TITLE: Why is the question valid? How can I make sense of it?
QUESTION [1 upvotes]: I am solving exercise from Terence Tao's Analysis- Vol 1. One of the exercise was-
Let $x$ and $y$ be real numbers. Show that $x\leq y+\varepsilon$ for
all real numbers $\varepsilon > 0$ if and only if $x\leq y$. Show that $|x- y|\leq \varepsilon$ for all
real numbers $\varepsilon > 0$ if and only if $x = y$.
How can $x\leq y+\varepsilon$ be true when $x\leq y$, wont it be $x<y+\varepsilon$? And the second part too, if $x=y$ won't it be $|x-y|=0$?
What am I not understanding here?
Is it because $x$ and $y$ are real numbers? because he defines real numbers as-
A real number is defined to
be an object of the form $\lim_{n\to \infty} a_n$, where $(a_n)^\infty_{n=1}$ is a Cauchy
sequence of rational numbers.

REPLY [2 votes]: If $x \leq y$, then certainly $x \leq y + \epsilon$ for every $\epsilon > 0$.
If $x > y$, then set $\delta = x - y > 0$. Then if $\epsilon = \delta/2 > 0$, say, we see that $$y + \epsilon = (x - 2 \epsilon) + \epsilon = x - \epsilon < x.$$
Therefore we proved

$x \leq y$ if and only if $x \leq y + \epsilon$ for all $\epsilon > 0$.

Let $x = |u - v|$ and $y = 0$. Then we also see as a corollary:

$|u - v| \leq 0$ if and only if $|u - v| \leq \epsilon$ for all $\epsilon > 0$.

Since $|u -v| \geq 0$,  we see that $|u - v| \leq 0$ is equivalent to $|u - v| = 0$ or equivalently $u = v$. Therefore, we can equivalently state:

$u =v$ if and only if $|u - v| \leq \epsilon$ for all $\epsilon > 0$.
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