The discussion revolves around the requirement that epsilon must be greater than zero in the limit definition of a sequence. Participants explore the implications of this condition and its relevance to convergence in sequences, touching on both theoretical and practical aspects.
Participants express differing views on the necessity of epsilon being strictly greater than zero, with some arguing for its importance in maintaining the definition of convergence, while others suggest that it may not be strictly necessary in all contexts. The discussion remains unresolved regarding the broader implications of allowing epsilon to be zero.
Participants note that allowing epsilon to be zero leads to a restriction where only constant sequences would converge, highlighting the limitations of this approach in the general definition of limits.
It doesn't really matter with sequences of real numbers. You could take both as you can always find another epsilon that is slightly smaller. In general, however, one speaks of open neighborhoods around the limit point as they are the defining element of general (topological) spaces. And open translates to smaller than, whereas smaller or equal includes the boundaries, and as such are closed sets. So the restriction to smaller than is somehow simply consequent, even if not needed (and it's available on the keyboard).Shlomi93 said:in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?
thanks in advance.
Yes, you're right. I confused it with the condition ##\,\vert \,a_n -L\,\vert \, < \varepsilon## where you could take ##\leq## instead.pwsnafu said:Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
So under this definition the only sequences that converge are those that are eventually constant.
fresh_42 said:Yes, you're right. I confused it with the condition ##\,\vert \,a_n -L\,\vert \, < \varepsilon## where you could take ##\leq## instead.
Of course ##\varepsilon = 0## would make no sense as there would be only constant sequences left over.
pwsnafu said:Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
So under this definition the only sequences that converge are those that are eventually constant.
Consider ##a_n = \frac {n - 1} n, n \ge 1##. It's easy to show that ##\lim_{n \to \infty}a_n = 1##. However, if ##\epsilon = 0##, it's not possible to find a specific number N for which ##|a_n - 1| = 0##, for all ##n \ge N##.Shlomi93 said:in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?