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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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4 Answers 4

up vote 5 down vote accepted

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 $x_n<k$. 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. –  ftiaronsem Jan 4 '11 at 16:53
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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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Thank you for your answer. However I am still not 100% convinced. Why can you be so sure that this not changing anything in the many possible proofs? After all you are adding an element which is strictly greater than zero. The whole point of $\epsilon > 0$ is $\epsilon$ approaching 0. However in case of adding a $\eta>0$, i would approach that $\eta$ wouldn't I? –  ftiaronsem Jan 4 '11 at 14:30
    
@ftiaronsem, the whole point of $\varepsilon>0$ is that it is arbitrary, not that it approaches zero. Only due to arbitrary nature the strict inequality versus non-strict inequality does not matter. Note that in all proofs you have for each $\varepsilon$, not for some small $\varepsilon$. Usually we concern ourselves with small $\varepsilon$ since for large $\varepsilon$ the proofs are very easy. –  mpiktas Jan 4 '11 at 14:51
    
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. –  ftiaronsem Jan 4 '11 at 16:51
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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. Thanks –  ftiaronsem Jan 4 '11 at 16:56
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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 you –  ftiaronsem Jan 4 '11 at 16:55
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