Why the definition of limit have a strict inequality rather than not?

Question

My question is why the definition of limit of sequence is for $\epsilon >0$, there exist $N$ st for $n\ge N$ that $|x_n-L|<\epsilon$ instead of $|x_n-L|\le \epsilon$

Answer 1

The two conditions are equivalent if you require them to be true for all $\epsilon>0$: if $|x_n-L|<\epsilon$ eventually, it's certainly true that $|x_n-L|\leq \epsilon$ eventually. And vice versa, if for any $\epsilon>0$ it's true that $|x_n-L|\leq \epsilon$, then it must also be true that $|x_n-L|$ is eventually at most $\epsilon/2$, which is strictly smaller than $\epsilon$.

So why is $|x_n-L|<\epsilon$ standard and not $|x_n-L|\leq\epsilon$? Because once you leave the familiar setting of metric spaces, the former generalizes properly but the latter does not. Topologies are defined with open sets, and a sequence $x_n$ converges to $L$ if $x_n$ eventually stays inside any open set containing $L$ (called "neighborhoods" of L). In a metric space like $\mathbb{R}$, that's equivalent to saying that for any $\epsilon>0$, $x_n$ has to eventually stay within the open ball of radius $\epsilon$ around $L$. In contrast, the closed balls of radius $\epsilon$ generalize to something like "closed neighborhoods", which is a less primitive notion combining the ideas of open and closed sets.

Answer 2

The definitions are equivalent; the quantifier absorbs the distinction. Here's a proof. It's a matter of logic that the strict inequality implies the soft one. Conversely, if the latter definition holds, for any $\epsilon >0,$ choose some positive $\epsilon ' < \epsilon $ and $N$ such that $|x_n - L| \le \epsilon '$ for $n\ge N.$ Then $|x_n - L| <\epsilon .$

Answer 3

Both could be taken as a definition, because the $\epsilon$ can be arbitrarily small.