Proving $∀ε>0:x<y+\epsilon ⇒ x<y$

Question

If $x, y \in \Bbb R$ and $x<y+\epsilon$ for every $\epsilon >0$, then $x<y$.

Okay so I went about this by proving the contrapositive.

Proof: Let $x,y\in\Bbb R$ and let $\epsilon>0$. Suppose $x\ge y$. Take $\epsilon = x-y$. This implies that $x=y+\epsilon$, as needed.

Is this a valid proof or not?

Answer 1

The statement you want to prove is not true.

Take $x = y =0$. You have, for all $\varepsilon > 0$, $x < y + \varepsilon$ (because $\varepsilon > 0$), but of course you don't have $x < y$.

Answer 2

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.

The correct statement is “if, for all $\varepsilon>0$, $x<y+\varepsilon$, then $x\le y$”.

Now your proof works! Suppose $x>y$ (that is, “not $x\le y$”) and take $\varepsilon=x-y$; then $\varepsilon>0$ and $x=y+\varepsilon$.

Answer 3

1) The statement is false.

Counter example: Let $x = y$. Then $x = y < y+\epsilon$ for all $\epsilon > 0$ but $x \not < y$.

2) Your error is in claiming setting $\epsilon = x - y$. If $x =y$ then that $\epsilon \not > 0$.

Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.

3) But you have proven that $x>y$ is impossible. That would mean if $\epsilon = x - y > 0$ then $x = y + \epsilon$ which contradicts our hypothesis that $x < y + \epsilon$ for all $\epsilon>0$.

4) So a TRUE statement would be: if $x < y + \epsilon$ for all $\epsilon > 0$ then $x \le y$ and you successfully have proven that.