# 'Obvious' theorems that are actually false

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!

-
The Banach-Tarski paradox is a good one. – mjqxxxx Jun 4 '14 at 18:08
– Will Jagy Jun 4 '14 at 18:29
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 '14 at 19:53
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 '14 at 7:04
"Every true statement can be proved" – sinelaw Jun 9 '14 at 17:27

Theorem (false):

One can arbitrarily rearrange the terms in a convergent series without changing its value.

-
The series $\sum \frac{(-1)^i}{i}$ is a counterexample; see this Wikipedia article for a discussion. – MJD Jun 4 '14 at 18:39
@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 '14 at 18:55
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 '14 at 21:25
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 '14 at 22:42
@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 '14 at 21:36

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.

-
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 '14 at 18:37

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.

-
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 '14 at 20:07

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.

-
Can you show a counterexample for the third claim? (The one about linear functions) – Alessandro Jun 4 '14 at 19:28
@alessandro see this math.stackexchange.com/questions/131197/… – LeGrandDODOM Jun 4 '14 at 19:30
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 '14 at 19:50
@user18921 jstor.org/stable/2318996?__redirected – LeGrandDODOM Jun 5 '14 at 11:12

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.

-
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 '14 at 19:37
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 '14 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.

-
@jpmc26 It does always get bigger as you keep adding numbers; it just doesn't necessarily do so without limit. – Mike Scott Jun 5 '14 at 7:26
On the other hand if the family is uncountable, it holds. – user87690 Jun 5 '14 at 13:27
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 '14 at 17:31
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 '14 at 21:10
Ooooh! I read it as "strictly positive integers" which I believe would diverge. Right? (Too much programming) – Almo Jun 5 '14 at 22:30

A subgroup of a finitely generated group may not be finitely generated and there are up to isomorphism at most two $pq$ groups, where $p$ and $q$ are prime.

To strengthen your example, $\sum_{p \ prime} 1/p$ diverges.

The continuum hypothesis also seems like it has to have in answer in ZFC, which it doesn't.

On another page, mathematicians thought that cyclotomic fields "obviously satisfy the unique factorization theorem" leading to some false proof attempts to Fermat's last theorem.

Next, one might think that the "angle trisection" is possible or that any set of analytic functions $\{f_\alpha\}$, such that for every $z \in \mathbb C$, the set $\{f_\alpha(z)\}$ is countable, itself has to be countable.

These are just some random examples that came to mind and since the term "obvious" is subjective, you may very well disagree with items on my list. I guess it heavily depends on your mathematical background.

-
I disagree that there is anything intuitively obvious about the continuum hypothesis, and I find even less that is intuitively obvious about what can or cannot be proved in ZFC, which is so complex that you have to study really hard just to understand what the axioms are saying. – MJD Jun 4 '14 at 18:31

EDIT: The counterexample I had in mind is incorrect; I've asked another question to try and clarify the matter one way or the other. Regardless, I think it's safe to say that this is no longer obvious! But I'll go ahead and update (or delete) my answer accordingly once I've gotten a bit more clarity.

Here's a topological example that takes some thought to falsify: roughly, 'for every non-intersecting curve between two opposite corners of a square, there's a curve between the other two corners that only intersects it once'. Formally:

Let $f: [0, 1]\mapsto [0,1]^2$ be a non-self-intersecting curve with $f(0) = (0,0)$, $f(1) = (1,1)$, and $f(t)\in (0,1)^2$ for $t\in(0,1)$. Then there exists a non-self-intersecting curve $g: [0, 1]\mapsto [0,1]^2$ with $g(0) = (1,0)$, $g(1) = (0,1)$, and $g(t)\in (0,1)^2$ for $t\in(0,1)$ such that there are unique $t_0$ and $t_1$ with $f(t_0) = g(t_1)$.

This seems obvious (at least to me) on first glance, and even on second glance, the example of the Jordan Curve theorem suggests that it should be true; after all, we get a 'left side' and a 'right side' of our curve by the JCT, and doesn't the Schoenflies theorem mean that we should be able to find an inverse mapping of our curve to the circle? But it's false; there are curves $f()$ that can't be intersected only once by any curve $g()$. Finding a counterexample makes a nice exercise...

-
@mjqxxxx: No, that does not work: the straight line segment might not be contained in $\Phi([0,1]^2)$ (since you do not know what the image of the rest of the boundary of the unit square looks like). One needs more work to get a real proof. – studiosus Jun 4 '14 at 21:30
Here is a stronger (straightening) theorem: Let $S$ be a surface with some fixed triangulation (say, a triangulated plane) and $h: G\to S$ a topological embedding of a finite graph. Then $h$ is isotopic to a piecewise-linear embedding $h': G\to S$. This reduces problems of the above type to piecewise-linear ones which are easily solvable. One can give, however, a more direct solution using only Schoenflies and a bit of algebraic topology. – studiosus Jun 4 '14 at 22:03

A shape with finite volume must have finite surface area.

-
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 '14 at 20:31
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 '14 at 20:41
@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 '14 at 0:00
I dunno either, I'm a physics ignoramus. Are you sure clay is made of atoms? – MJD Jun 5 '14 at 1:11
Mathematically this may be true, but physically impossible, as matter is discrete. The limit would be a row of atoms. – Alex Jun 5 '14 at 4:54

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 $n>7$.

-
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 '14 at 18:55
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 '14 at 4:47
What does "face" even mean in 7 dimensions? Is it a 6-dimensional thing or a 2-dimensional thing? – Mehrdad Jun 5 '14 at 7:04
What precisely do we mean by "share a face" in this context? – goblin Jun 5 '14 at 10:20
what do you mean by "tiling"? – electronp Jun 5 '14 at 12:04

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).

-
Your "another" was already mentioned......... – user55315 Jun 4 '14 at 19:34
Caould you give a $\mathcal{C}^\infty$ monotonic, counterexamples? Thank you very much! – 2016 Jun 5 '14 at 13:49
Yes, you can get one by tinkering with the responses here a bit: math.stackexchange.com/questions/788813/… – Kaj Hansen Jun 6 '14 at 2:35

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 $x$”. For others, see Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? $\{x:P(x)\}$ and Paradox of General Comprehension in Set Theory, other than Russell's Paradox.)

-
+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 '14 at 22:39
Why is this wrong? – mattecapu Jun 5 '14 at 6:45
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. – Stella Biderman Jun 5 '14 at 6:52
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 '14 at 15:22
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 '14 at 15:11

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.

-
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 '14 at 19:29
@Christopher: No, the tree has $\omega+1$ levels. The last is the infinite sequence $0.333\ldots$. – Asaf Karagila Jun 5 '14 at 6:27
@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 Sep 30 '14 at 19:32

An analytic function with compact support vanishes identically.

-
That is actually true, as long as we stay in $\mathbb{C}^n$ resp. $\mathbb{R}^n$. It becomes false when one takes a compact analytic manifold, but then it's obvious that it's false. – Daniel Fischer Jun 4 '14 at 19:45

Here are some "obvious" statements to which Richard's paradox can apply to:

1. For a given predicate $P$, there exists a set $S$ of $x$ for which $P(x)$ is true. (Russell's paradox)
2. The set of integers and the set of real numbers are the same infinite size. (Cantor's diagonal argument.)
3. There exists a formalization of arithmetic in which all true statements are exactly those which are provable. (Gödel's theorem(s))
4. There exists a computer program (Turing machine) that can effectively determine if any other computer program doesn't halt. (Halting problem)
-
– MJD Jun 4 '14 at 19:53
I think #4 is not merely not obviously true, but obviously false, because if it were true, then we wouldn't have to actually run our computer programs to find out what they did, instead we could use this hypothetical procedure to tell us, without running them. But then what would be the point of running the programs at all? The universe just doesn't work that way; if you want to find out what happens, then, at least some of the time, you just have to try it and see. – MJD Jun 4 '14 at 19:55
It doesn't matter; my comment is the same either way. I have a blog post drafted that claims that Rice's theorem is the most obvious theorem in mathematics, and the undecidability of the halting problem is a special case of that. – MJD Jun 4 '14 at 23:59
This is not about general predictions of the future; this has to do with time only insofar as countable infinity can be mapped to discrete Turing machine steps that take place in time. E.g. A Turing machine that halts when it finds $a,b,c,n\in\mathbb{Z}_+:a^{n+2}+b^{n+2}=c^{n+2}$ we know doesn't halt thanks to Wiles's proof. If the halting theorem were false, then a "theorem-proving machine" based on a formalization of any area of mathematics would spit out all true statements eventually. This was "obvious" to Russell, Whitehead, and even Hilbert until the 1930's thanks to Gödel, Turing et al. – Matt Jun 6 '14 at 19:42
@MJD If the halting theorem were false, it doesn't imply that we would be able to tell the outcome of any computation. It merely would imply that an algorithm exists. There is still a substantial gap between the existence of such an algorithm $U$, and both (1) our physical ability to run $U$ (it may require more memory than the number of atoms known to exist) and (2) more importantly our knowledge that a given algorithm is in fact $U$. (1) and (2) do not automatically follow from the halting theorem's falsity. The reasons for it being true are more subtle than you are framing it. – Matt Jun 6 '14 at 19:53

Cauchy's theorem implies that:

if one makes a physical model of a convex polyhedron by connecting together rigid plates for each of the polyhedron faces with flexible hinges along the polyhedron edges, then this ensemble of plates and hinges will necessarily form a rigid structure.

However, there are counter-examples if you allow a general polyhedron (not convex).

-

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

-
@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 '14 at 3:28

There are a lot of examples in (extremal) graph theory, where an obvious argument shows that a statement is true, except that there are a number of small counterexamples which are easy to overlook.

Consider the following statement: Let $G$ be a graph with $n$ vertices and the largest number of edges subject to the condition that $G$ does not contain a pair of disjoint edges (i.e. $K_2 + K_2$). Then $G$ is a star (i.e. $K_{1,n-1}$).

This is obviously true, if you think about it for a moment. But for $n=3$, a better solution is to take $G = C_3$. And for $n=4$, taking $C_3$ plus an isolated vertex is just as good as taking $K_{1,3}$.

-

Here is one thing commonly thought to be true but is quite horribly wrong on many accounts:

There is a notion of mathematics where we can say things are
"actually true" or "actually false".


An example of making this error: the OP. Other examples: the many responses.

There are several reasons why this is wrong. First, in the system most mathematicians assume when not being explicit, we have no standard model (we have no models within that system because any model would show the system consistent which we know we cannot show in that system through blah blah Godel blah.. I know you don't want details, just explaining what I'm getting at). Truth and falsity are semantic - they exist in models and so without one, we don't make claims of truth or falsity.

But also, mathematics is not "the system most mathematicians assume when not being explicit" - it is formalisation in general. There are many systems seriously investigated by mathematicians that make numerous "counterintuitive" derivations. For instance, these are all obvious and wrong in different systems:

• A statement cannot be both true and false. (In paraconsistent logics, statements can be both true and false and the system does not collapse to trivial - in fact, a number of dialetheists argue this is a much more accurate logical system for real world reasoning).
• There are discontinuous total functions. (In a number of constructive systems it is not possible to prove the existence of discontinuous total functions. Some are even strong enough to prove that all total functions are continuous.)
• Every infinite set A has the same cardinality as AxA. (This is not necessarily true in systems without the axiom of choice. Famously, Tarski tried to publish his result on this implication and was rejected by both Frechet and Lebesgue. Frechet thought the paper was obvious and well-known and had no mathematical merit. Lebesgue thought both the axiom of choice and the implication from it were both wrong, so the paper had no mathematical merit.)

I only bring these examples up not as answers to the OP but simply to illustrate my real answer that the question itself demonstrates an extremely common assumption in mathematics that is in fact wrong.

EDIT

This is an area that I think is often a place of common misunderstanding, and discussion in the comments makes it clear I should elaborate. Modern mathematics separates out the domains we make statements on into syntax and semantics.

Syntax

The syntax is the theory - the formal language, axioms specified as sentences in the formal language, and some metalogical rules of inference. In the syntax, we talk about sentences, propositions, terms, derivations, and proofs. It is a place of symbol manipulation.

Semantics

The semantics is the model - it is the meaning we ascribe to the statements of the theory. An interpretation of a theory is a model that assigns to each formula of the theory a meaning value - typically truth. Truth is semantic and is specific to a model.

The "problem"

A model is a consistent interpretation of the truth meaning of a theory. If a theory has a model, it has almost trivially been shown to be consistent. But... it is well known that a theory strong enough to express the Gödel diagonalisation can never prove it's own consistency. For these theories, we will never have a model and cannot make statements about the meaning of any formula.

In these theories, it is wrong to talk about truth or falsity. We don't have a model giving meaning to that. We will never have a model.

That's not really a problem. For centuries, mathematicians had loosely combined derivation and truth and had mostly discussed them as one thing. Derivation and proof were seen as the important part of mathematics and formalization. You still have that.

Also, it is perfectly meaningful to derive results that say "if this theory is consistent and has a model, then...". Model theory has been doing that for nearly a century.

But people seem to want more. They want to talk about truth, as that is a form of meaning that holds a special place. They often go to great lengths to try to continue to assign truth and falsity. One common approach is to form truth predicates - predicates in the syntax that have the property that asserting the predicate on a formula corresponds to asserting the validity of the statement (that it is true in all models).

Note the switch - a truth predicate is syntactical. We still aren't talking about true or false here - the context of their use is still whether statements including the predicate "are derivable" or "obtain". Theories may have multiple models - most theories are not categorical just from things like Löwenheim–Skolem, so predicates cannot talk about truth. They can talk about validity - and that's really what is going on here - but even that is extremely problematic.

Incomplete theories cannot actually derive anything about validity on the total theory. And actually, this is where Tarski's theorem on nondefinability comes in and it is shown that such a predicate doesn't actually exist. So others keep at it with a hierarchy and reflection extensions of the base theory, seeking out some approximation of a fixed point for validity.

But this doesn't actually buy anything to do with insight into truth. It cannot. There is nothing you can do to reach truth because you cannot know if the theory is consistent or not and whether truth exists. And no attempts to reach beyond derivability actually give a predicate that can be used and say "this is true". The predicate is only useful to say "this is provable".

But there are already provability predicates, and that investigation is much more profitable. Truth predicates are a voiceless oracles. They do not help anyone make assertions on truth. They are simply reformulations of "if we knew that X was consistent, and we had some platonic sight that could see the truth values in all models, and we could collate the infinite possibilities and see the validities forever hidden, then this predicate applied to this class of statements would agree with those assertions that are valid". But if we had that supernatural sight, we could more easily just say "hey, that's true in that model - and that's false over there." Without that, we can use the predicate to say "truth is preserved in this derivation". Which doesn't add anything.

A truth predicate doesn't talk about truth. It is irrelevant to the point.

So...

So.. life goes on. My whole point in posting this answer was to illustrate that the initial question was making a common obvious assumption that is actually wrong. You should not talk about truth in the commonly used ambient theory - just talk about what is provable and you are fine. If you want to talk about truth, ensure you specify the ambient theory and it is one where such discussions are meaningful. Or talk about conditional models as model theorists do.

It may not be intellectually satisfying to some people. Clearly, as of writing this, my answer has received 3 downvotes and two upvotes, so it doesn't sit right with some anonymous readers of a math web site. But there is nothing controversial about the point. It has been known for almost 100 years and it is still a common mistake.

-
Your third paragraph, in its most precise form, is Tarski's undefinability theorem. But being aware of what it is, I still disagree with the preceding two paragraphs. Relative to another system, you can talk about truth, for example arithmetical truth is definable in ZFC (but not in PA). Truth in the set theoretic universe is not definable in ZFC, but so what? – Burak Jun 4 '14 at 23:02

The following is obviously false, but it is actually true, as shown in the Wikipedia article about Vitali sets.

The exists a countable collection $\left\{V_n\right\}$ of subsets of the unit circle such that:

1. Any two distinct $V_n$ are disjoint.
2. Any $V_n$ can be obtained from any other by a rotation.
3. The union of all the $V_n$ is the whole circle.

All the $V_n$ must have, from property two, the same "size" (for any reasonable definition of "size"), but if the above fact was true, the sum of their (equal) sizes would be the size of the circle (positive, but finite). But if the size was zero, the the sum should be zero, and if the size was positive, the sum should be infinite.

A consequence of this is that the following is false (although we all would like it to be true):

There exists a function $\mu$ that, given a bounded subset of $\mathbb{R}$, tells you its "size". Precisely:

1. If $A\subset\mathbb{R}$ is bounded, then $\mu\left(A\right) \in \left[0,\infty\right[$.
2. If $\left\{A_n\right\}_{n\in\mathbb{N}}$ is a sequence of bounded disjoint subsets of $\mathbb{R}$ (that is, $A_n \cap A_m=\emptyset$ whenever $n\neq m$) with bounded union (that is, $\bigcup A_n$ is bounded), then $\sum\mu\left(A_n\right)=\mu\left(\bigcup A_n\right)$.
3. If $A$ is bounded, $x$ is a real number, and we define $A+x=\left\{a+x:a\in A\right\}$, then $\mu\left(A+x\right) = \mu\left(A\right)$.
4. $\mu\left(\left[0,1\right]\right) = 1$

Indeed, rewriting the first fact exchanging the circle by the half-open interval $\left[0,1\right[$ and exchanging rotations for cyclic shifts "mod $1$", we realize that if $\mu$ satisfies the first three conditions above, then $\mu\left(A\right)=0$ for all $A$.

-

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.

-
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 '14 at 8:27
@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 '14 at 12:23
@JpMcCarthy ok I changed the format of my post to follow exactly how the question was asked. – DanZimm Jun 5 '14 at 13:03
+1 for you after all that! – Jp McCarthy Jun 5 '14 at 13:04
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 W Jun 11 '14 at 2:43

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

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

-
See this interesting answer by Dave L Renfro, and the posts it links to. – Andrés Caicedo Jun 6 '14 at 3:02

The following is a very well-known example, though probably slightly outside of the world of mathematics, rather physics. A great many people would 'intuitively' consider the following to be true:

The heavier the object, the faster it falls down.


In fact, the story goes that this was supposed to be common knowledge until Galileo Galilei disproved it (as the story goes, by dropping two balls of the tower in Pisa, which never happened though).

One of the first physics classes many people have (I'm talking primary school here) aims at showing this theorem is false, and actually everything falls with the same acceleration (ignoring resistance by air) regardless of weight.

-
Everything falls with the same acceleration not at the same speed... big difference. – Dan Jun 5 '14 at 8:29
Aristotle's claim was actually that objects fall with speed proportional to their weight, which is much more specific, and so much less obvious-seeming. It is incredible that Western civilization swallowed this whopper for as long as it did. – MJD Jun 5 '14 at 11:47
@MJD Universally people tend to believe things according to popularity rather than evidence or experience. This is as true today as it was in the past, and studies show that (counterintuitively) well-educated people are better at doing it than poorly-educated people. – AndrewC Jun 5 '14 at 16:22
But a gas balloon is lighter (relative to its volume) than eg a rock and it even rises instead of falling down. – Robert Jun 8 '14 at 20:57

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.

-
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 '14 at 16:09
How about the statement that 1/n > 0 which holds for any integer n but fails in infinity? – user102184 Jun 6 '14 at 16:37
@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 '14 at 16:40
@Hurkyl I agree and was trying to argue from Mark's position. – Jp McCarthy Jun 8 '14 at 22:17
This is Leibniz's Law of Continuity, right? – Tanner Swett Jun 9 '14 at 4:29

If you start thinking about the rigidity of thin shells in $\mathbb{R}^3$, you quickly encounter a slew of counterintuitive results.

For instance, it is obvious that a spherical shell is ($C^2$) rigid, and this is in fact true. A smooth, closed, compact surface with everywhere positive Gaussian curvature is likewise rigid. One might imagine these results generalize to

• Any closed surface;
• Any closed surface with positive Gaussian curvature everywhere but at finitely many points;
• Any surface with boundary with everywhere positive Gaussian curvature;

and all of these are false.

Moreover, after thinking about reflections, or poking and prodding at a ping-pong ball, it is intuitively obvious that a spherical shell is not $C^0$ rigid. But you can't really "see" any difference between a $C^1$ and a $C^2$ deformation of the sphere, so surely the sphere is $C^1$ rigid? Far from it -- given any arbitrary closed surface that is topologically a sphere, and distance $\epsilon$, it is possible to $C^1$-isometrically embed a sphere $\epsilon$-close to the target surface!

-

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!

-
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 '14 at 9:14
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. – Justin Jun 6 '14 at 2:51
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 '14 at 3:14

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

-
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 '14 at 2:30
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 '14 at 4:15
@TannerSwett More straightforwardly, consider Dedekind cuts of rational numbers. – Dustan Levenstein Jun 12 '14 at 3:56
@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 '14 at 4:12

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

-
Believing $i^i$ is well-defined is illusory. – Marc van Leeuwen Jun 6 '14 at 13:26
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 '14 at 13:40
$i^i\approx 0.207879$ – Anonymous Pi Jun 6 '14 at 14:24
@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 '14 at 3:04
@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 '14 at 16:50

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

-
+1 Love this one! A counterexample is $\frac{\sin{x^2}}{x}$. – Gamma Function Jun 5 '14 at 10:30
A sufficient condition for this to be true is for the second derivative to be bounded. – A.P. Jun 5 '14 at 12:38
@A.P. or even just that $\lim f'(x)$ exists. – Ant Jun 5 '14 at 17:50
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 '14 at 20:41
Ah, we have almost-duplicate responses. The interesting thing is that you can add the following restrictions to make it even more bizarre: $f$ is $C^\infty$, bounded, and monotonic. – Kaj Hansen Jun 6 '14 at 9:28

Here is a proposition, when put in layman's terms, an average person would claim to be true.

Every subset of real numbers has a measure.

How can this be false, when you mark a region, say in two dimensions, of course, it has an area?! Unless you constructed a Vitali set at some point, we tend to think that the concept of length/area/volume should extend to all possible subsets.

Here is another such false proposition.

Axiom of Determinacy

If we are playing a two-player infinite game where we create a real number in $[0,1]$ by choosing decimal digits in turns and one of us tries to land the resulting number in a pre-determined payoff set that is known to both of us and the other tries to avoid it, how could it be that there is a game where neither of us have a winning strategy? We both have complete information about the payoff set what numbers to avoid and what numbers to hit, one of us should be able to come up with a strategy. Well, unfortunately no.

Both of these propositions are inconsistent with the axiom of choice, with which you can construct the counterexamples that will not "nicely behave".

Fact: The latter proposition implies the former. (in ZF, with which AD is believed to be consistent).

-

## protected by Community♦Jun 6 '14 at 13:08

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.