The problem that I'm presented with is as follows:

Let $(x_k)$ be a bounded sequence in $\mathbb{R}^n$, and let $a$ be a vector in $\mathbb{R}^n$. It is given that $(x_k)$ does not converge to $a$. (However, there is the possibility that there exist some subsequences of $(x_k)$ which converge to $a$.). Prove that there exists a vector $b \neq a$ in $\mathbb{R}^n$, and indices $1 \leq k(1) < k(2) < \cdots < k(p) < \cdots$ in $\mathbb{N}$, such that $\lim\limits_{p \to \infty} x_{k(p)} = b$.

If my understanding is correct then what I need to prove is that even if there exists a subsequence of $(x_k)$ which converges to $a$ then there necessarily exists a vector $a\in \mathbb{R}^n$ such that at least one subsequence of $(x_k)$ also converges to $a$.

Here's my proof:

Since $(x_k)$ is a bounded sequence, $\exists m,M\in \mathbb{R}$ such that $m<\|x_k\|<M$ for all $k$. Now, let $m\le x_{k(1)}:=x_1$, $x_{k(2)}:=\{x_i\in(x_k):\|x_i\|=\max\{\|x_{k(1)}\|$, $\|x_2\|\} \text{ and } 1\le i\le 2\}, \dots$, $x_{k(p)}:=\{x_i\in (x_k):\max\{\|x_{k(p-1)}\|, \|x_p\|\}\}, \dots \le M$. By Bolzano-Weierstrass, $(x_{k(p)})$ converges to some vector in $\mathbb{R}^n$. Suppose that this vector is $a$. Now let $M\ge x_{k(1)}:=x_1$, $x_{k(1)}:=x_1$, $x_{k(2)}:=\{x_i\in(x_k):\|x_i\|=\min\{\|x_{k(1)}\|$, $\|x_2\|\} \text{ and } 1\le i\le 2\}, \dots$, $x_{k(p)}:=\{x_i\in (x_k):\min\{\|x_{k(p-1)}\|, \|x_p\|\}\}, \dots \ge m$. Then this subsequene will converge to a vector $b\in\mathbb{R}^n$, with $b\ne a$ since $\|b\|$ can be taken arbitrarily closely to $m$.

Please let me know what you think about my proof. I'm not very confident that it is quite rigorous. Maybe I misunderstood what actually needs to be proved. Your insight would be much appreciated.

Update: I think the easiest would be to consider subsequences which converge to $\liminf$ and $\limsup$, so there are at least two such subsequences in a bounded sequence.

  • $\begingroup$ The problem in the title is different from the problem in italics: here the sequence might be convergent (just not to $a$) and have only one limit point. Choose one of the versions. $\endgroup$ May 14 '17 at 21:49
  • $\begingroup$ @MichałMiśkiewicz I've corrected the title. Please check. $\endgroup$
    – sequence
    May 14 '17 at 22:12
  • $\begingroup$ The title is more clear now, but still different from what you're describing here. The assumption and the claim on the sequence $(x_n)$ here are different. $\endgroup$ May 15 '17 at 8:56


The sequence must have at least two limit points. Bolzano-Weierstrass guarantees at least one, say $b.$ Since $(x_n)$ is not convergent show there must be an $\epsilon > 0$ and a subsequence $x_{n_j}$ such that $\|x_{n_j} - b \| > \epsilon$ and notice that $(x_{n_j})$ is another bounded sequence.


To construct the above subsequence note that the total sequence is not convergent and, in particular, not convergent to $b$. Negating the definition of convergence, there exists $\epsilon > 0$ such that for every $N \in \mathbb{N}$ there exists an integer $n \geqslant N$ such that $\|x_n - b \| > \epsilon$.

With $N = 1$ there exists $n_1 \geqslant 1$ such that $\|x_{n_1} - b\| > \epsilon$. Using induction, suppose we have $n_1 < n_2 < \ldots < n_m$ such that $\|x_{n_j} - b \| > \epsilon$ for $1 \leqslant j \leqslant m$. There must exist an integer $n_{m+1} \geqslant n_m +1 > n_m$ such that $\|x_{n_{m+1}} - b \| > \epsilon$. By induction, we have constructed a subsequence $(x_{n_j})$ that fails to converge to $b$. This subsequence is bounded and thus has a convergent subsequence (which also is a subsequence of $(x_n)$) that converges to some $b' \neq b$. Hence, there are at least two limit points.

  • $\begingroup$ How do we know that there exists a subsequene of $(x_{n_j})$ which does not converge to $b$? $\endgroup$
    – sequence
    May 14 '17 at 22:19
  • 1
    $\begingroup$ @sequence: Explained above. $\endgroup$
    – RRL
    May 14 '17 at 23:13
  • $\begingroup$ Can we just say that there exists a subsequence of $(x_k)$ which diverges and thus does not converge to $b$? I'm not quite sure what your induction exactly shows. How does it follow that if $\|x_{n_j}-b\|>\epsilon$ for $1\le j \le m$ then $\|x_{n_{m+1}}-b \|>\epsilon$? I mean $(x_{n_i})$ is an arbitrary subsequence, so we don't know how it behaves. Also, how can we know whether or not the subsequence of the above subsequence does not converge to $b$? Why not? Thanks @RRL $\endgroup$
    – sequence
    May 14 '17 at 23:41
  • 1
    $\begingroup$ If you don't like the inductive statement just follow this. Given $1$ we can find $n_1$ such that $\|x_{n_1} - b \| > \epsilon$. Now given $n_1$ we can find $n_2 > n_1$ such that $\|x_{n_2} - b \| > \epsilon$. Given $n_2$ we can find $n_3 > n_2$ such that $\|x_{n_3} - b \| > \epsilon$. And so on to produce the subsequence that does not converge to $b$ and for which every subsequence (including the convergent ones) does not converge to $b$. $\endgroup$
    – RRL
    May 15 '17 at 0:01
  • 1
    $\begingroup$ Not that $(x_{n_j})$ is convergent but that it is (1) staying away from $b$, (2) bounded since it is a subsequence of $(x_n)$, and (3) as a bounded sequence it has its own convergent subsequence (Bolzano-Weierstrass) which converges to something that can't be $b$. So the second part of your reasoning is correct. $\endgroup$
    – RRL
    May 15 '17 at 0:07

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