1. ## Topology help

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

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

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