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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!

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    $\begingroup$ 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. $\varepsilon = (x-y)/2$. $\endgroup$
    – StackTD
    Feb 26, 2015 at 10:22
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    $\begingroup$ @StackTD But proving something WLOG is just a proof by cases which gets swept under the rug. $\endgroup$
    – Git Gud
    Feb 26, 2015 at 10:29
  • $\begingroup$ 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") $\endgroup$
    – StackTD
    Feb 26, 2015 at 10:32
  • $\begingroup$ Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon&gt;0$ then $a\leq b$? $\endgroup$ May 27, 2023 at 15:20
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    $\begingroup$ Does this answer your question? Intuition about : $a = b \iff | a − b| &lt; \epsilon$, for every $\epsilon &gt; 0$ $\endgroup$
    – user1211588
    Sep 22, 2023 at 7:45

3 Answers 3

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$( \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$

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

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  • $\begingroup$ Note that it doesn't need to be a contradiction, contrapositive and direct proof work with the same idea. $\endgroup$
    – Git Gud
    Feb 26, 2015 at 10:29
  • $\begingroup$ Git Gud, I don't see a way to find a direct proof for the right implication. Can you explain it? $\endgroup$
    – Lin
    Apr 27, 2015 at 17:55
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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.

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