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As an exercise I was doing a proof about equicontinuity of a certain function. I noticed that I am always choosing the limits in a way that I finally get:

$\text{Expression} < \epsilon$

However it wouldn't hurt showing that

$\text{Expression} \leq \epsilon$

would it, since $\epsilon$ is getting infinitesimally small? I have been doing this type of $\epsilon$ proofs quite some time now, but never asked myself that question. Am I allowed to write $\text{Expression} \leq \epsilon$? If so, when?

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Suppose for every $\epsilon >0$, there is an $N$ such that $n>N$ implies $x_n\leq\epsilon$. Let $k>0$. Then $\frac{k}{2}>0$, so there is an $M$ such that $n>M$ implies $x_n\leq\frac{k}{2}$ which implies that $x_n. This corresponds to the original definition of continuity, doesn't it?

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    thank you very much. This is actually very clear, without having to argue about anything. Since this is the answer I understand best, I accept it.2011-01-04
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Since $\varepsilon>0$ is arbitrary it does not matter. If you have a non strict inequality for each $\varepsilon>0$ you can get strict inequality by adding arbitrary $\eta>0$.

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    ok, thanks for your answe. I hadn't fully comprehended that arbitrary point before. (Comes from always writing lim ^^) But as far as I can judge you are right. Thanks for your answer.2011-01-04
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You are right, it doesn't matter.

Given an arbitrary $\epsilon > 0$, you can make your expression strictly less than $\epsilon$ by choosing your $\delta$ such that the expression is less than or equal to $\epsilon/2$, for example. (This is possible since you know that you can choose $\delta$ so that your expression is less than or equal to any positive number that you wish.)

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    Thank you too, for your answer, you all have helped me a lot in understandig that. Thanks2011-01-04
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There are already some good answers here, so I won't bother repeating them. However, I'd like to add a cautionary note that there are instances where using the wrong inequality gives nonsense — for example, it is essential in the statement of the Banach fixed point theorem that the Lipschitz constant is strictly less than one. If I remember correctly, even $\dfrac{\| f(x) - f(y) \|}{\| x - y \|} < 1$ for all $x \ne y$ is not good enough; we must have $\sup \left\{ \dfrac{\| f(x) - f(y) \|}{\| x - y \|} : x \ne y \right\} < 1$.

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    ahh, thought that it couldn't be that simple in a general case. However since I don't know anything about Banach, I'll for the moment believe math being so nice as to allowe me using both the strict and the non strict variant ;-). But nevertheless, thank you2011-01-04