Let $E$ be a set and {$x_{n}$} a sequence of points not...

Question

not necessarily elements of $E$. Suppose that $\lim_{n\to \infty}x_{n} = x$ and that $x$ is an interior point of $E$. Show that there is an integer $N$ so that $x_{n} \in E$ for all $n \geq N$.

Answer 1

Let $x$ be an interior point of $E$. Then there is an $\varepsilon$-neighborhood of $x$, $U_\varepsilon$, which is contained in $E$, $U_\varepsilon\subset E$. Since $\lim\limits_{n\to\infty}x_n=x$ we have that $\forall \delta>0$, $\exists N_\delta$ such that $|x_k-x|<\delta$ for all $k\geq N_\delta$. Choose $\delta=\varepsilon$ and then for all $k\geq N_{\varepsilon}$ we have $|x_k-x|<\varepsilon$, i.e. $x_k\in U_\varepsilon\subset E$ for all $k\geq N_\varepsilon$.

Answer 2

If $x $ is an interior point of $E $, then

$$\exists \epsilon>0 \;\;:\; (x-\epsilon,x+\epsilon)\subset E $$

but $\lim_{n\to+\infty}x_n=x,$

so, $$\exists N\in \Bbb N \;:\; \forall n\ge N \; x_n\in (x-\epsilon,x+\epsilon) $$

Done.

Answer 3

If $x \in$ int$(E)$, theres exist $\delta > 0$ such that $(x - \delta, x + \delta) \subset E$.

If $x_{n} \longrightarrow x$, for all $\epsilon > 0$, theres exist $n_{0} \in \mathbb{N}$ such that $n > n_{0, \epsilon} \Longrightarrow x_{n} \in (x - \epsilon, x + \epsilon)$.

Take $\epsilon = \delta$, and for all $n > n_{0, \delta}$:

$$x_{n} \in (x - \delta, x + \delta) \subset E.$$