Theorem – x <= e + y – Math-reference.net

Contents
1. Theorem
2. Proof

PAGE NEEDS CHANGING DUE TO ERROR IN THE PROOF

Theorem

Let $x\leq \epsilon + y$ be true for all $x,y,\epsilon \in \mathbb{R}$ and $\epsilon > 0$ . Then, $x \leq y$ .

Proof

We will prove the theorem by contradiction. The statement is that

$\forall \epsilon >0, ~x \leq \epsilon + y \implies x \leq y$ .

Thus, the negation of this statement is that

$\forall \epsilon >0, ~(x \leq \epsilon + y) \wedge (x > y)$ .

So that $x-y>0$ and $\forall \epsilon>0, x-y \leq \epsilon$ . Restating these equations clearly, we have

$0<x-y$ , and therefore $0<kx-ky$ where $k>0$ .
$x-y \leq \epsilon$

Adding these two equations together gives

$x-y \leq \epsilon +kx -ky \Leftrightarrow (k+1)x-(k+1)y = (k+1)(x-y)\leq \epsilon$ . Therefore, since $\epsilon > 0$ , and $(k+1)>0$ , it follows that $x-y\leq 0 \implies x \leq y$ . Thus we have the desired contradiction, and so it must be the case that the theorem holds. $\square$