# Show that if $\{x_n\}_{n=1}^\infty$ converges to $x\in \mathbb{R}$ then $\overline{\{x_n:n\in\mathbb{N}\}}=\{x_n:n\in\mathbb{N}\}\cup\{x\}$

Show that if $$\{x_n\}_{n=1}^\infty$$ converges to $$x\in \mathbb{R}$$ then $$\overline{\{x_n:n\in\mathbb{N}\}}=\{x_n:n\in\mathbb{N}\}\cup\{x\}$$

My definition of the closure is that $$x\in \overline{A}$$ if and only if for any open set $$U$$ containing $$x$$, $$U\cap A\neq \emptyset$$

So showing $$\{x_n:n\in\mathbb{N}\}\cup\{x\}\subseteq\overline{\{x_n:n\in\mathbb{N}\}}$$

Let $$A=\{x_n:n\in\mathbb{N}\}$$

I have $$A\subseteq \overline{A}$$

Need to show $$\{x\}\subseteq \overline{A}$$

let $$U$$ be an open set containing $$x$$. Since $$U$$ is open there exists an epsilon ball $$B_\epsilon(x)\subseteq U$$. And since $$x_n$$ converges to $$x$$, there exists an $$N$$ such that $$x_n\in B_\epsilon(x)\subseteq U$$ for any $$n>N$$. thus $$x_n\in U$$ for any $$n>N$$ and $$U\cap A\neq \emptyset$$. then $$\{x\}\subseteq \overline{A}$$

Showing $$A\cup \{x\}\supseteq \overline{A}$$

Let $$y\in \overline{A}$$ and $$y\not\in A$$ then for any open set $$U$$ containing $$y$$ $$U\cap A\neq \emptyset$$.

Then for any $$\epsilon>0$$, since $$B_\epsilon(y)$$ is open, $$B_\epsilon(y)\cap A\neq \emptyset$$, thus there is an $$x_m\in B_\epsilon(y)$$

So I believe I just need to show that if $$n>m$$ then $$x_n\in B_\epsilon(y)$$ so that $$A$$ converges to $$y$$ which means $$y=x$$ but I'm not sure how to show that.

• You can use the fact that all the subsequences of $\{ x_n \}_{n \in \mathbb{N}}$ converge to the same limit $x$, thus, the closure of the set should just contain $x$ in addition to $\{x_n\}$. – sudeep5221 Dec 21 '19 at 15:29
• Yes, if a sequence converges, it only has one limit point: the limit of the sequence. – bjorn93 Dec 21 '19 at 15:48
• @bjorn93 But how do I show that if $x_m\in B_\epsilon(y)$ that any $n>m$, $x_n\in B_\epsilon(y)$? – AColoredReptile Dec 21 '19 at 16:01
• @AColoredReptile If $y\in\overline{A}$ and $y\not\in A$, then since every open set $U$ containing $y$ must contain element of $A$, it follows that $y$ is a limit point. The only such point is $x$ by convergence. Perhaps you don't want to use limit points? – bjorn93 Dec 21 '19 at 16:13
• $x$ is the only limit point of the set. – monoidaltransform Dec 26 '19 at 18:13

The problem in your argument is that, having one term $$x_m$$ in the open ball $$B_{\epsilon}(y)$$ does not imply all successive terms $$x_n \in B_{\epsilon}(y)$$, where $$n > m$$. It needn't even be the case that the next term $$x_{m+1}$$ is in the open ball. I would rethink your approach, but this time utilize convergence $$x_n \to x$$. Said convergence is going to guarantee you that, eventually, successive terms $$x_N,x_{N+1},...$$ are all going to be close to $$x$$. For any point $$y$$ other than $$x$$, we know that a tail of successive terms will be pulled away from $$y$$ and towards $$x$$. That leaves you with only finitely many terms $$x_1,...,x_N$$ to deal with...

Given any positive integer $$N$$, let $$S_N \; = \; \Big\{x_1,...,x_N\Big\}$$ and $$T_N \; = \; \Big\{x_n \; : \; n > N \Big\},$$ so that clearly $$A = S_N \cup T_N.$$

Now suppose $$y \in \bar{A} \backslash A$$, but $$y \neq x$$. We may choose radii $$r,s > 0$$ such that $$B_{r}(x) \cap B_s(y) = \varnothing.$$ Due to convergence $$x_n \to x,$$ there exists a positive integer $$N$$ such that $$T_N \subseteq B_r(x).$$ In particular, $$T_N \cap B_s(y) = \varnothing.$$ Moreover, $$\mathbb{R} \backslash S_N$$ is an open set of the real line (being the compliment of a finite set) that contains $$y$$ (as otherwise $$y \in S_N \subseteq A$$). It follows that there exists a radius $$t > 0$$ for which $$B_t(y) \subseteq \mathbb{R} \backslash S_N$$, or rather, $$S_N \cap B_t(y) = \varnothing.$$

All that is left to do is take $$\varepsilon = \min\{s,t\}$$. With this radius we have, $$T_N \cap B_{\varepsilon}(y) = \varnothing$$ and $$S_N \cap B_{\varepsilon}(y) = \varnothing.$$ Combining both results tells us that $$A \cap B_{\varepsilon}(y) = \varnothing.$$ Here is where the contradiction arises. We supposed $$y$$ was in the closure of $$A$$. It could not be the case that an open ball around $$y$$ is completely disjoint from $$A$$, but under the assumption that $$y \neq x$$ we found such an open ball. Our assumption that $$x,y$$ were distinct must have been incorrect.

• I don't understand the part about $\mathbb{R}\setminus S_N$. Doesn't that contain $T_N$? – AColoredReptile Dec 21 '19 at 17:08
• Up till that part, we managed to separate $T_N$ from $y$. Now we want to separate $S_N$ from $y$. Also, note that $S_N$ and $T_N$ aren't necessarily disjoint. It is true that members of $T_N$ are close to $x$. But who knows, so may be some members of $S_N$. – joeb Dec 21 '19 at 17:29
• I think maybe it's a typo then, you said $\mathbb{R}\setminus S_N$ is an open set that does not contain $y$. But doesn't it have to contain $y$? – AColoredReptile Dec 21 '19 at 17:32
• Ah yes it was a typo. I corrected it – joeb Dec 21 '19 at 20:01

It's clear that all points $$\{x_n: n \in \Bbb N\} \cup \{x\}$$ are points of $$\overline{\{x_n: n \in \Bbb N\}}$$, all $$x_n$$ because always (by the definition) $$A \subseteq \overline{A}$$ (if $$x \in A$$, then any open point containing $$x$$ intersects $$A$$ (as $$x \in A \cap U$$)), and $$x$$ because each neighbourhood of $$x$$ contains infinitely many $$x_n$$ (all $$n$$ with $$n \ge N$$ for some $$N$$) by definition of convergence (note that these need not be distinct points, but there is at least one, say $$x_N$$).

Suppose for the converse that $$y \in \overline{\{x_n: n \in \Bbb N\}}$$. Then there is a sequence $$(y_n)$$ from $$\{x_n: n \in \Bbb N\}$$ such that $$y_n \to y$$. So each $$y_n = x_{m_n}$$ for some $$m_n \in \Bbb N$$. If the $$m_n$$ are cofinal in $$\Bbb N$$ it's easy to see we get a subsequence of the sequence $$(x_n)$$ so $$y=x$$ by unicity of limits. Of not, some $$m_n$$ occurs infinitely often and we have an eventually constant sequence, so converging to some term of $$\{x_n: n \in \Bbb N\}$$. In either case $$y \in \{x_n: n \in \Bbb N\} \cup \{x\}$$ and we're done.