Conditions for a real number not to be a limit point

[monkeybird]

Homework Statement

Given a sequence <x_n>_n of real numbers.

Give the conditions for a real number a not to be a limit point of the sequence. (lim x_n not equal to a.)

The Attempt at a Solution

I'm really not sure if this is the whole answer or if it's only a part of it:

For all e>0 there exists an n that belongs to the real numbers s.t. |x_n - a| >= e.


Is there more to this or do I have it correct?

 

[VeeEight]

Consider a closed interval of the form [a,b] where a,b are some reals. Every point in the set is a limit point because for some x in [a,b], the sequence x + 1/n converges to x as n approaches infinite.

 

[HallsofIvy]

Homework Statement

Given a sequence <x_n>_n of real numbers.

Give the conditions for a real number a not to be a limit point of the sequence. (lim x_n not equal to a.)

The Attempt at a Solution

I'm really not sure if this is the whole answer or if it's only a part of it:

For all e>0 there exists an n that belongs to the real numbers s.t. |x_n - a| >= e.


Is there more to this or do I have it correct?

It's not a matter of "more to this", that's not at all correct. For example, the sequence {(-1)ⁿ} does not converge to 0 (it doesn't converge at all) but if $\epsilon= 2$ there is NO n such that |(-1)ⁿ- 0|= 1> $\epsilon$. On the other hand, {1/n} converges to 0 but if $\epsilon= 1/4$, for n= 2, |1/2- 0|= 1/2> 1/4.

The definition of "{a_n} converges to a" is "For all $\epsilon> 0$, there exist N> 0 such that if n> N, |a_n- a|< $\epsilon$". The opposite of that is "There exists $\epsilon> 0$ such that for all N, there exist n> N such that |a_n-a|> $\epsilon$. Do you see how, in taking the "opposite", "for all" changes to "there exist" and vice-versa?