\infty$ also has a $C^\infty$ border (perhaps thinking something like "the border of the sum is parametrized by a smooth function of the borders of the summands"). But this is false: in fact, the border of the sum is always $C^{20/3}$ (meaning six times differentiable and with a sixth derivative which is appropriately Hölder) and no more in general. A simple counterexample is given by the epigraphs of $x^4/4$ and $x^6/6$. For details, see Kiselman, "Smoothness of Vector Sums of Plane Convex Sets", Math. Scand. 60 (1987), 239–252.

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$20/3$? What kind of magic is that!? +1 interesting example. –  Olivier Bégassat Jun 5 at 23:24

I really like "wrong proofs" as typically the insight why the proof is wrong gives you some understanding of the topic. One very simple version is this one, which I threw at my first semesters when I was a tutor:

Each binary relation which is symmetric and transitive is also reflexive and therefor an equivalence relation.

"Proof":

Let $\sim$ denote a symmetric and transitive relation and let $x$, $y$ be two elements with $x \sim y$. As $\sim$ is symmetric, it holds that $y \sim x$. Since $x \sim y$ and $y\sim x$ it follows by the transitivity of $\sim$ that $x \sim x$, which is the definition of reflexivity.

Edit: Since I was asked, here's why the proof is wrong (move your mouse there to show):

Take a look at the empty relation on a non-empty set $S$, so that there are no $x, y \in S$ so that $x \sim y$. This relation is symmetric and transitive, but it is not reflexive. Reflexivity needs $x \sim x$ to hold for all $x$. The proof assumes that there is a y so that x ~ y, which isn't necessarily the case for all $x$.

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+1 for set theory! –  Michael Zalla Jun 10 at 8:29

The probability that you hit any single point on a dart board is $0$ but the probability that you hit the dart board is $1$ (as long as you're not as bad as I am at throwing darts ;D).

EDIT:

As @JpM pointed out I didn't follow the format of these posts albeit the idea can (easily in my opinion) be understood from what I've said above.

Pseudo-Claim: The probability of hitting a single point on a dart board is greater than $0$ since the probability of hitting it at all (assuming that you will hit the dart board) is $1$.

Seems obvious in the sense that a bunch of $0$ can't add up to be $1$ so each point must have some probability. Actually false because of some properties of measures.

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In particular, there's an "obvious" intuition that the point you actually did hit must have had probability greater than $0$, "since probability $0$ events are impossible". I guess that this example is so obvious that it has to be made axiomatically false in the theory in order to get it out of the way and do some work :-) –  Steve Jessop Jun 5 at 8:27
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@DanielV I would say one of two things based on my mood: a) A lay-person would not claim such a thing, i.e. they wouldn't say the probability of hitting a single point is an infinitesimal. b) Generally (in fact I have not seen otherwise, then again I'm young and naive) measures are defined with values in the extended real numbers, meaning without infinitesimals. –  DanZimm Jun 5 at 12:23
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@JpMcCarthy ok I changed the format of my post to follow exactly how the question was asked. –  DanZimm Jun 5 at 13:03
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+1 for you after all that! –  Jp McCarthy Jun 5 at 13:04
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I think the essential thing some people simply can't get their head around is that an event with probability 0 can still happen. –  Neil Jun 11 at 2:43

I'm surprised noone gave this answer already, so here it is:

There are more integers than there are natural numbers.

It's obvious, isn't it?

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I hesitate with this one, because it depends on (somewhat unsound) definition of "more". –  DanielV Jun 7 at 16:24
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There absolutely are more integers than natural numbers (by inclusion). And cardinality is a slightly contrived notion if you think about it for a moment. –  Tibor Jun 8 at 9:51
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@Tibor: No, cardinality is not a contrived notion if you think about it for a long moment. –  Asaf Karagila Jun 8 at 21:09
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All of mathematics is contrived, by definition. :) –  Ryan Jun 9 at 1:32

There are a good number of counterintuitive probability situations out there. One of my favorites is nontransitive dice:

There are 3 dice, A, B and C. The dice have numbers from 1-9 on their sides (repeats possible). If die B beats (higher number) die A more than half the time and die C beats die B more than half the time, then die C will beat die A more than half the time.

This is not necessarily a true statement. Dice can be designed such that the "x beats y" property is not transitive. A beats B, which beats C, which beats A.

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There are a number of similarly counterintuitive statements in multiway voting, of which this could be seen as a special case. For example, in figure skating competitions at the championship level there are nine judges. It would sometimes happen that a majority of judges would rank skater $A$ above skater $B$, a majority would rank $B$ above $C$, and a majority would rank $C$ above $A$. The scoring rules had to have conditions in place to deal with this kind of situation. –  MJD Jun 5 at 21:35

Hypothesis: Every infinitely-differentiable function is real-analytic somewhere.

This is false, as shown by (for example) the Fabius function.

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See this interesting answer by Dave L Renfro, and the posts it links to. –  Andres Caicedo Jun 6 at 3:02

Here's one of my favorites: Let's assume playing with a fair coin.

Theorem (false) In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

The following is from the classic of Chung & Feller Introduction to Probability Theory and It's Applications, Vol 1:

According to widespread beliefs a so-called law of averages should ensure the Theorem above. But, in fact this theorem is wrong and contrary to the usual belief the following holds:

With probability $\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.

In fact this leads to the Arc sine law for last visits (see e.g. Vol 1, ch.3, section 4, Theorem 1).

Note: Please note their remarkable statements cited from Chapter III: Fluctuations in Coin Tossing and Random Walks:

(Chung & Feller): For example, in various applications it is assumed, that observations on an individual coin-tossing game during a long time interval will yield the same statistical characteristics as the observation of the results of a huge number of independent games at one given instant. This is not so.

and later on:

(Chung & Feller): Anyhow, it stands to reason that if even the simple coin-tossing game leads to paradoxical results that contradict our intuition, the latter cannot serve as a reliable guide in more complicated situations.

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In my opinion, the most interesting (but also sometimes not intuitive) results in mathematics are those that state a theorem that ends up being false because it actually holds in many cases, except for very few or very strange cases. In other words, the most "obvious" false theorems to me are those that have very difficult counterexamples.

Some examples:

  • Banach-Tarski: There exists a strict subset $A$ of the Euclidean $n$-ball $B$ such that one can partition $A$ and $B$ into an equal number of further subsets that can be mapped to each other by isometries. This shows that not all sets are measurable, and that it's possible to perform partitions that do not preserve measure.

  • Non-finiteness of differentiable structures: For $\mathbb{R}^n$ with $n = 4$, there are an uncountable number of distinct differentiable structures.

  • Divergence of Fourier series: There exists an integrable function on $[-\pi, \pi]$ whose Fourier series diverges everywhere. This is extremely unusual because for any typical function we might write down, usually its Fourier series might diverge at one or a finite number of points, but will probably converge everywhere else.

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You don't need a weird example to find a subset of $\Bbb R^n$ that is neither open nor closed; just take a closed ball, and remove a boundary point; or take a sequence of points that converges to a limit, but omit the limit. The terminology may lead one to think that sets are either open or closed, but there is nothing about the actual concepts that would make it appear so. –  MJD Jun 4 at 20:07

Consider a function $f:(0, \infty) \rightarrow \mathbb{R}$ that is $\mathcal{C}^\infty$ on that interval. At first glance, one might think that, if $\lim(f) = 0$ as $x \rightarrow \infty$, then $\lim(f') = 0$ as $x \rightarrow \infty$. However, this is false. Here is but one counterexample:

$$f(x) = \frac{1}{x}\sin(x^2)$$

Further, if we add the stipulation that $f$ also be monotonic, counterexamples can still be found (though they are quite pathological).

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Your "another" was already mentioned......... –  user55315 Jun 4 at 19:34
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Caould you give a $ \mathcal{C}^\infty$ monotonic, counterexamples? Thank you very much! –  2016 Jun 5 at 13:49

The Hauptvermutung states that there is essentially only one PL structure on a manifold. More precisely, it states that any two triangulations have a common subdivision. The reason why this seems "obviously true" is that you can take both triangulations and superimpose them one on top of the other, subdividing the manifold into a bunch of cells, and then taking the barycentric subdivision to get a triangulation. It turns out that this is false and one needs some pretty subtle invariants to detect it. The problem with the argument that I gave is that one triangulation could be very wild with respect to the other (fractally wiggly) so that their union does not subdivide the manifold into a nice collection of cells.

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Theorem: Let $f_1(x,y)$ and $f_2(x,y)$ be two joint probability densities, each having its $x,y$ components positively correlated ($Cov_1(x,y)>0$, $Cov_2(x,y)>0$). Let $f_3=\alpha f_1 + (1-\alpha) f_2$ be the mixing density, for some $0\le \alpha\le 1$. Then $Cov_3(x,y)>0$.

In words: mixing populations preserves the correlation sign. In other words: if the average MSE male user is brighter than the mean, and if the average MSE female user is brighter than the mean, then the average MSE user is brigther than the mean. Obviously true.

False. See Simpson's paradox.

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Perhaps the following example will make the obviousness more obvious. Baseball team $A$ has a better win-lose ratio than team $B$ in the first part of the season. Then there is a strike, and some games are missed. When the season resumes, team $A$ also has a better win-lose ratio than team $B$ in the second portion of the season. Therefore, team $A$ has a better win-lose ratio than team $B$ overall. (Wrong!) Baseball did play such a split season in 1981, but I don't know if the paradox actually occurred then. Probably not, as it requires $A$ and $B$ to play very different numbers of games. –  MJD Jun 6 at 0:43

The following statement I once believed to be "obvious":

If $f:\mathbb{R} \rightarrow [0,\infty)$ continuous is such that $\int_{-\infty }^{\infty }f(x)\text{d}x<{\infty } $, then $\lim \limits_{x \to \pm\infty} f(x) = 0$

which is actually false.

(Note: It is true if $f$ is uniformly continuous!)

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Can you provide a counterexample? –  hallaplay835 Jun 11 at 22:26
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This is pretty easy to give a counterexample to: let $f$ have bumps as far down as you like, just with sufficiently small widths so that the sum of their integrals goes to zero. You can make $\limsup_{x \to \infty} f(x) = \infty$ this way. –  Dustan Levenstein Jun 12 at 3:43

A simple arc (homeomorphic image of the closed unit interval) in the plane has $2$-dimensional Lebesgue measure zero.

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@athos It should be obvious, otherwise it wouldn't be a valid answer to this question. Are you asking why it's actually false? There is probably a simpler direct construction of a "fat arc" in the plane, but it follows from a more general theorem of R. L. Moore and J. R. Kline (On the most general plane closed point-set through which it is possible to pass a simple continuous arc, Ann. of Math. (2) 20 (1919), 218-223) that every (homeomorph of the standard) Cantor set in the plane is contained in an arc; apply this to a fat Cantor set in the plane. –  bof Jul 7 at 3:28

Something I used to be seduced by in my mathematical immaturity (which is sadly still existing):

Suppose that $P_n$ are a family of statements indexed by $n\in\mathbb{N}$ and we can assign meaning to $P_{\infty}$. Then if $P_n$ is true for all $n\in\mathbb{N}$, then $P_{\infty}$ is true also.

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The example I had in mind when I wrote this was $P_n$: sum of $n$ differentiable functions is differentiable; $P_\infty$: infinite sum of differentiable functions is differentiable. –  Jp McCarthy Jun 6 at 16:09
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How about the statement that 1/n > 0 which holds for any integer n but fails in infinity? –  Jón Áskell Þorbjarnarson Jun 6 at 16:37
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@JónÁskellÞorbjarnarson Mark would claim that $1/\infty>0$ is not meaningful because $1/\infty$ is not defined. An easier example is just a sequence of rationals converging to an irrational. –  Jp McCarthy Jun 6 at 16:40
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@Hurkyl I agree and was trying to argue from Mark's position. –  Jp McCarthy Jun 8 at 22:17

Here are some of the false statements popping into my mind that made me raise at least one eyebrow when I first realized they were not true.


Every linear function between two vector spaces is continuous.

True only as long as the domain is finite-dimensional. If it is not, then there exists a linear function that is not continuous—at any point!


The set of real numbers can in no way be (totally) ordered in such a way that every non-empty set in it has a least element.

False if choice is assumed, by the well-ordering theorem.


$\mathbb Q$ is not countable.

I am still tempted to believe it sometimes...


If the derivative of a continuous real-to-real function exists almost everywhere and (wherever it exists) vanishes almost everywhere, then the function must be constant.

False. In fact, there exists a function that satisfies the premise and it is strictly [sic!] increasing!


Any compact set is closed.

The name “compact” would suggest this, but this can be guaranteed only in Hausdorff spaces.


A set is compact if and only if every sequence in it contains a convergent subsequence.

While true in metric spaces, not only is it false in some more general topological spaces, but also neither condition implies the other!

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I totally believe that $\mathbb{Q}$ is countable. Did you mean $\mathbb{R}$? For the next, do you assume that the derivative exists everywhere, or only almost everywhere? –  Daniel Fischer Jun 5 at 9:14
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I've convinced myself that $\mathbb Z^2$ is countable; obviously it is possible to devise an algorithm to count them. And I've convinced myself that this means that $\mathbb Q$ is countable, because $\mathbb Q$ is ordered pairs of elements in $\mathbb Z^2$, so there are "less" elements in $\mathbb Q$ than $\mathbb Z^2$ (thinking of graphing elements, $\mathbb Q$ leaves holes because $\frac 24 = \frac 12$), so $\mathbb Q$ is definitely not uncountably infinite. –  Quincunx Jun 6 at 2:51
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I guess the one feature of $\mathbb Q$ that throws me off balance is that it just somehow doesn't feel “right” to me for a countable set to be dense in an uncountable one. –  triple_sec Jun 6 at 3:14

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