Let $(X_n)$ be a bounded sequence, and let $E$ be the set of subsequential limits of $(X_n)$. Prove that $E$ is bounded and contains $\sup E$ and $\inf E$.
Does this ask us to prove limsup and liminf exists?
Could you help me ?
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1If you can prove that E is bounded then the sup of E and the inf of E exist by the least upper bound property. What you really want is to prove that E is bounded and that sup E and inf E are subsequential limits of ($X_n$) – John Douma Mar 06 '14 at 16:09
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@user69810 I agree with you. So how can I prove there is subsequence converges to supE? – ZHJ Mar 06 '14 at 16:15
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What is the largest element in E to which a subsequence of ($X_n$) converges? – John Douma Mar 06 '14 at 16:20
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@user69810 Since Xn is bounded, so it has sup(Xn), but there is a monotone subsequence converges to sup(Xn), so the largest element of E is sup(Xn)? – ZHJ Mar 06 '14 at 16:27
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Not quite. There is nothing that says sup($X_n$) is a limit of a subsequence. Consider the sequence consisting of $1$ followed by all zeros. Then sup($X_n$) is $1$ but there is no subsequence that converges to it. – John Douma Mar 06 '14 at 16:35
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@user69810 I still don't know how to do it, could you help me ? – ZHJ Mar 06 '14 at 17:37
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If sup E is not a limit point for any subsequence of $(X_n)$ there will be a radius around it that will contain no subsequence. Show that any point in that interval shares that property and therefore you can find one less than sup E which is an upper bound for E. That gives you your contradiction. – John Douma Mar 06 '14 at 17:46
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Since $(x_n)$ is a bounded sequence by the Bolzano–Weierstrass Theorem contain a convergent subsequence, then $E$ is non-empty.
Now we have to show that $E$ is bounded. Let $M$ be a bound for the entire sequence, i.e., $|x_n|< M$ for all $n\in \mathbb{N}$, and suppose for sake of contradiction that $E$ is unbounded. Then we must have $|x|>M$ for all but finitely many $x\in E$. Choose one, say $x_0\in E$ and $|x_0|>M$, then since $x_0\in E$ and by definition of $E$, we can find a subsequence $(x_{n_i})$ which converges to $x_0$ and in particular for $i\ge n_0$ we have $|x_{n_i}-x_0|< |x_0|-M$, so $|x_{n_i}|\ge |x_0|-|x_{n_i}-x_0|>M$, a contradiction. Hence $E$ is non-empty and bounded.
Let $s= \sup E$, we shall show that lies in $E$. Only we check the claim for the least upper bound, the other part is similar.
First we will show that is a point of accumulation (a limit point). Given $\varepsilon>0$ and $n_0\ge 0$. Then $s-\varepsilon/2$ is not the least upper bound and so there is $x\in E$ such that $s-\varepsilon/2<x\le s$, but since $x\in E$, we have $|x_{n_i}-x|< \varepsilon/2$ for some $n_i\ge i\ge n_0$. Thus $|x_{n_i}-s|\le |x_{n_i}-x|+|x-s|< \varepsilon$. Hence $s$ is a limit point.
Define recursively the sequence: $n_0=0$ and $n_k = \min \{n>n_{k-1}:|x_n-s|<1/k \}$. Notice that since $s$ is a limit point $\{n>n_{k-1}:|x_n-s|<1/k \} \not= \varnothing$ for all $k$, since otherwise contradicts what we have shown in the above paragraph. Then $(x_{n_k})$ is a subsequence of $x_n$, and $s-1/k<x_{n_k}<s+1/k$, by the squeeze theorem we conclude $x_{n_k} \to s$. Thus $s\in E$.
as was to be shown.
Jose Antonio
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No you are not required to prove the existence of the limsup or liminf. You are required to prove that $E$ is bounded and the supremum and infimum of the set $E \subseteq \Bbb R$ is contained in $E$. Here is my humble suggestion:
Suppose $x \in E$. Then $x = \lim (a_n)$ where $(a_n)$ is a subsequence of $(x_n)$. $x \gt x_n$ for every $n \in \Bbb N \implies x \gt a_n \;\; \forall n \in \Bbb N$. Then there is a neighbourhood of $x$ which contains no elements of the sequence $(a_n)$ leading to a contradiction. Therefore $x \le x_m$ for some $m \in \Bbb N$. A similar proof shows that $x_r \le x$ for some $r \in \Bbb N$. Therefore $|x| \le |x_t|$ for some $t \in \Bbb N$. But since $(x_n)$ is a bounded sequence there is a positive real number $M$ such that $|x_n| \le M$ for every $n \in \Bbb N$, $|x| \le M$ for every $x \in E$. This proves that $E$ is bounded.
Now, $\sup E$ and $\inf E$ exist since $E$ is bounded and non-empty. Let Consider the sequence $(r_n)$ such that $r_n \in \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 n\} \cap (x_n)$ for each $n \in \Bbb N$ such that $r_n \neq r_i$ for $i \in \{1,2,.. n - 1\}$. Suppose $r_m$ does not exist for some $n \in \Bbb N$. Then either there are no elements of $(x_n)$ in $ \{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies $ No subsequence in $(x_n)$ can converge to a limit in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \implies c - \frac 1 m$ is an upper bound for $E$ leading to a contradiction.
OR
For some $m \in \Bbb N$ Every element in $\{y\in \Bbb R \ | \ |y - c| \lt \frac 1 m\} \cap (x_n)$ is in $\{r_1, r_2,.. r_{m - 1}\}$. Let $M = \text{Max} \{r_1, r_2,.. r_{m - 1} \}$. If $M \lt c$ then picking $ l = \frac {c - M} 2 \gt 0$ we have that $c - l$ is an upper bound for $E$ leading to a contradiction. If $M \ge c$ then either the elements in $\{r_1, r_2,.. r_{m - 1}\}$ form a subsequence of $(x_n)$ or they do not. If they do then their limit is either equal to $c$ or is greater than it. If they do not form a subsequence of $(x_n)$ they are irrelevant.
This is the best I could do. I apologise for the immense number of cases and for clustered language.
The proof for $\inf E \in E$ is similar.
Hope this helped.
Ishfaaq
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Thanks for your feedback! But I do not understand why there is nerghbourhood V of supE s.t, no element of (Xn) are in V? and finally, what is the contradiction? – ZHJ Mar 06 '14 at 16:50
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Think I screwed up. Gimme a second to come with an alternative argument. – Ishfaaq Mar 06 '14 at 17:03
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I found the solution of this problem, but I still do not understand it. https://www.calvin.edu/~rpruim/courses/m361/F03/hw/sols/sol11-12-print.pdf ( first problem) – ZHJ Mar 06 '14 at 17:36
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Well they've proved it directly and in a much prettier fashion. A subsequence $(a_n)$ of $(x_n)$ is defined such that $a_k \in (c - \frac 1 k, c]$ where $c = \sup E$. The justification is the interval $(c - \frac 1 {2k}, c)$ must contain a limit of a subsequence $(d_n)$ of $(c_n)$ (or else $c - \frac 1 {2k}$ is the supremum of $E$ or there is a subsequence which converges to $c$) and there are infinitely many terms of $(d_n)$ in a neighbourhood which can be created inside the open interval $(c - \frac 1 {2k}, c)$. And $(a_n)$ clearly converges to $c$. – Ishfaaq Mar 06 '14 at 18:09
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why "No subsequence in (xn) can converge to a limit in {y∈R | |y−c|<1/m}" imply "c−1m is an upper bound for E" ? You mean in this neighborhood of C, there is no subsequence, but why it implies there is a upperbound? – ZHJ Mar 06 '14 at 18:39
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@user127534: No subsequence converges to a limit in the said neighbourhood and hence contains no elements of $E$ which are limits. Hence the smaller upper bound. Hence the contradiction. – Ishfaaq Mar 07 '14 at 00:46