Prove : $|x - y| \leqslant \epsilon$ $\forall \epsilon > 0$ iff $x = y$

Question

I have to do this exercise for my math study, and I'm having trouble with doing the second part of it.

Let $x, y, \epsilon \in \mathbb{R}$ and $\epsilon > 0$. Prove: $|x - y| \leqslant \epsilon$ $\forall \epsilon > 0 \Leftrightarrow x = y$

I think I have the left implication $\Leftarrow$:

Assume $x = y \Rightarrow x - y = 0 \Rightarrow|x - y| = 0 = |0| \Rightarrow|x - y| = 0 < \epsilon \Rightarrow |x - y| \leqslant \epsilon$ $\forall \epsilon > 0 \in \mathbb{R}$

Is this argument correct?

For the right implication, I do only have an idea of how to prove it. I think I have to assume first that $x > y$ and then $x < y$ and get a contradiction form both.

Could you please explain me the right implication and tell me if my left implication is correct?

Thanks in advance!

Instead of looking at two cases, you could assume WLOG that $x>y$, that means $x-y>0$. Choose $\varepsilon$ positive, but smaller than $x-y$, e.g. $\varepsilon = (x-y)/2$. — StackTD, Feb 26 '15 at 10:22
@StackTD But proving something WLOG is just a proof by cases which gets swept under the rug. — Git Gud, Feb 26 '15 at 10:29
Of course... What I means was: you can avoid having to show both cases explicitly, if you use the symmetry/WLOG-argument (as a response to "I have to assume first that $x>y$ and then $x<y$ and get a contradiction form both") — StackTD, Feb 26 '15 at 10:32
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, May 27 '23 at 15:20
Does this answer your question? Intuition about : $a = b \iff | a − b| < \epsilon$, for every $\epsilon > 0$ — , Sep 22 '23 at 07:45

score 8 · Answer 1 · answered Apr 24 '15 at 04:14

$( \Rightarrow ) $ I will prove by contraposition. Our affirmative states that:

$$ \forall \epsilon > 0 \; ( | x - y | < \epsilon ) \implies x = y $$

The contrapositive is

$$ \lnot ( x = y ) \implies \lnot \forall \epsilon > 0 \; ( | x - y | < \epsilon ) \\ x \neq y \implies \exists \epsilon > 0 \; \lnot ( | x - y | < \epsilon ) \\ x \neq y \implies \exists \epsilon > 0 \; ( | x - y | \ge \epsilon ) $$

Therefore, if we prove the contrapositive, we have proved the affirmative. We need to find an $ \epsilon \le | x - y |$.

Let $ \epsilon = \frac{|x-y|}{2} $, since $ x \neq y $, without loss of generality, let $ x > y $, that implies $ x - y > 0 $, since we have $ | x - y | = x - y > 0 \implies | x - y | > 0 $.

Therefore $ | x - y | \ge \epsilon = \frac{|x-y|}{2} \implies 2 |x-y| \ge |x-y| $. That is true for all positive values, since we have proved our contraposition our initial hypothesis is valid.

$ ( \Leftarrow ) $ Direct proof. Let $ x = y \implies x - y = 0 \implies | x - y | = | 0 | < \epsilon \;\;\; \square$

score 7 · Accepted Answer · answered Feb 26 '15 at 10:22

7

Your left implication is OK. For the right implication, if $x\neq y$, just take $\epsilon=\frac{|x-y|}{2}$ to obtain a contradiction.

answered Feb 26 '15 at 10:22

Olivier Oloa

122,789

Note that it doesn't need to be a contradiction, contrapositive and direct proof work with the same idea. – Git Gud Feb 26 '15 at 10:29
Git Gud, I don't see a way to find a direct proof for the right implication. Can you explain it? – Lin Apr 27 '15 at 17:55

score 5 · Answer 3 · answered Feb 27 '15 at 12:12

5

Let $x\neq y$. Then show that there exists an $\epsilon>0$ such that $|x-y|>\epsilon$. (For example $\epsilon =\frac{|x-y|}{2}$, which satisfies $\epsilon>0$ because of our starting assumption.) The result follows from contraposition.

answered Feb 27 '15 at 12:12

Nesta

343

Prove : $|x - y| \leqslant \epsilon$ $\forall \epsilon > 0$ iff $x = y$

3 Answers3

Linked

Related