Show that there is a $N \in \mathbb{N}$ such that $x_n \in A$ for all $n \geq N$. 
Question:
Let $A$ be an open subset of $\mathbb{R}^p$.
Suppose than $(x_n)$ is a sequence in $\mathbb{R}^p$ such that $x_n \to x$, where $x \in A$.
Show that there is a $N \in \mathbb{N}$ such that $x_n \in A$ for all $n \geq N$.

If my understanding is correct, we have to prove that every term of the sequence $(x_n)$ will lie within some open ball $B(x,\epsilon)$ after some $N^{th}$ term.
Is this correct? If so, how can I go about proving this?
 A: In some books this is an alternative definition of convergence: instead of open balls we say $(x_n)$ converges to $x$ if for every open $A\ni x$, there exists $N\in\Bbb N$ such that $(x_n)\subset A$ for every $n\geq N$.
I'll prove that this is equivalent to the old definition. First, the easy direction (by taking the new definition): since every open ball $B(x,\varepsilon)$ is, well... open, so our definition means that there exists $N\in\Bbb N$ such that etc.
The converse, which is what you want to prove, is proved by noting that, since $A$ is open there exists $\varepsilon > 0$ such that $B(x,\varepsilon)\subset A$. Remember that our hypothesis is now the "old definition", so there exists $N\in\Bbb N$ such that $$(x_n)\subset B(x,\varepsilon) \subset\,...$$
Fill in the last inclusion and you're done.
A: Yes that is correct.
Let me explain you how this works.
SinceA is an open set and x $\in$ A, therefore exists an $\epsilon$ such that the the ball B(x,$\epsilon$)$\subset$ A. Therefore by the definition of convergence for that $\epsilon$ exists a N $\in$ $\mathbb{N}$ such that $x_{n}$ $\in$ A.
