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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 $|x_n-L|\le \epsilon$

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up vote 7 down vote accepted

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.

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but why they don't use close neighborhood instead of open? – Mathematics Oct 11 '12 at 16:19
Because "closed neighborhood" means "a closed set containing a smaller open set containing the point", so you might as well use the simpler "open neighborhood" meaning "an open set containing the point." If you try to define "closed neighborhood" to mean "a closed set containing the point", you get an entirely wrong notion of convergence, because in many topological spaces, including metric spaces, the point itself is a closed set, and you'd be requiring the sequence to eventually be constant. – Owen Biesel Oct 11 '12 at 16:26

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 $|x_n - L| <\epsilon .$

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then why don't use the former one? It is more general – Mathematics Oct 11 '12 at 15:55
No - they're equally general. That's what this argument shows. – user17794 Oct 11 '12 at 16:32

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

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