Definition: limit of a sequence What is the purpose not to choose $|x_n-a|\leq\epsilon$ instead of $|x_n-a|<\epsilon$ in the definition of convergence? Is their a substancial difference (or a practical one)?
Thanks in advance.
 A: There is no difference. It is easy to prove that the statement


*

*For each $\epsilon>0$, there exists $N\in\mathbb N$ so that for each $n>N$, $|x_n-a|<\epsilon$


is equivalent to the statement


*

*For each $\epsilon>0$, there exists $N\in\mathbb N$ so that for each $n>N$, $|x_n-a|\leq\epsilon$.

A: It's clear that $|x_n-a|<\varepsilon$ implies $|x_n-a|\leqslant\varepsilon$. To see it conversely, note the condition "for all $\varepsilon>0$". Thus, since the statement "for all $\varepsilon>0$ ...  $|x_n-a|\leqslant\varepsilon$" holds for arbitrary $\varepsilon$, it holds also when we replace $\varepsilon$ by $\varepsilon/2$ throughout. That is, "for all $\varepsilon/2>0$ (i.e. for all $\varepsilon>0$), there is an integer $m$ such that $|x_n-a|\leqslant\varepsilon/2$ whenever $n>m$"; and from this it follows that "for all $\varepsilon>0$, there is an integer $m$ such that $|x_n-a|<\varepsilon$ whenever $n>m$". The reason for preferring $<$ to $\leqslant$ is simplicity and ease of application. 
A: For me, the main point of using open definitions $(\lt)$ rather than closed ones $(\le)$ is because similar definitions arise in dealing with continuity, and this is a concept naturally generalised to topological spaces via open sets.
So some people would want to use open definitions rather than closed ones for pedagogical reasons - because that avoids confusion later, and one simply uses open definitions for everything.
