Definition: limit of a sequence

Question

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.

Answer 1

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.

Answer 2

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 $n>N$, $|x_n-a|\leq\epsilon$.

Answer 3

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.