Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Choose a sequence $(x_n)_{n\in\mathbb N}$ of nonnegative real numbers with finite sum $x=\sum\limits_{n\in\mathbb N}x_n$ and consider the set $X=\{x_I\mid I\subseteq \mathbb N\}$ where, for every $I\subseteq \mathbb N$, $x_I=\sum\limits_{n\in I}x_n$. Thus, $X\subseteq[0,x]$.

Question: Can the set $X$ fail to be closed?

A motivation from probability theory is explained there. Note that $X=[0,x]$ if $x_n=1/2^n$ and that $X\neq [0,x]$ if $x_n=a^n$ for some $a$ in $(0,1/2)$ (yielding measure zero Cantor sets), but that in all these cases $X$ is closed.

share|improve this question
    
Since $x_{(\cdot)}:2^\Bbb N\to [0,\infty)$ is a finite positive measure, it is interesting whether a more general result hold. For any finite positive measure $\mu$ on a measurable space $(X,\mathfrak B_X)$ its range $\mu(\mathfrak B_X)$ is a closed subset of $\Bbb R$. –  Ilya Mar 16 '13 at 17:18
6  
The downvote boggles the mind. –  cardinal Mar 16 '13 at 17:38
    
@Ilya Endow the sigma-algebra $\mathfrak F$ with the metrics $d(A,B)=\mu(A\Delta B)$ (or rather, the quotient of $\mathfrak F$ by the negligible sets). Is $(\mathfrak F,d)$ a compact metric space? –  Did Mar 16 '13 at 17:57
    
@cardinal: Yup. –  Did Mar 16 '13 at 17:59
    
@Did: I guess, since $\mu$ is continuous w.r.t. $d$, compactness of $\mathfrak F$ implies the positive answer to the general answer. Sequential compactness, using boundness of $\mu$, perhaps holds indeed - though I don't see it now. Before I left the first comment, I was rather thinking about the following, related problem: if $\lim_n \mu(A_n) = m$, can we assume that $\{A_n\}$ is a monotone sequence? (for example, w.l.o.g. $\{\mu(A_n)\}$ is monotone) I tried to work with $\liminf/\limsup$ but didn't come up with a proof or found counterexamples. –  Ilya Mar 16 '13 at 19:14
add comment

2 Answers

up vote 13 down vote accepted

Consider the compact metric space $(A,d)$ where $A$ is the set of binary sequences with metric $$d(a,b) = \sum_{n=1}^{\infty} \frac{ |a_i - b_i|}{2^n}.$$ Define $f:A\to [0,x]$ by $f(a) = \displaystyle \sum_{n=1}^{\infty} a_n x_n.$ Then $f$ is continuous so $f(A)=X$ is compact.

share|improve this answer
1  
@DavideGiraudo Suppose $a^{(m)}\to a$ and let $\epsilon>0.$ There exists an $N$ such that $\sum_{n>N} x_n < \epsilon.$ So $$| f( a^{(m)} ) - f(a) | < \epsilon+ \sum_{n=1}^N | a^{(m)}_n - a_n | x_n .$$ Letting $m\to\infty$ gives $\limsup_{m\to\infty} |f(a^{(m)}) - f(a)| \leq \epsilon.$ –  Ragib Zaman Mar 16 '13 at 11:15
    
@RagibZaman Thanks for this nice answer. Any idea about the more general question asked in a comment? –  Did Mar 16 '13 at 18:10
1  
@Did I've had no luck showing that space is compact, sorry. –  Ragib Zaman Mar 17 '13 at 4:52
3  
@RagibZaman: perhaps, because it's not compact –  Ilya Mar 17 '13 at 21:54
add comment

The range of an $n$-dimensional vector measure is always closed. See the paper (and references therein)

P. R. Halmos (1948), The range of a vector measure, Bull. Amer. Math. Soc. 54, 416–421.

So the set will always be closed.

share|improve this answer
1  
Thanks for the (freely available) reference. –  Did Mar 20 '13 at 14:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.