Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.
My attempt is: Since the sequence is bounded , there exists $M>0$ such that $x_n \in [-M,M]$ for all $n \in \mathbb{N}$. Since the sequence does not converge to $x$, there exists $\epsilon_0>0$ such that $ \forall N \in \mathbb{N}$, there exists $n \geq N$ such that $|x_n-x| \geq \epsilon_0$.
Then we have $x_n \in [-M,x-\epsilon_0] \cup x_n \in [x+\epsilon_0,M]$. By Bolzano-weierstrass theorem, there exists a convergent subsequence in the two intervals.
Is my proof valid? ${}{}$
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$\begingroup$Use Bolzano-Weierstrass to extract a subsequence $x_{i_1}, x_{i_2}, \dotsc$ that converges to some $a$. Since $x_1, x_2, \dotsc$ does not converge to $a$, there exists some $\varepsilon > 0$ such that for each positive integer $N$, there exists some $j(N) > N$ such that $|x_{j(N)} - a| \geq \varepsilon$. Use Bolzano-Weierstrass to extract from $x_{j(1)}, x_{j^2(1)}, x_{j^3(1)}, \dotsc$ a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $b$. Clearly, $a \neq b$.
By the way, this method works for any $x_1, x_2, \dotsc$ in $\mathbb{R}^n$, which goes beyond what the OP intended.
$\endgroup$ 11 $\begingroup$- If a sequence $(x_n)$ is bounded, $a\le x_n \le b$ say, then it has at least one limit point $x$ with $a\le x\le b$ (Bolzano-Weierstraß) and
- a bounded sequence with exactly one limit point $x$ converges towards that limit point.
Therefore there must exist at least two distinct limit points and we can extract a converging sequence for each.
Just in case the second bullit point above is not clear: If $a\le x_n\le b$ for all $n$ and $x$ is not the limit of $x_n$, then there exists $\epsilon>0$ such that infinitely many $x_n$ are outside $(x-\epsilon,x+\epsilon)$, hence there are infinitely many $x_n>x+\epsilon$ or infinitely many $x_n<x-\epsilon$, leading to at least one limit point $x'$ in $[x+\epsilon,b]$ or $[a,x-\epsilon]$. Thus if $x$ is a limit point but not the limit, there is another limit point $x'\ne x$.
$\endgroup$ 1 $\begingroup$Yes, valid, but could write it even clearer.
For each $N$ there is an $n$ such that $|x_n-x|\ge \epsilon_0$. Let such an $n$ be chosen for all $N$ and denote it by $k(N)$, for example. Thus we get a subsequence $(x_{k(1)},x_{k(2)},\dots)$ of $(x_n)$, which lies in $[-M,x-\epsilon_0]\cup [x+\epsilon_0,M]$, so this subsequence has a convergent subsequence, and the limit of that cannot be $x$.
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