How do I prove that a sequence is open?

[lolalyle]

Homework Statement

Let B be an open set. Let x \in A. Let sn be a sequence such that lim sn=x. Then there exists an N such that sn \in A for all n>N.

Definition of Open Set: S is open if for every x \in S, there exists an E>0 such that (s-E,s+E) \subset S

Homework Equations

Prove that the sequence is open.

The Attempt at a Solution

Let E>0 be given.
Let x \in A be given.
Let lim sn=x.
Choose N such that ... for all n>N.
...

 

[Inferior89]

The first sentence of your post should be "Let A be an open set" I guess..
$$.\quad A$$
(--------x-)

And then you have a sequence of number that converge to x. What is the definition of a sequence converging to a number?

 

[lolalyle]

For all E>0, there exists a real number N such that for all n in the natural numbers, n>N implies that abs(sn-s)<E.

 

[Inferior89]

So basically you can get as close to x as you want as long as you chose a N big enough.

Do you see that this together with that A is open will give you the required result?

 

[lolalyle]

Yes! Thank you. I definitely do not know why I didn't see that in the first place.

 

[Nirjhor]

If a sequence is not bounded does it have a maximum?