# Convergent rearrangement of convergent sequence has limit in the sequence or equal to the original limit?

I'm working on a proof for analysis: "Let $\{x_n\}$ be a convergent sequence in $\mathbb R^d$ with limit $x$. Set $$A=\{x_1,x_2,x_3,\cdots\}\cup\{x\},$$ that is, $A$ is the set consisting of all the points occuring in the sequence together with the limit $x$. Show that $A$ is a closed set."

I feel like I need to use the theorem that states that a set $A$ is closed if and only if every covergent sequence in $A$ converges to a point of $A$, but to do this I essentially need to show that every convergent "rearrangement" of the sequence $\{x_n\}_{n\ge0}$ (where $x_0$ is taken to be $x$) converges either to one of the points of the sequence (or to $x$ itself). By a "rearrangement" I mean a sequence of the form $\{x_{n_k}\}$ where $\{n_k\}$ is itself a sequence in $\mathbb N\cup\{0\}$ (which may be non-injective or non-surjective).

However, I really don't know where to even start trying to show this. Am I even on the right track?

You probably know the following fact.

If $x_n$ is a sequence converging to $x$, then any subsequence $\{x_{m_k}\}_{k=1}^\infty$ also converges to $x$.

Above, the definition of subsequence requires $m_1 \le m_2 \le \cdots$, so it is not the same as your "rearrangement" definition.

Let us return to your dilemma. Suppose you have a rearrangement $\{x_{n_k}\}$ where $\{n_k\}$ is some sequence in $\mathbb{N}\cup \{0\}$. There exists a nondecreasing subsequence of $\{n_k\}$: just drop anything that violates the nondecreasingness. For example if $\{n_k\} = \{1,3,2,4,5,7,6,8,\ldots\}$, then one subsequence (among many) you can pick is $\{1,3,5,7,\ldots\}$. Let us call this nondecreasing subsequence $\{m_k\}$.

Then $\{x_{m_k}\}$ is a subsequence of the original sequence $\{x_n\}$, so it must converge to $x$. It is also a subsequence of your rearrangement sequence $\{x_{n_k}\}$ which by assumption is convergent, so $\{x_{n_k}\}$ must also converge to $x$.

• This is an excellent answer! Thank you! I feel I should point out though that this new subsequence could easily end up being constant after a certain time, so the limit might be some other point in the sequence. – themathandlanguagetutor Nov 28 '17 at 8:02
• @themathandlanguagetutor Yes you are absolutely right, that was a bit sloppy on my part. – angryavian Nov 28 '17 at 8:08

If you can show that $A$ is compact, you are done. To this end let $(A_i)_{i \in I}$ be an open cover of $A$.

There is $i_0 \in I$ with $x \in A_{i_0}$. Since $A_{i_0}$ is open, there is $m \in \mathbb N$ such that

$x_n \in A_{i_0}$ for all $n>m$.

To each $n \in \{1,2,...,m\}$ there is $i_n \in I$ with $x_n \in A_{i_n}$.

Therefore $A_{i_0}, A_{i_1},...,A_{i_m}$ is an open subcover of $A$.

• Compactness isn't covered until the next section, unfortunately – themathandlanguagetutor Nov 28 '17 at 7:58
• That said, now that I've gotten to that section, this is rather a nice approach – themathandlanguagetutor Dec 4 '17 at 2:57