# Analysis convergence definition

When proving convergence the part where $|a_n-a|<\varepsilon$, is it necessary to use a strict inequality? Could I use $\leq$ instead? Thanks.

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HINT If $0 \leq x \leq \epsilon$, then $x$ is strictly less than $2 \epsilon$. – Srivatsan Sep 15 '11 at 14:11
Aha, thank you! – leonard Sep 15 '11 at 14:13

Yes we can use $\leq$ instead, but we usually don't because we set an arbitrary $\epsilon>0$. So if some quantity $M$ has the property that $M \leq \epsilon$ for any $\epsilon > 0$, then certainly if we replace the $\epsilon$ by $\epsilon/2$, we see that we can also say that $M \leq \epsilon/2 < \epsilon$ for any $\epsilon > 0$. In the other way, if $M< \epsilon$ for any $\epsilon > 0$, then certainly $M \leq \epsilon$ as well. Thus, allowing the equality is somewhat redundant and because we would rather be "cleaner" in our definitions, we exclude it.
+1 I like the point about arbitrary $\epsilon$. But I want to point out that certain things do change when $\epsilon$ is replaced by $\epsilon/2$. For e.g., as in Levon's answer, the $N(\epsilon)$ threshold in the definition of limits will increase to $N(\epsilon/2)$. However, I agree that this shouldn't be a big deal in most cases. – Srivatsan Sep 15 '11 at 14:20
No, it isn't. If for every $\epsilon$, you can find $N(\epsilon)$ such that $\forall n \ge N(\epsilon), |a_n - a| \le \epsilon$, then just pick an $\epsilon' < \epsilon$ and $N(\epsilon')$ will do the job.