Let $a_n$ a sequence such that $\{0,2\}\subseteq P(a_n)$ and $\forall n \in \mathbb{N} |a_{n+1}-a_n|<1.

Show that the sequence has another subsequential limit, i.e $\exists L \in P(a_n) \ s.t \ L \notin \{0,2\}$.

I tried to solve the question, but my answer is probably wrong... I would like to know what's is the problem.

My solution: Assume there is no such limit.

0 is a subsequntial limit, so there is subsequence such that $\forall \epsilon_1>0 \ , \ \forall N \in \mathbb{N} \ \exists n_1>N: \ |a_{n_1}|<\epsilon_1$

Same way, 2 is a subsequential limit, so there is subsequence such that $\exists n_2>n_1: \ |a_{n_2}-2|<\epsilon_2$

let $\epsilon_1=\epsilon_2=0.5$ and $\exists M \in \mathbb{N} \ M>n_2: |a_M|<0.5 \ and \ |a_{m+1}-2|<0.5$. We can deduce that $-0.5<a_m<0.5 \ and \ 1.5<a_{m+1}<2.5 \Rightarrow 1<a_{m+1}-a_m<3$ in contradiction to what is is given. Eventually, there is another subsequential limit.

For some reason, my answer wasn't accepted...

And there is another part to the question which I couldn't solve:

Is it possible that $a_n$ has exactly 3 subsequential limits? If yes - demonstrate, if not - prove.

Please help, thank you!


Your first part is incorrect because the two subsequences do not necessarily have to contain elements that are adjacent to each other in the original series (in fact, you rather proved that they won't).

Here is a rough proof of that part: Because of the condition $|a_{n+1}-a_n|<1$, and because the sequence converges to both 0 and 2, there must be infinitely many elements between 0 and 2 (and not close to either). Take all those elements. Then this new bounded sequence (which is more or less (original sequence) - (subsequence at 0) - (subsequence at 2) - (stuff not between 0 and 2)) will have a convergent subsequence, which is a sub-sub-sequence - and hence sub-sequence, of the original sequence. I apologize about the ridiculous language, I didn't make it up.

For the second part, consider the sequence $$1/n,1,2-1/n$$ for each $n\ge2$, which has only three subsequential limits 0, 1, and 2.


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