# Famous vacuously true statements

I am interested to know other examples vacuously true statements that are non-trivial. My starting example is Turan's result in regards to the Riemann hypothesis, which states

Suppose that for each $N \in \mathbb{N}$ the function $\displaystyle \sum_{n=1}^N n^{-s}$ has no zeroes for $\mathfrak{R}(s) > 1$. Then the function $$\displaystyle T(x) = \sum_{n \leq x} \frac{\lambda(n)}{n}$$ is non-negative for $x \geq 0$. In particular, this would imply the Riemann hypothesis.

Here $\lambda(n) = \lambda(p_1^{a_1} \cdots p_r^{a_r}) = (-1)^r$ is the Liouville function.

The interesting thing about this statement is that both the hypothesis and the consequence can be proven false independently. In particular, Montgomery showed in 1983 that for all sufficiently large $N$ the above sums have zeroes with real parts larger than one, and Haselgrove showed in 1958 that $T(x)$ is negative for infinitely many values of $x$. Boreweint et.al found the smallest such $x$ in 2008.

I find this result fascinating because it relates to a well-known conjecture, and both the hypothesis and consequence were plausible. Are there any other mathematical facts of this nature, perhaps in other areas?

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If you stop one of the standard proofs of the irrationality of $\sqrt2$ halfway through, you have a proof (of the vacuously true statement) that if $\sqrt2=a/b$ then $a$ is even. The hypothesis, at least, was considered more than plausible by Pythagoras. –  Gerry Myerson Aug 2 '13 at 0:50

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While vacuously true, your statement does not fulfill the criterion (yet) of "famous" :-) but +1 for the humor. –  Suvrit Aug 1 '13 at 23:13
But is it really true? –  François G. Dorais Aug 1 '13 at 23:13
It is definitely not “non-trivial”. –  The User Aug 1 '13 at 23:28
I reformatted this answer in accordance with this meta.mathoverfow thread: meta.mathoverflow.net/questions/462/… I hope its meaning was not changed in the process. :) –  Ricardo Andrade Aug 1 '13 at 23:28
The fact that this was edited just adds to the silliness :). –  Daniel Rust Aug 2 '13 at 13:43

Every prime for which the first case of FLT fails is a Wieferich prime.

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By making various reductions, Brown was able to narrow the invariant subspace question for $P^2(\mu)$ to the case where $P^\infty(\mu)$ is just $H^\infty$ of the unit disk, and $P^2(\mu)$ admits no bounded point evaluations. He showed in that case that the evaluation functionals on $H^\infty$ at the points of $\mathbb D$ have spatial representations of a certain simple kind in $P^2(\mu)$, from which the existence of nontrivial invariant subspaces follows immediately... It was quickly realized that Brown’s basic ideas, including his method for constructing spatial representations, apply far beyond the realm of subnormal operators.
"Thomson’s result shows, paradoxically, that the situation in which Brown originally applied his technique ($P^\infty(\mu)= H^{\infty}$, yet $P^2(\mu)$ has no bounded point evaluations) is in fact void. Even theorems about the empty set, it seems, can contain interesting ideas."