$\exists \implies \forall$

Question

I want to see some example theorem, when existence implies universality, so $\exists \implies \forall$ is true.

I think matematical induction is a related technique, but I just don't see that induction covers the whole topic. On the other hand, there are some situation, when existence implies that the statemant is true for "some" element. For example in the fundamental theorem of algebra the existence of one root is equivalent of the existence of exactly n roots.

So I expect some example. Thanks.

Answer 1

This's an example from linear algebra:

Let $f\colon\Bbb R^n\rightarrow R^n$ a linear transformation.

these statements are equivalent

• $f$ is orthogonal
• $f$ maps an orthonormal basis to an orthonormal basis
• $f$ maps every orthonormal basis to an orthonormal basis

Answer 2

$\exists n \in \mathbb{Z} ( 2n=1 ) \Rightarrow \forall n \in \mathbb{Z} ( 2n=1 )$

Answer 3

For a homogeneous DE $y'' + p(x)y' + q(x)y=0$ on some interval $J$, Let $y_1,y_2$ be two solutions. We have the Wronskian $W(y_1,y_2)(x) = y_1y_2' - y_2y_1'$. Then it is a result of Abel's identity that if $\exists x \in J $ such that $W(y_1,y_2)(x) =0 \implies \forall x\in J$ we have $W(y_1,y_2)(x) =0$.

Answer 4

Here's a simple but widely used theorem.

For any polynomial $p$ of degree $n$:

  $\exists( n+1 \text{ distinct values for }x )( p(x) = 0 ) \Rightarrow \forall x( p(x) = 0 )$