Shlomi93
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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.
thanks in advance.
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?