1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $x$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – themathandlanguagetutor Nov 28 '17 at 8:02
  • $\begingroup$ @themathandlanguagetutor Yes you are absolutely right, that was a bit sloppy on my part. $\endgroup$ – angryavian Nov 28 '17 at 8:08
2
$\begingroup$

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 $A$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.