# Show that $x \in \bar A$ iff there exists a sequence of elements of $A$ that converges to $x$ (In $\mathbb{R}_{usual}^n$)

I need to show:

Let $A$ be a subset of $\mathbb{R}^n$ with its usual topology. Show that $x \in \bar A$ iff there exists a sequence of elements of $A$ that converges to $x$.

$\text{int}(A)$ are interior points of $A$. $\partial(A)$ are boundary points of $A$.

The definition of convergence I'm working with is:

A sequence $\{x_n\}^{\infty}_{n=1} \subseteq \mathbb{R}^n$ is said to converge to a point $x \in \mathbb{R}^n$ if for every $\epsilon > 0$ there is a $N \in \mathbb{N}$ such that $n > N$ implies $||x_n −x|| <$ $\epsilon$

This is what I have so far (for $\implies$):

We have that $\bar A = \text{int}(A) \cup \partial(A)$.

If $A$ is empty, the statement is vacuously true. We continue assuming that $A$ is nonempty.

For the following part, I realize I could have simply used a constant sequence instead. I had already come up with this approach, so I kept it.

For any $\textbf{x} \in \text{int}(A) \subseteq \mathbb{R}^n$, we have that $\exists \epsilon > 0, \textbf{x} = (x_1, x_2,...,x_n) \in B_{\epsilon}(\textbf{x}) \subseteq \text{int}(A) \subseteq A$, we have the sequence $\{\textbf{x}_m\}_{m=1}^{\infty} = (x_1 - \frac{1}{m}, x_2 - \frac{1}{m},...,x_n - \frac{1}{m}), m \in \mathbb{N}$ that converges to $\textbf{x}$ and for all $m > \frac{1}{|\textbf{x}-\epsilon|}$, $\{\textbf{x}\}_{m=1}^{\infty} \in B_{\epsilon}(\textbf{x})$.

For any $\textbf{x} \in \partial(A) \subseteq \mathbb{R}^n$, we have that any open set $B_{\epsilon}(x) \subseteq \mathbb{R}$ containing $\textbf{x}$ has a non-empty intersection with $A$. If this non-empty intersection contains $\textbf{x}$, we have the constant sequence $\{\textbf{x}_{m}\}_{m=1}^{\infty} = \textbf{x}$ and we are done.
If it does not, then $B_{\epsilon}(\textbf{x})$ contains some other point $\textbf{x}_{1} \in A$. ???

I'm stuck here, on the case $x \in \partial(A) \backslash A$. I have tried to use $\epsilon_n = \frac{1}{n}$, but the problem here is there is no guarantee that $x_n \in B_{\epsilon_n}(\textbf{x}) \backslash B_{\epsilon_{n+1}}(\textbf{x})$.
I have tried to take $\epsilon_{n+1} = ||\textbf{x} - x_n||$, but this only gurantees that I can have a countable infinite number of points from $\{x_n\}$ inside any open ball around $\textbf{x}$ but not that only finitely many points of $\{x_n\}$ are outside of the ball.

• Shouldn't $B_{\epsilon}(x) \subseteq A$ instead of int($A$)? Commented May 31, 2017 at 19:57
• Since $\text{int}(\text{int}(A)) = \text{int}(A)$, the ball should also be in $\text{int}(A)$, no? Although, in this case, I suppose it doesn't make a difference either way. Commented May 31, 2017 at 20:00

If $x\in \partial A$, then any open ball centered at $x$ contains points both in $A$ and not in $A$. For $x\in \mathrm{int}(A)$, every open ball contains points in $A$. So, in either case, any open call centered around $x$ contains points in $A$.
Thus, assume $x\in \overline A$. Then for each $n\in \mathbb N$, $B(x,1/n)\cap A \neq \varnothing$, so let $x_n \in B(x,1/n)\cap A$. Then the sequence $(x_n)$ has values in $A$ and converges to $x$.
Conversely, suppose $x\notin \overline A$. Then there exists $\varepsilon > 0$ such that $B(x,\varepsilon) \cap A = \varnothing$. If $(x_n)$ is a sequence converging to $x$, then there exists $m\in \mathbb N$ such that $d(x,x_m) < \varepsilon$, and so $x_m \notin A$. So, there is no sequence in $A$ converging to $x$.
• In the first part, isn't there the possibility that $x_n$ could be the same for infinitely many $n$? Since $B_{1/n+k} \subseteq B_{1/n}$ or does that not matter? Commented May 31, 2017 at 20:09
• Since $d(x,x_n)<1/n$, the sequence gets arbitrarily close to $x$ (and stays that close), so in order for $x_n$ to be equal for infinitely many $n$, they would have to all be $x$. Commented May 31, 2017 at 20:11