Prove that if $| x - y | < \varepsilon$ then $x = y$

Question

Prove that if $x, y \in \mathbb{R}$ and $| x - y | < \varepsilon$ for all $\varepsilon \in \mathbb{R}$ where $\varepsilon > 0$, then $x = y$.

I have gotten this so far. Any suggestions would be helpful:

$$| x - y | < \varepsilon \implies |x| < |y| + \varepsilon$$

If we assume that $x \neq y$, then $x < \varepsilon$ or $x > \varepsilon$

Therefore, $x > y$, not $x = y$.

First, not $\epsilon \in \Bbb R$, just $\epsilon > 0$. Now, assume not and choose $\epsilon = |x-y|/2$. What happens? — Ivo Terek, Jan 24 '22 at 01:04
An interesting corollary of this is that $.999999999999\ldots \ = \ 1$. — Mike, Jan 24 '22 at 01:11

user829347 · Accepted Answer · 2022-01-24T01:24:09.900

2

Let $x,y\in\mathbb{R}$ with $|x-y|<\epsilon$ for all $\epsilon>0$. Suppose (for a contradiction) that $x\neq y$. Then $|x-y|>0$ so that $|x-y|<|x-y|$, a clear contradiction. It then follows that $x=y$ and we are done.

edited Jan 24 '22 at 01:24

answered Jan 24 '22 at 01:09

user829347

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Prove that if $| x - y | < \varepsilon$ then $x = y$

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