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In the definition of the limit of a sequence, why can't we replace $ \lt \epsilon $ with $ \le \epsilon $?

Here is a proof I thought for the equivalence of the two definitions, is there anything wrong with it?

Let $ \{a_n\}_{n=1}^\infty \, $ be a sequence in $ \mathbb{R} $.

Assume $ \forall \epsilon \gt 0 \, \exists N : \forall n \gt N, \left|a_n - L\right| \lt \epsilon. $ Then, of course $ \left| a_n - L \right| \le \epsilon, \forall \epsilon \gt 0 $, because $ \lt \subseteq \le $.

Assume $ \forall \epsilon \gt 0 \, \exists N : \left|a_n - L\right| \le \epsilon $. Given $ \epsilon \gt 0 $, $ \exists N : \forall n \gt N,\left|a_n - L\right| \le \frac{\epsilon}{2} \lt \epsilon.$

talopl
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    Yes that's right. And you can play the same game with $N$. – peek-a-boo Jul 23 '22 at 16:32
  • @peek-a-boo thrn why, when negating the definition, we say $ \ge \epsilon $ and not just $ \gt \epsilon $? – talopl Jul 23 '22 at 17:03
  • @talopl: When negating the definition, it is entirely acceptable to write $>\epsilon$ rather than $\ge \epsilon$. Same difference. – Joe Jul 23 '22 at 17:05
  • These statements are equivalent, so their negations are also equivalent. But until you have shown the equivalence (like you just did) you should of course not mix up strict vs weak inequalities. – peek-a-boo Jul 23 '22 at 17:15
  • oh and I just remembered: here is a link where I write down a list of several equivalent (though some more useful in practice than others) conditions for limits of functions (you can easily adapt it to sequences). Since all those statements are logically equivalent, so are their negations; but again, don't mix up statements before proving their equivalence. – peek-a-boo Jul 24 '22 at 00:56

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