# '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!

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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

## 65 Answers

Someone else mentioned "there are more rational numbers than integers". Along the same lines, I had a hard time accepting that

There are more integers than there are real numbers between 0 and 1

is false. I mean, I get it now, but it seemed intuitively very wrong to me before I studied transfinite numbers.

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An analytic function with compact support vanishes identically.

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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

For me a nice example of all of "evidence" suggesting it was true is

$$\pi(x) < \operatorname{li}(x)$$

until Skewes showed that $\pi(x) - \operatorname{li}(x)$ changes sign infinitely often

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The question asks for obvious but false statements. Are you seriously claiming that $\pi(x) < \operatorname{li}(x)$ is obvious? – MJD Jun 5 '14 at 21:32
hard to tell whether that is supposed to be obvious... What are $\pi(x)$ and $\operatorname{li}(x)$? – example Jun 5 '14 at 22:03

Lebesgue once stated that the projections of Borel sets in $\mathbb R ^2$ on to one of its axes are also Borel sets. This fact is actually false, the realization of which is attributed to the short-lived mathematician Mikhail Yakovlevich Suslin.

Unfortunately it is very difficult to find a counterexample. The only one I've ever seen takes a result in descriptive set theory about $\mathbb N ^ {\mathbb N}$ and uses the fact that it's homeomorphic to $\mathbb R$.

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I hope you are misremembering or misunderstanding the arguments you have seen, because $\mathbb N^{\mathbb N}$ is not homeomorphic to $\mathbb R$. For example, one is connected and the other one is not. One is $\sigma$-compact and the other one is not. – Andrés E. Caicedo Jun 6 '14 at 2:54
That depends on the topology, but it might be that the argument I'm thinking of used the space of all sequences (both finite and infinite) of natural numbers or developed the homeomorphism with the exclusion of $\mathbb Q$. – jjfunk Jun 6 '14 at 4:04
"That depends on the topology". Obviously. "The space of all sequences (both finite and infinite) of natural numbers." Perhaps, but that is not $\mathbb N^{\mathbb N}$. "With the exclusion of $\mathbb Q$." But then that is not $\mathbb R$. Anyway, a word or two clarifying the imprecision would not be superfluous. – Andrés E. Caicedo Jun 6 '14 at 4:51
@jjfunk: Is it possible that the map you are talking about is the homeomorphism between the Baire space and the irrationals given by infinite continued fractions? – Burak Apr 9 at 0:58

The following statement is wrong:

The inner angles of a triangle always sum to 180 degrees.


While it sounds plausible that the sum of the angles is a constant, it is actually a property of the space. In Euclidean space the inner angles of a triangle always sum to 180 degrees.

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But it is fairly obvious and it is true. – bjb568 Jun 9 '14 at 1:48
@bjb568 I think Robert may be thinking of non-planar triangles. – MJD Jun 9 '14 at 5:16
@MJD Well, that makes more sense, but then it isn't obvious… – bjb568 Jun 9 '14 at 6:05
It doesn't seem obvious to me that the angles of a plane triangle always have the same sum. In fact, it seems astounding, and I'm sure you could find other places on this web site where other people said it seemed astounding. – MJD Jun 9 '14 at 14:14
For example, this highly-voted answer begins “I found it completely amazing that the angles in a triangle always added up to 180 degrees”. – MJD Jun 9 '14 at 21:22

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