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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.

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  • $\begingroup$ 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 $\{x_n\}$. $\endgroup$ – sudeep5221 Dec 21 '19 at 15:29
  • $\begingroup$ Yes, if a sequence converges, it only has one limit point: the limit of the sequence. $\endgroup$ – bjorn93 Dec 21 '19 at 15:48
  • $\begingroup$ @bjorn93 But how do I show that if $x_m\in B_\epsilon(y)$ that any $n>m$, $x_n\in B_\epsilon(y)$? $\endgroup$ – AColoredReptile Dec 21 '19 at 16:01
  • $\begingroup$ @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? $\endgroup$ – bjorn93 Dec 21 '19 at 16:13
  • $\begingroup$ $x$ is the only limit point of the set. $\endgroup$ – monoidaltransform Dec 26 '19 at 18:13
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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.

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  • $\begingroup$ I don't understand the part about $\mathbb{R}\setminus S_N$. Doesn't that contain $T_N$? $\endgroup$ – AColoredReptile Dec 21 '19 at 17:08
  • $\begingroup$ 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 $S_N$. $\endgroup$ – joeb Dec 21 '19 at 17:29
  • $\begingroup$ 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 $y$? $\endgroup$ – AColoredReptile Dec 21 '19 at 17:32
  • $\begingroup$ Ah yes it was a typo. I corrected it $\endgroup$ – joeb Dec 21 '19 at 20:01
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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.

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