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Let $\{x_n\}_{n=1}^{\infty}$ be a bounded sequence, and let $E$ be the set of subsequential limits of that sequence.

Prove that $E$ is bounded and contains both sup$E$ and inf$E$.

Here is my sad attempt:

Let $\{x_n\}_{n=1}^{\infty}$ be a bounded sequence, and let $E$ be a set of subsequential limits of that sequence. $E$ is not empty since every bounded sequence of real numbers has at least one convergent subsequence.

  1. Suppose $\{x_n\}_{n=1}^{\infty}$ converges to a real number $L$. Then $E=\{L\}$, since every subsequence of $\{x_n\}_{n=1}^{\infty}$ must also converge to $L$. Therefore, $E$ is bounded and sup$E$=inf$E$=$L\in E$.

  2. Suppose $\{x_n\}_{n=1}^{\infty}$ diverges. As stated above, there is at least one convergent subsequence, so $E$ is not empty. Let $S$ be the range of $\{x_n\}_{n=1}^{\infty}$. Then $S$ is bounded since the sequence is bounded

    1. Suppose $S$ is finite; that is, $S=\{x_1,...,x_k\}$ for $k\in\mathbb{N}$. Then there exists a term $x_i$ with $1\leq i\leq k$, of $\{x_n\}_{n=1}^{\infty}$ that repeats infinitely many times, and thus the subsequence $\{x_i\}_{n=1}^{\infty}$ converges to $x_i$. This implies that, $E\subset S$, which also implies that $E$ is bounded, since $S$ is bounded. Furthermore, since $S$ is finite, then $S$ has a smallest and largest element and so does $E$. Hence, inf$E=$ min$E$ and sup$E=$ max$E$, which are both contained in $E$.

    2. Suppose $S$ is infinite. Then, by the Bolzano-Weierstrass Theorem, there is at least one accumulation point, $a_1$, and thus, at least one subsequence that converges to $a_1$. Let $A$ be the set of all accumulation points of $S$. Then $E\subset A$, since for each member $a_i$ of $A$ there exists a subsequence of $\{x_n\}_{n=1}^{\infty}$ that converges to $a_i$. Also, notice that since $S$ is bounded, then there exists $M_1,M_2\in\mathbb{R}$, with $M_1<M_2$, such that for all $x\in S$, $x\in(M_1,M_2)$.

      1. If $A$ is finite, then $A$ must have a smallest and a largest element, which implies that $E$ has a smallest and largest element. So, inf$E=$ min$E$ and sup$E=$ max$E$, which are both contained in $E$, which also implies that $E$ is bounded.

      2. Suppose $A$ is infinite. Every element of $S$ being contained in $(M_1,M_2)$ implies that every element of $A$ is contained in $[M_1,M_2]$. Thus, since $E\subset A$, $E$ is bounded. We also see that $A$ has a smallest and largest element.

Ok, that's as far as I can go. All I need to show now is that inf$E$ and sup$E$ are contained in $E$, but I just can't seem to figure that out. I'm sure that there is probably a better way to approach this whole proof. If you have any advice on this, please share. Thank you in advance for any and all assistance.

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    $\begingroup$ $|x_n|<M$ then we have the $x_n$ admits a convergent subsequence say $\{x_{n_k}\}$, which is also bounded. This is true for any subsequence of E since $\{x_n\}<M$ for all $n$. Thus for all $x \in E$ $x < M$, so the set is bounded. Use least upperbound property to get the latter results. E is nonempty and bounded above thus by L.U.B property the sup exists. $\endgroup$ – Mr.Fry Jun 27 '14 at 0:55
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We can proceed more simply in showing that $E$ is bounded. Since $\{x_n\}_{n=1}^\infty$ is bounded, then there is some $M>0$ such that $|x_n|\le M$ for all $n.$ Now, suppose that $L\in E,$ and show that $|L|<M+1,$ using the definitions. (Actually, you can show that $|L|<M+\epsilon$ for all $\epsilon>0,$ so that $|L|\le M,$ but that's more than we really need.) Since $L$ was an arbitrary element of $E,$ then $E$ is bounded.

Now, let's put $\alpha=\sup E.$ Since $\alpha-1$ isn't an upper bound of $E,$ then there is some $L_1\in E$ such that $\alpha-1<L_1,$ and so there is some least $n$ (say $n_1$) such that $x_n>\alpha-1.$ (Why?) Suppose we have found $n_k$ for some positive integer $k.$ Since $\alpha-\frac1{k+1}$ isn't an upper bound of $E,$ then there is some $L_{k+1}\in E$ such that $\alpha-\frac1{k+1}<L_{k+1},$ and so there is some least $n$ (say $n_{k+1}$) such that $x_n>\alpha-\frac1{k+1}$ and $n>n_k.$ (Why?) In this fashion, we construct a subsequence $\{x_{n_k}\}_{k=1}^\infty$ of $\{x_n\}_{n=1}^\infty$ such that $x_{n_k}\to\alpha.$ You'll need to justify that this actually is a (well-defined) subsequence, and that it converges to $\alpha.$ Let me know if you get stuck along the way. The proof that $\inf E$ is an element of $E$ is similar.

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  • $\begingroup$ is my argument valid in the comment? At least up until the sup. $\endgroup$ – Mr.Fry Jun 27 '14 at 1:03
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    $\begingroup$ As for the infinite case, you are correct that any accumulation point of $E$ will be a subsequential limit, but you haven't fully justified it here. (You can use a similar approach to the one I outline in showing that the supremum of $E$ lies in $E$.) Not knowing what results you have available to you, I can't really say for sure whether you need to show more or not. $\endgroup$ – Cameron Buie Jun 27 '14 at 1:13
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    $\begingroup$ To get |L|< M+1 we just say knowing $|x_n|< M$ for all $n$ then we know at most $|L| \leq M$ so we get strict inequality using $M+1$? And thanks. I am just getting ready for graduate analysis, so I'm trying my hand at every problem I see. $\endgroup$ – Mr.Fry Jun 27 '14 at 1:18
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    $\begingroup$ Yeah, that's pretty much it. You can use an $\epsilon$-$N$ argument with $0<\epsilon<1,$ if you really want to, but that works just fine. $\endgroup$ – Cameron Buie Jun 27 '14 at 1:18
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    $\begingroup$ Whoops! Sorry @Rod. I thought I was addressing the OP. My fault for not reading your comment question carefully enough. Your argument is perfectly valid, and is basically in the same vein as the one I outlined. $\endgroup$ – Cameron Buie Jun 27 '14 at 1:21

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