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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.


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The Banach-Tarski paradox is a good one. – mjqxxxx Jun 4 '14 at 18:08
related:… – 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

64 Answers 64

If a propositional calculus A contains all theorems of propositional calculus B under detachment and uniform substitution for propositional variables, but B does not contain all of the theorems of A, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B. Or one might more haphazardly say "if propositional calculus A is bigger than propositional calculus B, then one of the shortest single axioms of A is longer than any of the shortest single axioms of B."

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Good luck finding a layman whose eyes don't glaze at the term "propositional calculus", let alone one who finds something here obvious. – rumtscho Jun 13 '14 at 17:46

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

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 Caicedo Jun 6 '14 at 2:54
"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 Caicedo Jun 6 '14 at 4:51

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

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