Given the following conditions:
$x \in \Bbb R$ and $y\in (0,1)$
I was asked to prove that
inf $ |x-y |=0 $
My Attempt:
By the elementary properties of the modulus function , we know that $ 0 \leq |x-y|$ $\forall x \in X , y \in (0, 1)$. So clearly it bounded below by $0$.
Suppose there exist a $\epsilon > 0$ such that $ \epsilon \leq |x-y|$ $\forall x \in X , y \in (0, 1)$. However letting $x= \epsilon$ and $y= \frac{\epsilon}{\epsilon +1 } $ then $|x-y|< \epsilon $ So the exist no such $\epsilon>0$
Is this correct and sufficient enough?
Edit: Well the proof suggest is a lot simpler and easier. I still would like to know whether what I provide constituents a proof.
$\endgroup$ 42 Answers
$\begingroup$Your proof is valid. However, you would have to say why $|x-y|<\epsilon$.
$$|x-y|= \left |\epsilon - \dfrac{\epsilon}{\epsilon + 1}\right | = \left | \dfrac{\epsilon^2}{\epsilon + 1}\right | < \epsilon$$
Since $\epsilon > 0$.
However, you can proof it much easier:
You can say that clearly $\inf|x−y|\geq 0$, and taking for example $x=y=\frac{1}{2}$, then $|x−y|=0$, so $\inf|x−y|=0$.
$\endgroup$ $\begingroup$Your proof is fine. In this case, it can be made a little simpler.
Since $|z| \ge 0$ for all $z$, we have a lower bound of $0$.
Now choose $x=y = {1 \over 2}$. Then $|x-y| = 0$, hence the infimum is attained.
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