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As I procrastinate studying for my Maths Exams, I want to know what are some cool examples of where math counters intuition.

My first and favorite experience of this is Gabriel's Horn that you see in intro Calc course, where the figure has finite volume but infinite surface area (I later learned of Koch's snowflake which is a 1d analog). I just remember doing out the integrals for it and thinking that it was unreal. I later heard the remark that you can fill it with paint, but you can't paint it, which blew my mind.

Also, philosophically/psychologically speaking, why does this happen? It seems that our intuition often guides us and is often correct for "finite" things, but when things become "infinite" our intuition flat-out fails.

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Why does it happen? Because our intuition is developed by dealing with finite things: it is quite unsurprising that we are surprised by phenomena specific to infinite objects! This is exactly the same as the fact that our bodies are trained to move and act under the effect of gravity, so when we are in space we become clumsy and need to retrain. Intuition is not fixed: if you study phenomena associated to infinite objects, you develop an intuition for that, and presumably people working with large cardinals, (cont.) – Mariano Suárez-Alvarez May 2 '12 at 1:13
(cont) or strange objects like graphs with chromatic number $\aleph_8$ or Banach-Tarski partitions of a sphere, after a while find them just as intuitive as you and me find the formula for the area of a triangle. Intuition is, in most situations, just a name we put on familiarity. – Mariano Suárez-Alvarez May 2 '12 at 1:15
Philosophically / psychologically speaking, human brains weren't adapted for intuiting mathematical truths. The fact that we can repurpose our brains to do mathematics at all (beyond counting etc.) is astonishing. As for Gabriel's horn, I don't think this is a good example: see . – Qiaochu Yuan May 2 '12 at 1:20
I think remarks like "you can fill it with paint, but you can't paint it" are actually not helpful. In trying to appeal to our everyday intuition, they get in the way of mathematical understanding. Of course, you can't paint Gabriel's Horn (it's surface area is infinite) but you can't fill it with paint either (because paint molecules have a finite size, and Gabriel's Horn gets infinitely thin). Or, more prosaically, you can't fill Gabriel's Horn with paint because it's a mathematical idealisation that doesn't exist in the physical world. – Chris Taylor May 2 '12 at 7:35
"In mathematics you don't understand things. You just get used to them." ---John von Neumann. – Nate Eldredge May 2 '12 at 19:33

41 Answers 41

Seat $n$ dining people at a round table. Is it possible to re-seat these people $n-2$ times during that dinner so that at the end everybody has been seated to the left of everybody else?

Intuition suggests that this works (at least) for $n$ prime. I find it highly counterintuitive that this is possible if and only if $n \not\in \{4,6\}$. This is a special case of the Oberwolfach problem, see also this answer by Ben.

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A famous example of a counterintuitive fact in statistics is the James-Stein phenomenon. Suppose $X_1,\ldots,X_m$ are independent normally distributed random variables with expected values $\mu_1,\ldots,\mu_m$. One wishes to estimate $\mu_1,\ldots,\mu_m$ based on observation of $X_1,\ldots,X_m$. If instead of using $(X_1,\ldots,X_m)$ as the estimator of $(\mu_1,\ldots,\mu_m)$, one uses the James-Stein estimator $$ \left(1-\frac{(m-2)\sigma^2}{X_1^2+\cdots+X_m^2}\right)(X_1,\ldots,X_m) $$ (where $\sigma^2$ is the common variance) then the mean square error is smaller, regardless of the value of $(\mu_1,\ldots,\mu_m)$.

And the James-Stein estimator is demonstrably not even an admissible estimator, in the decision-theoretic sense. Thus the obvious estimator is inferior to one that is inferior to some admissible estimators.

One is "shrinking toward the origin", and it should be apparent that it doesn't matter which point you take to be the origin. In practice one should take the point toward which one shrinks to be the best prior guess about the value of $(\mu_1,\ldots,\mu_n)$.

The reason for the non-admissibility is that sometimes $(m-1)\sigma^2/(X_1^2+\cdots+X_n^2)$ is more than $1$, so that the sign gets reversed. That's too extreme by any standards. A piecewise-defined estimator that shrinks toward the origin but no further than the origin is superior in the mean-squared-error sense.

In the '80s and '90s, Morris L. Eaton showed that the fact that this works if $m\ge 3$ but not if $m\le2$ (apparent from the "$m-2$" in the numerator) is really the same fact as the fact that random walks are recurrent in dimension $\le2$ and transient in dimension $\ge 3$, which I think was discovered about a hundred years ago.

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If I remember correctly, that last fact about random walk was discovered by G Polya in the 1920's, so a little less than 100 year! – kjetil b halvorsen Dec 20 '12 at 0:13

I find it counterintuitive that the rational numbers have zero measure. This means that for any $\epsilon$, no matter how small, we can find a collection of intervals that includes every rational but whose total length is less than $\epsilon$.

Such a covering includes all the rationals, but somehow must omit nearly all the irrationals. The question has some up here several times about what irrationals are missed (1 2 3), and Asaf Kargila recently described the result as “baffling”. So I am not the only person who is surprised by this.

Von Neumann famously said that in mathematics you don't understand things, you just get used to them. This for me is one of those things.

(The countability of the rationals, or the uncountability of the irrationals, may be similarly baffling, but I have gotten so used to both that I no longer find either one baffling, and I am not sure which one should be considered counterintuitive.)

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This, is the most counterintuitive fact that I ever saw:

Blue Eyes Islanders Question:

This question was created by mathematician Terence Tao.

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And on the 101st day everyone else commits suicide. Tragically – Thomas Ahle May 25 '13 at 9:53
Not really, @Thomas, because no islander has reason to believe that their eye color is exactly one of blue or brown. – Student Jan 6 '14 at 0:26
I am unsure that this question should be attributed to Terry Tao. I don't know who it should be attributed to, but I have had a quick search about and I would be surprised if it was actually due to him... – user1729 Apr 22 '14 at 13:40

I am surprised nobody has given this example: It was once thought that continuous functions have derivatives at almost all points (as other answerers have used). Then comes the Brownian motion model that generates a random curve, which, with probability 1, is a curve without derivative at any points!

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The Weierstrass function. It showed that a function can be continuous everywhere but differentiable nowhere. This was (and still is) counterintuitive.

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Although Dan Brumleve and plm linked the list of paradoxes, the famous Braess's paradox deserves a special attention.

"For each point of a road network, let there be given the number of cars starting from it, and the destination of the cars. Under these conditions one wishes to estimate the distribution of traffic flow. Whether one street is preferable to another depends not only on the quality of the road, but also on the density of the flow. If every driver takes the path that looks most favorable to him, the resultant running times need not be minimal. Furthermore, it is indicated by an example that an extension of the road network may cause a redistribution of the traffic that results in longer individual running times."

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The concentration of measure phenomena on the sphere:

If $A\subset\mathcal{S}^{n-1}$ is a measurable set on the sphere with $\lambda(A)=1/2$ and $A_\epsilon$ is an epsilon neighborhood of $A$ on $\mathcal{S}^{n-1}$, then

$$\lambda(A_\epsilon)\geq 1-\frac{2}{e^{n\epsilon^2/2}}$$

So say you take $A$ to be a cap on the sphere and fix a small $\epsilon$. As the dimension of the sphere increases, eventually the $\epsilon$ enlargement of $A$ will have almost the entire area of the sphere! Playing with the upper and lower cap and the corresponding enlargements, one sees that area is concentrated around the equator.

Imagine you have a lawnmower and you cut the grass moving along the equator. What percentage of the sphere do you mow? Well, in 3 dimensions, not that much. But as you cut the grass on higher and higher dimensional spheres, moving centered along the equator, the surface area covered becomes almost 100% of the entire area of the sphere!

This result felt pretty counter-intuitive to me the first time I saw it.

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I think a puzzle at calculus level is the following: Given a real number $x$ and a conditionally convergent series, the series can be re-arranged so that its sum is $x$.

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Independence in Statistics. If I have a box with red & black socks, and I take one, and then another one the probability of both socks being red is sometimes different from the case when I, instead, pick a pair. (EX.If all socks are connected to every other with a string, like a complete graph, and then I pick some string, automatically choosing the socks at either end as the pair selected).

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The existence of countable countably infinite connected Hausdorff spaces is (to me) counterintutive. (Just one example; I could think of others . . . . .)

Later edit: A Hausdorff space is a topological space in which, for every pair of points $x$ and $y$, there are open neighborhoods of $x$ and $y$ that do not intersect each others, i.e. $x$ and $y$ can be in a certain sense separated from each other.

A connected space is a topological space that cannot be broken into separate components having no proximity to each other. Imagine two disks remote from each other. No sequence of points in one disk can approach a point in the other as a limit. That's a space that is not connected.

Countable means either finite or countably infinite, as opposed to uncountably infinite, and that means one can list all the point in a sequence: $x_1,x_2,x_3,\ldots$. The sequence may be infinite, but each term in the sequence has only finitely many terms before it.

So figure out what a countable connected Hausdorff space is based on all that.

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It would be interesting if you expanded. However, it sounds quite advanced theory. – Pedro Tamaroff May 2 '12 at 2:30
@Peter: see here – t.b. May 2 '12 at 6:50
It doesn't required any background beyond a semester of point-set topology. – Michael Hardy May 2 '12 at 17:07

Another elementary one. There is a configuration of 30 convex bodies in 3-dimensional space with disjoint interiors that "cannot be taken apart with two hands". That is, it's impossible to split up the set of bodies into two nonempty subsets and, by a rigid motion, move one of the subsets away to infinity without disturbing a member of the second subset. See and

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For quite a while, when I was very young, I thought that the solution to the famed brachistochrone problem (which asks about the shape of a wire where a bead sliding frictionlessly on it under the influence of gravity will take the least amount of time to finish) was not a straight line was pretty counterintuitive, after being repeatedly told that the straight line is the shortest path between two points. (To spoil: the true brachistochrone is an inverted cycloid, which is the curve traced by a fixed point on the rim of a rolling circle.) I went through the derivation a number of times before finally being convinced.

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How about the Löwenheim–Skolem theorem?

One of its consequences is that the field of real numbers has a countable model.

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Not exactly. The completeness property is second order. The countable model will be a real-closed field, but not complete. – Asaf Karagila May 8 '12 at 4:38

There are a number of results of the form "Proposition P fails in dimension $d$" where P holds in lower dimensions, many of which can seem counterintuitive until you understand higher dimensional phenomena.

Here's an elementary one, which many people on this site won't find counterintuitive but some might. Consider the question "What is the maximum number of vertices a polyhedron in $\mathbb{R}^d$ can have such that there is a segment joining every pair of points which is an edge of the polyhedron?" For $d=2$, the answer is obviously 3, with a triange. It's not difficult to see that a tetrahedron is optimal for $d=3$. Intuition suggests that the $d$-simplex is optimal based on this.

But for $d=4$, in fact, there is no maximum number. There are polyhedra in $\mathbb{R}^4$ with arbitrarily many vertices and an external edge joining each pair of vertices. If you take any finite collection of points on the moment curve $\{(t,t^2,t^3,t^4)\, | \, t>0\}$, the segment joining any two of the points is a face of the convex hull of the collection. Once you have an intuition for higher dimensional geometry, this is obvious, but it can seem counterintuitive.

A more advanced example, that I still find counterintuitive at times, is this: In $\mathbb{R}^d$ for $d=2,3$, given any polyhedron, one can move each of the vertices a small amount to obtain a combinatorially equivalent polyhedron with rational vertices. But in $d=4$ and higher there are polyhedra which can not be realized with rational coordinates.

EDIT: I was asked to provide a reference. This is a well-known result in some circles, particularly in computational geometry, so it's covered in a number of locations. Marcel Berger's Geometry Revealed covers both of the above so-called counterintuitive statements, as well as the surprisingly nonobvious case $d=3$, in chapter 8, roughly page 550, and is a pretty easy read. If you don't have access to Springer, the paper Realization spaces of polytopes by Richter-Gebert is the most comprehensive treatment I know of, and probably any book citing this paper is quoting the result.

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I never found this one to be as counterintuitive. Comparing hypervolumes of n-spheres is geometrically more meaningfully thought of as comparing the ratio of their hypervolumes to those of unit hypercubes (via dimensional analysis). But for me, the more natural thing was to compare the ratio of their hypervolumes to that of their circumscribing cubes, which then gives a monotonically decreasing sequence... – Logan Maingi May 2 '12 at 7:02
The question then becomes whether the sequence decreases faster than $2^{-n}$, and you can probably convince yourself that the sequence should decrease super-geometrically based on geometric intuition. If you look at it that way, then there's nothing mysterious about the volume formula. Unfortunately, this is how I first considered the problem, and so I never had the opportunity to be surprised by this result. – Logan Maingi May 2 '12 at 7:06
That is a good point. When I first discovered this I thought it was really weird (maybe because I'm not a very visual thinker). Here is another one: a 2-dimensional random walk returns to the origin almost surely, but in 3 or more dimensions it may not! – Dan Brumleve May 2 '12 at 7:24

Here are a few counter-intuitive results that have surprised me at one point or another:

  1. Impossible Constructions using Straightedge and Compass. Not all regular $n$-gons are constructible with straightedge and compass.
  2. Godel's Incompleteness Theorems. Certain non-trivial arithmetics cannot be both complete and consistent.
  3. Exotic spheres. In certain dimensions there are spheres which are homeomorphic but not diffeomorphic to the standard sphere.
  4. Kuratowski's Closure-Complement Theorem. The largest number of distinct sets obtainable by repeatedly applying closure and complement to a given starting subset of a topological space is 14.
  5. Dehn's Answer to Hilbert's Third Problem. The cube and regular tetrahedron are not scissor-congruent.
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I didn't think of this until today, but it's an important thing that I, and many other people, find completely mindboggling.

Let's consider properties, like "is red" or "has kidneys" or "has a heart". Now there's a certain sense in which two properties might be the same even though they don't look the same, which is that they might be true of exactly the same entities. For example, it might turn out that everything that has kidneys also has a heart and vice versa, so that even though the two properties have different meanings (kidneys are not the same as hearts), they amount to the same thing in practice.

Mathematics is of course full of such properties; consider for example the property ${\mathcal O}_1$ of being expressible in the form $2n+1$ for some integer $n$, and the property ${\mathcal O}_2$ of being expressible in the form $S_{n+1} - S_n$ for some pair of consecutive squares. Many theorems are of this type, that two seemingly different properties are actually the same.

So let's try to abstract away the senses of properties, leaving only the classes of things that possess them. We'll say that there are these entities called sets which are abstractions of properties. Things belong to a set exactly if they possess the property of which the set is the extension:

  1. For every property $P(x)$, there is a corresponding set $\{x : P(x)\}$ of exactly those entities $x$ for which $P(x)$ is true.

  2. An entity $y$ is a member of a set $\{x : P(x)\}$ if, and only if, $P(y)$ is true.

That seems utterly straightforward and utterly unexceptionable, and yet, it is utterly wrong.

There are completely mundane properties for which there is no corresponding set of all the entities with the property.

What? Who ordered that?

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@jake: I am talking about Russell's paradox. Take $P(x) = x\not\in x$, $P(x) = (x\in x\implies 2+2=5)$, or $P(x) = \lnot\exists y: y\in x\wedge x\in y$. None of these properties has an extension. – MJD May 6 '12 at 19:23

Choose a natural number, for example $n=8$. Then pick a base, for example $b=2$, and finally select another natural number called the bump factor, for example $B=1000$. Then construct a sequence of natural numbers as follows: The first term of the sequence is simply $n$ written in expanded base $b$. $$m_{0}=2^{2+1}=8$$ The second term is obtained from the first by bumping the base $b$ by a factor of $B$ and then subtracting $1$ from the result. $$m_{1}=2000^{2000+1}-1=\sum_{k=0}^{2000}1999\cdot2000^{k}>10^{10^3}$$ The third term is obtained from the second by bumping the new base ($2000$) by a factor of $B$ and then subtracting $1$ from the result. Denoting $d=2\cdot 10^{6}$ we have $$m_{2}=1999d^{d}+1999d^{1999}+\cdots+1999d+1998>10^{10^7}$$ Continuing in this fashion we denote $e=2\cdot10^{9}$ and the next term is $$m_{3}=1999e^{e}+1999e^{1999}+\cdots+1999e+1997>10^{10^{10}}.$$ The next term $m_{5}$ has over 24 trillion decimal digits.

Intuition tells us that the sequence $(m_{r})$ goes to infinity, and very fast. However, this is not the case. Surprisingly, the sequence will reach $0$ in finitely many steps. That is, there is an $r\in \mathbb{N}$ for which $m_{r}=0$.

The sequence we constructed is an example of a Goodstein sequence, and the fact that it terminates is a very particular case of Goodstein's Theorem. This theorem is counterintuitive for two reasons. First because of what the theorem concludes. Roughly speaking, it states that any sequence of natural numbers of the type constructed above (i.e. a Goodstein sequence) will always terminate. Second, because of what it is required to prove it. Goodstein's theorem is a fairly elementary statement about natural numbers (i.e. formulated within the Peano Axioms of Arithemtic) and yet its proof cannot be carried out using only these axioms. It requires infinite ordinals.

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I always found the main correspondence of class field theory counter-intuitive. Somehow the (Abelian) field extensions of a number field $K$ correspond to arithmetic objects in $K$ itself.

You wouldn't think that the possible ways of extending a field (albeit in a nice way) would just depend on the field you started with.

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I also Think The Kakeya Needle Problem is worth mentioning (see To me it is counter-intuitive that there is no smallest set, in which a needle of unit length can be freely rotated. Unless it has to be convex, of course.

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This is great; I hadn't heard of that problem before. – joriki May 6 '12 at 7:27

The fact that one can easily prove the existence of uncountably infinite (as opposed to countably infinite) sets is counterintutive to me. Not that fact that uncountably infinite sets exist, but the fact that the proof is so simple. I was astonished when I first learned of it. I was in ninth grade. I think it was in a book by Vilenkin that I read the proof.

Similarly the fact that one can easily prove that the square root of $2$ is irrational. I hadn't expected that to be so simple. And the mere existence of irrational numbers seems counterintuitive: why shouldn't fractions be enough to fill up all the space between integers?

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As some other people said, "intuition is highly subjective". Different people think about problems in different ways.

That said, there are many, many counter-intuitive results in mathematics. This is why people demand rigorous proof! ;-)

  • Almost any result involving probability. Humans suck at probability! (E.g., the birthday paradox: The probability that anyone in the room shares the same birthday as you is very small, unless you have a lot of people. But the probability that anybody in the room shares a birthday is very high. Way higher than you'd imagine...)

  • Almost any result involving infinite sets. Infinity doesn't behave how you'd expect at all! ("Infinity" actually comes in different sizes. $\mathbb{Q}$ is the same size as $\mathbb{N}$, despite being a superset of it. Subtracting an infinite set from an infinite set can yield a result of positive finite size. Etc.)

  • Several results about things which are impossible to compute. (E.g., the halting problem looks like it should be really, really easy, but it's actually impossible. Rice's theorem also sounds completely ludicrous. The busy beaver function is non-computable, regardless of how easy it looks. And so forth.)

  • Fractal geometry contains a few results which break people's minds. (E.g., polygon which has infinity perimeter and zero area. A Julia set where every point simultaneously touches three basins of attraction. A connected curve with no derivatives...)

I could probably think of more, given enough time...

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Really interesting question, I have some examples that many people find counterintuitive.

The set $\mathbb Q$ of rational numbers as the same cardinality of the set of natural numbers $\mathbb N$, although $\mathbb N$ is strictly contained in $\mathbb Q$. Similarly many people find it to be counterintuitive that even numbers are equal in cardinality to the naturals (i.e. the sets $\{2n \mid n \in \mathbb N\}$ and $\mathbb N$ have the same cardinality).

The set $\mathbb R$ has cardinality strictly greater than the set $\mathbb N$ (and so also of the set $\mathbb Q$) (so there's not just one type of infinity).

Another good example of a counterintuitive fact is the Banach-Tarski paradox stating that a ball can be decomposed in a finite number of pieces which can be glued together to build up two balls identical to the first one (I say that this is a paradox because the axiom of choice is clearly true :D).

If other examples come to my mind I'll add them later.

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+1 for the Banach-Tarski paradox, it's the first that came to mind when read the question. I think that it is counter-intuitive because intuition would tell that any 3d object has volume. But no well-defined volume can be assigned to these pieces. – ypercubeᵀᴹ May 2 '12 at 21:31
"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?" (According to Wikipedia, this is a quote from someone called Jerry Bona). – user1729 May 10 '12 at 13:33

I think the following has (suprisingly) not been pointed out already:

As a general rule paradoxes (counterintuitive truths) are very important in mathematics and there are many books dedicated to them. 1 and 2 are famous examples. The Monty Hall problem and Banach-Tarski paradox even have books dedicated to them, and each is the subject of ongoing research.

Paradoxes arise when simplification does not work, when usual assumptions do not hold. Of course this will depend on the person thinking about the phenomenon, on her experience. A topologist is well aware of counterexamples in her field so she would not find them paradoxical anymore.

Also I am not sure the Blue-eyed Islanders Paradox has been mentioned here. It has received much internet attention recently, foremost thanks to Terence Tao, c.f. also xkcd.

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Here's a counterintuitive example from The Cauchy Schwarz Master Class, about what happens to cubes and spheres in high dimensions:

Consider a n-dimensional cube with side length 4, $B=[-2,2]^n$, with radius 1 spheres placed inside it at every corner of the smaller cube $[-1,1]^n$. Ie, the set of spheres centered at coordinates $(\pm 1,\pm 1, \dots, \pm 1)$ that all just barely touch their neighbor and the wall of the enclosing box. Place another sphere $S$ at the center of the box at 0, large enough so that it just barely touches all of the other spheres in each corner.

Below is a diagram for dimensions n=2 and n=3.

enter image description here

Does the box always contain the central sphere? (Ie, $S \subset B$?)

Surprisingly, No! The radius of the blue sphere $S$ actually diverges as the dimension increases, as shown by the simple calculation in the following image,

calculation of inner sphere radius

The crossover point is dimension n=9, where the central sphere just barely touches the faces of the red box, as well as each of the 512(!) spheres in the corners. In fact, in high dimensions nearly all of the central sphere's volume is outside the box.

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But the volume of the box diverges just as well. As you increase dimensions shouldn't you expect everything to just keep growing? – Steven-Owen Nov 3 '12 at 17:02
1) This is not counterintuitive, one can see what happen comparing cases $n=2$ and $n=3$, relative difference in volumes between blue sphere and box is less. 2) $2^n$ spheres always has radious 1 when diagonal of box increases. 3) The fact that a sphere bounded by the vertex of a box can get out of the box in any dimension. 3 facts that makes this result perfectly logic!. – Gastón Burrull May 5 '13 at 6:48
@Steven-Owen but notice that the distance from the origin to the center of each cube face remains constant. – Thomas Ahle Jan 6 '14 at 8:19
This example shows how important is that we think outside the box! :-) – Asaf Karagila Jan 5 '15 at 16:16

The topological manifold $\mathbb{R}^n$ has a unique smooth structure up to diffeomorphism... as long as $n \neq 4$.

However, $\mathbb{R}^4$ admits uncountably many exotic smooth structures.

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The only dimension for which $\mathbb{R}^n$ admits exotic smooth structures is $n = 4$... I just can't get over it. – Jesse Madnick May 4 '12 at 7:47
@JessMadnich: Why is this? When or how does "4" enter the proof? – NikolajK May 6 '12 at 21:10
Interesting coincidence, the only dimension for which $\mathbb{R}^n$ admits a (non-comutative) skew filed structure, compatible with the multiplication of $\mathbb{R}$ is also $n=4$. – N. S. May 8 '12 at 16:38
There are no coincidences in mathematics - only reasons too abstract for us to have spotted yet :) – Chris Taylor May 11 '12 at 10:35
@NikolajK There are very distinct techniques used for $n \leq 3$ and $n \geq 5$. The key to the former is that in low dimensions, smooth manifolds are the same as topological manifolds; every top. man. has a unique smooth structure. In high dimensions this is a largely algebraic story; relevant parts were written by Kirby and Seibenmann. See this answer. In low dimensions we can't use algebra as Whitehead's trick doesn't work, and the special cases for $n \leq 3$ depend on certain homotopy groups vanishing, which only happens in small dimensions. – Mike Miller Jan 5 '15 at 16:43

Intuition is a really subjective and personal matter. To go even further with the problem of such list is that there are many proof requiring some use of the axiom of choice. On the other hand, not assuming the axiom of choice can be equally reasonable, and here is a short list of how things might break down completely:

  1. The real numbers can be a countable union of countable sets.
  2. There might be no free ultrafilters, at all (on any set).
  3. The rational numbers might have at least two non-isomorphic algebraic closures.
  4. The natural numbers with the discrete topology might not be a Lindelof space.

Some results in ZFC which are completely unintuitive the first time you hear them:

  1. While being perfectly definable, the set $\mathcal P(\mathbb N)$ can differ greatly between models of ZFC; or an even worse formulation:
  2. There are models $M\subseteq N\subseteq V$ such that $N$ has more reals than $M$ and $V$ has more reals than $N$, but the amount of real numbers of $M$ and $V$ is the same.
  3. There is a polynomial with integer coefficients which has a rational root if and only if ZFC is inconsistent.
  4. Every model of ZFC is one class forcing from being $L[a]$ where $a$ is a real number; and every model is one set forcing away from being $HOD[A]$ for some set $A$.
  5. The union of countably many disjoint open intervals might have uncountably many boundary points (e.g. the complement of the Cantor set in $[0,1]$).

Both lists are infinitely long, and I can probably ramble about the first list for several days. The point, as I say at first, is what we take for "granted" as intuitive which can change greatly between two people of different mathematical education; mathematical culture; and what is their usual axiomatic system (which is essential for "results").

One strange result on mathematicians is a direct corollary of the first result in the second list:

People are used to think that there is only one universe, only one fixed way to handle sets. While it is true that for the working mathematician this is often a reasonable approach, set theorists deal with models of set theory, much like group theorists deal with models of group theory.

Somehow everyone is flabbergasted when they are being told (for the first time, if not more) that there are many models of ZFC with different number of sets of natural numbers in each model; but no one falls of their chair when they are told that some fields have more irrational numbers than others...

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It's fairly straightforward to prove that the Cantor set is uncountably infinite (i.e. it has the same cardinality as the real numbers). You would think that such a set would have at least positive Lebesgue measure. WRONG-IT ACTUALLY HAS MEASURE ZERO! This blew my mind when I first learned about in undergraduate real analysis and it's why it puzzles me why Paul Halmos famously described measure theory as "a generalized kind of counting". This example seems to show that measure has nothing to do with cardinality, which is what most of us think of as counting in abstract sets!

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The standard Lebesgue measure definitely has something to do with cardinality. If a set has positive measure (or is unmeasurable) it has the same cardinality as the reals. Measure is a more refined sense of counting-so refined, in fact, that it is only really interesting on a class of sets all of which have the same cardinality. – Logan Maingi May 4 '12 at 7:56
The existence of "counting measure" on any set seems to give one indication that measure theory is a generalized kind of counting. – Pete L. Clark May 5 '12 at 17:52
@Pete: I think he was trying to ping you. – The Chaz 2.0 May 8 '12 at 4:00
If instead of taking away the middle 1/3 at each iteration, you instead take away the middle 1/4th, you end up with a "fat" Cantor set which, surprisingly, has positive measure. – Nick Alger May 16 '12 at 2:07
@LoganMaingi: Solovay showed that it is independent of ZFC whether there are nonmeasurable sets with cardinality less than $2^{\aleph_0}$. See this answer of JDH for a survey of related results. – Jonas Meyer May 17 '12 at 3:40

Löb's or Curry's paradox:

If this sentence is true, then Germany borders China.

Logic says this means Germany borders China (or anthing you want to put after the then).

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This got a lot more interesting after I thought about it for a minute! It's different from "this sentence is false". – Nick Alger May 4 '12 at 8:08
@NickAlger Not really, instead of just "paradox" it is "If true fact, then paradox." as in: If Germany does not border China, then this sentence is false. – Phira May 6 '12 at 9:59
What does it mean for the sentence to be true? Sentences of the form if p then q, are true (or provable) in my naive sense if I can get you from p to q using "logic," however, "if this sentence is true, then Q" is confusing. – Steven-Owen May 6 '12 at 19:06

Although well-known, I feel compelled to note the remarkable equation

$$ e^{i\pi} + 1 = 0. $$

That five of mathematics most well-known quantities are related in such a pleasantly simple way is astonishing and, to the the uninitiated, is certainly not intuitive. Of course, once one knows about infinite series, their basic properties and how to define the trigonometric and exponential functions with them, deriving this equation is routine. But, without this knowledge, the above equation seems almost mystical. In fact, this equation is what first piqued my own interest in mathematics.

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protected by J. M. May 6 '12 at 4:09

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