Many mathematical definitions almost used $\epsilon >0$ to define any mathematical notion, for example if we want to give definition to the convergence of sequence we say " $\forall \epsilon >0 , \cdots $ which means $\epsilon \in (0,\infty)$ than $\epsilon$ can take any large value since its belong to this domain $(0,\infty)$ , I have read here that late mathematician P. Erdős also used the term "epsilons" to refer to children (Hoffman 1998, p. 4). But how I can convince my self that epsilon must refer to small quantity since we assumed it greater than $0$ ?

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    $\begingroup$ There is no reason to assume $\varepsilon$ is small in a given mathematical context unless otherwise explicitly stated (that is, you might see $0<\varepsilon<1/100$ for someone that wants to emphasize the size of $\varepsilon$. In most contexts, it does seem to be the case, though. $\endgroup$ – Clayton Apr 13 '20 at 19:20
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    $\begingroup$ Most times when you're proving something, you look at $\epsilon$ being small, because the statement is trivially true when $\epsilon$ is large. $\endgroup$ – Paul Apr 13 '20 at 19:25
  • $\begingroup$ Just $\epsilon>0$ means $\epsilon$ is any positive quantity, but $\forall \epsilon>0$ implies even for very small $\epsilon$ $\endgroup$ – J. W. Tanner Apr 13 '20 at 19:26
  • $\begingroup$ The Greek letter “iota” is often used in English By non-mathematicians to mean a small quantity. “I don’t care one iota that...” for example. It is in English dictionaries this way. $\endgroup$ – Thomas Andrews Apr 13 '20 at 19:38

In a typical analysis context, the statement to be proved is of the form

For all $\epsilon > 0$ there exists ... such that $ \ldots < \epsilon$

There is no need for $\epsilon$ to be small, but if the statement is true for some $\epsilon_0$, then it is automatically true for all larger $\epsilon$. So the hard part is to deal with small $\epsilon$.


You cannot, and really must not. If you are given that $\varepsilon>0$, that's really all you can assume about it. For instance, consider the following start of an argument:

Let $\varepsilon>0$ be given. Let $\delta:=\varepsilon^2$. Since $\delta<\varepsilon$,...

Well, no. Even though this is a delta-epsilon proof and your attention is focused on small values of epsilon, the statement is not true if $\varepsilon>1$. In cases like this, you must take $\delta:=\min\{.1,\varepsilon^2\}$ or something similar to have a valid argument.


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