- #1
jaejoon89
- 195
- 0
In class we defined convergence as
\forall\varepsilon>0\;\;\;\exists\mathrm{N}\epsilon\mathbf{N}\;\;\;\forall\mathrm{n}\geq\mathrm{N}\;\;\;\left|a_{n} \right|<\varepsilon
So if a sequence {a_n} of real numbers converge to 0 if for every ε > 0 there is N s.t. |x_n| < ε for n ≥ N...
Is the natural numbers implicit in the statement? or do you have to change it somehow for the reals? How would you use the symbolic logic to negate it? (i.e. if it did not converge to 0)?
So... to understand better... when it is converging to a particular number, is that whatever the epsilon is greater than?
\forall\varepsilon>0\;\;\;\exists\mathrm{N}\epsilon\mathbf{N}\;\;\;\forall\mathrm{n}\geq\mathrm{N}\;\;\;\left|a_{n} \right|<\varepsilon
So if a sequence {a_n} of real numbers converge to 0 if for every ε > 0 there is N s.t. |x_n| < ε for n ≥ N...
Is the natural numbers implicit in the statement? or do you have to change it somehow for the reals? How would you use the symbolic logic to negate it? (i.e. if it did not converge to 0)?
So... to understand better... when it is converging to a particular number, is that whatever the epsilon is greater than?