What happens to inequalities when $\epsilon$ $\to$ $0$?

Question

I recently came across a proof in topology. There was an inequality in which $A$$<$$B$ + $\epsilon$ (strictly less than),But when $\epsilon$ $\to$ $0$ , then it was inferred that $A$<=$B$.

Here $B$ is the infimum of a sequence and we are using the basic definition of infimum.(There is always a number between infimum and infimum + $\epsilon$)

How is this possible?

Answer 1

If $A<B+\epsilon$ for all $\epsilon > 0$, then we can conclude that $A\leq B$. In fact, this is the most you can conclude. It is in fact rather easy to prove the following statement:

$A$ is less than or equal to $B$ if and only if, for all $\epsilon>0$, $A$ is less than $B+\epsilon$. In symbols, this is: $$(A\leq B) \iff (\forall \epsilon > 0: A<B+\epsilon)$$

We cannot conclude that $A<B$, and we cannot conclude $A=B$.

Answer 2

I think you're misguided.

In general, for any $a,b\in\Bbb R$ such that $a<b+\varepsilon$ for all $\varepsilon>0$, we can deduce only that $a\le b$. Indeed, to prove the claim we can assume that $a>b$ and take $\varepsilon=\frac {a-b}2$ to get a contradiction.

Since I've seen you used the tag measure theory I'll assume further about how the proof you've seen conclude $a=b$ . Normally, when you want to prove that $a=b$ you'd need to show that $a\ge b$ and $a\le b$. In measure theory, it is often the case that the $a\ge b$ direction is obvious or easy to show, hence you only need to do the other half. Proving $a\le b$ directly could still be difficult, hence we instead prove that $a<b+\varepsilon$ for all $\varepsilon>0$ and then deduce that $a\le b$.