1. ## Topology help

Q: Let $x\in{O}$, where $O$ is an open set. If $(x_{n})$ is a sequence converging to x, prove that all but a finite number of the terms of $(x_{n})$ must be contained in $O$.

I am not sure where to begin with for this proof. Looking at it, i'd assume we need to use the definition of a cluster (limit point) somewhere in the proof. However, I am not sure. I am having a trouble breaking down the statement, so that I know what my goal really is.

2. Originally Posted by Danneedshelp
Q: Let $x\in{O}$, where $O$ is an open set. If $(x_{n})$ is a sequence converging to x, prove that all but a finite number of the terms of $(x_{n})$ must be contained in $O$.

I am not sure where to begin with for this proof. Looking at it, i'd assume we need to use the definition of a cluster (limit point) somewhere in the proof. However, I am not sure. I am having a trouble breaking down the statement, so that I know what my goal really is.

O is open. So there exists an open ball B(x,epsilon) in O containing x, for some epsilon > 0. Pick such an epsilon. Now note that for some natural N and all n>N, |x_n - x|<epsilon and hence x_n is in B(x,epsilon) and also in O for all n>N.

Hence. . .

(obviously if we assume that x is not a limit point of O, that is it is in the interior of O, the assumption that O is open is unnecessary)

3. Originally Posted by harbottle
O is open. So there exists an open ball B(x,epsilon) in O containing x, for some epsilon > 0. Pick such an epsilon. Now note that for some natural N and all n>N, |x_n - x|<epsilon and hence x_n is in B(x,epsilon) and also in O for all n>N.

Hence. . .

(obviously if we assume that x is not a limit point of O, that is it is in the interior of O, the assumption that O is open is unnecessary)
So, since $(x_{n})$ $\in(x-\epsilon,x+\epsilon)$ for every $n\geq\\N$, only a finite number of the terms of $(x_{n})$ reside in the neighborhood centered around x.

Thanks.

4. Originally Posted by Danneedshelp
So, since $(x_{n})$ $\in(x-\epsilon,x+\epsilon)$ for every $n\geq\\N$, only a finite number of the terms of $(x_{n})$ reside in the neighborhood centered around x.
Exactly
Only the terms $x_1,~x_2,\cdots,x_{N-1}$ may be outside that open set.