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Is the following proof correct? Can it be improved?

Let $x\in \mathbb{R}$ and suppose that $\{x_n\}$ is a sequence that does not converge to $x$. Does there exist a subsequence $\{x_{n_i}\}$ which has no subsequence that converges to $x$?

Call a sequence $\{y_m\}$ perfect if

  1. every subsequence $\{y_{m_i}\}\subseteq \{y_m\}$ converges to the same limit, or

  2. every subsequence $\{y_{m_i}\}\subseteq \{y_m\}$ is divergent.

Note that the assertion is clearly true for perfect subsequences. Therefore, it suffices to prove that $\{x_n\}$ contains a perfect subsequence that does not converge to $x$.

If $\{x_n\}$ is convergent, let $E$ denote the set of all subsequential limits of $\{x_n\}$, pick $L\in E\setminus\{x\}$, and define $\{x_{n_i}\}$ by $$n_i=\operatorname{max}\left\{m:\left| x_{m}-L \right|=\operatorname{min}\{\left| x_j-L \right|:1\leq j < i\}\right\}.$$
If $\{x_n\}$ is divergent, define $\{x_{n_i}\}$ by $$n_i=\operatorname{max}\left\{m:\left| x_{m} \right|=\operatorname{max}\{\left| x_j \right|:1\leq j < i\}\right\}.$$

In both cases, it is clear by construction that $n_i\leq n_j \Leftrightarrow i<j$, so $\{x_{n_i}\}$ is a subsequence of $\{x_n\}$. In the convergent case, $x_{n_i}$ is perfect with every subsequence converging to $L\ne x$. In the divergent case case, every subsequence of $\{x_{n_i}\}$ is divergent. This exhausts all possibilities, so the assertion is true.

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  • $\begingroup$ Note, $\mathbb{R}$ is Hausdorff (i.e. limits are unique, so the first case is meaningless). In the second case, its somewhat unclear how you're defining $n_i$, note that divergent doesn't mean going off to infinity ($1,-1,1,-1,\cdots$ diverges). I also think its better not to worry about the perfectness of the subsequence you choose, and simply choose a sequence which stays away from $x$. $\endgroup$ – Deven Ware Oct 7 '13 at 20:00
  • $\begingroup$ If $(x_n)$ is divergent and bounded, you can't find a subsequence such that every further subsequence diverges. Instead, if $(x_n)$ does not converge to $x$, choose $\epsilon>0$ such that for every $n$, there is an $m>n$ with $|x_m-x|>\epsilon$. Use this to construct a subsequence of $(x_n)$ that is "bounded away" from $x$. $\endgroup$ – David Mitra Oct 7 '13 at 20:01
  • $\begingroup$ @DavenWare You are right- what I meant to say in case $1$ was that the set of all subsequential limits of $\{x_n\}$ has cardinality $1$, and in case $2$ was that $\{x_n\}$ is unbounded. $\endgroup$ – Samuel Handwich Oct 7 '13 at 20:04
  • $\begingroup$ @Samuel: Even those two corrected cases don't cover everything. Daven's example gives a sequence that is bounded, having $2$ subsequential limits. Neither of the two cases covers it. Take a look at the answers given below, and see if they help you. $\endgroup$ – Cameron Buie Oct 7 '13 at 20:21
  • $\begingroup$ @CameronBuie Read the rest of my answer. $\endgroup$ – Samuel Handwich Oct 8 '13 at 16:46
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It is simpler to note that if $x_n$ does not converge to $x$, then for all $\epsilon >0$, and for all $N$, there exists some $n \ge N$ such $x_n \notin B(x, \epsilon)$. Let $\nu(N)$ denote the first index greater than or equal to $N$ for which this is true.

Let $n_1 = \nu(1)$. Then let $n_{k+1} = \nu(n_k+1)$. Then $x_{n_k} \notin B(x, \epsilon)$ for all $k$.

In particular, $|x_{n_k} -x| \ge \frac{1}{2} \epsilon$ for all $k$, hence $x_{n_k}$ cannot contain as subsequence that converges to $x$.

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In the divergent case, your argument does not work. Consider $x_n=(-1)^n,$ for an example. Since $|x_n|=1$ for all $n,$ then your definition of $n_i$ fails, as the natural numbers have no greatest element.

A better approach is as follows: Since the sequence $\{x_n\}$ does not converge to $x,$ then by definition, there is some $\epsilon>0$ such that for all $N\in\Bbb N,$ there is some $n\ge N$ such that $|x_n-x|\ge\epsilon.$ Let $A=\{n\in\Bbb N:|x_n-x|\ge\epsilon\},$ so that $A$ is an infinite subset of $\Bbb N$ by assumption. Enumerate the elements of $A$ in increasing order by $i\mapsto n_i.$ Show that no subsequence of $\{x_{n_i}\}$ converges to $x$.

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