# How I convince my self that if we say $\epsilon >0$ we must refer to small quantity?

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 $$0$$ ?

• 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. – Clayton Apr 13 '20 at 19:20
• Most times when you're proving something, you look at $\epsilon$ being small, because the statement is trivially true when $\epsilon$ is large. – Paul Apr 13 '20 at 19:25
• Just $\epsilon>0$ means $\epsilon$ is any positive quantity, but $\forall \epsilon>0$ implies even for very small $\epsilon$ – J. W. Tanner Apr 13 '20 at 19:26
• 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. – Thomas Andrews Apr 13 '20 at 19:38

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.