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Let $\{u^n\}$ be a bounded sequence in $\mathbb{R}^n$ satisfying $$ \lim_{n\rightarrow\infty}\|u^{n}-u^{n+1}\|=0. $$ Prove that the set of accumulation points of $\{u^n\}$ is nonempty, closed and connected.

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2 Answers 2

up vote 1 down vote accepted

Note that since the set $X$ in question is bounded, by the Bolzano-Weierstrass theorem it must have some accumulation point.

To see that the set of accumulation points is closed, let $x=\lim\limits_{k\to\infty} x_k$ be the limit of a convergent sequence of accumulation points, each of which is the limit of a subsequence $\{u_k^n\}$ of $\{u^n\}$. Then the sequence $\{u_n^n\}$ is a subsequence of $\{u^n\}$ which converges to $x$ (prove this), hence $x\in X$. Thus $x$ is closed.

Edit: As David Mitra pointed out, taking the diagonal subsequence may not suffice if convergence is sufficiently slow. Instead, take the subsequence $\{u_{n}^{k^n}\}$ where $k_n$ is chosen large enough so that $$i\geq k_n\implies \|u_n^i-x_n\|<2^{-n}$$ which will then necessarily converge to $x$.

For connectedness, suppose $X$ is disconnected so $X=A\cup B$ where $A,B$ are disjoint closed subsets of $X$ (hence, by closure of $X$, of $\mathbb R^n$). Consider the function $d(x,y):A\times B\to\mathbb R$ which is just the Euclidean distance between two points $x,y\in X$. Since $A$ and $B$ are closed and bounded, so is $A\times B$ hence by the Heine-Borel theorem $A\times B$ is compact. Hence by $d(x,y)$ achieves its infimum at some $(x,y)\in A\times B$, which must be nonzero as otherwise $A$ and $B$ would intersect (prove this). Let $\{u_x^n\}$ be a subsequence of $\{u^n\}$ converging to $x$, and $\{u_y\}$ one converging to $y$. Note that by discarding at most finitely many terms, we can assume $\{u_x^n\}$ is in $A$. Consider the sequence of real numbers $\{\|u^n-x\|+\|u^n-y\|\}$. Since $(x,y)$ are an infimum for $d$, there are at most finitely many terms $u^k$ such that $\|u^k-x\|+\|u^k-y\|<d(x,y)-\epsilon$ for any fixed $\epsilon>0$, as otherwise we would have an accumulation point $z$ satisfying $\|z-x\|+\|z-y\|<d(x,y)$ by Bolzano-Weierstass. Thus $d(x,y)$ is the limit infimum of this sequence. If you can show that the infimum is $0$, you are done.

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Minor qualm: I don't think the diagonal of the sequences $(u_k^n)$ is necessarily a subsequence of $(u_n)$? –  David Mitra Apr 2 '12 at 2:10
    
Dear Alex Becker. Thank you for your interesting solutions. –  drmath Apr 5 '12 at 10:41

Hint: if a compact set $K$ is not connected, it is the union of two nonempty compact subsets $A,B$ with $\text{dist}(A,B) > 0$.

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Dear Robert Israel. Thank you for your comments. –  drmath Apr 5 '12 at 10:42

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