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In any metrical space $(M,d_M)$, consider $n$ bounded subsets $S_i\subset M$. Then, is $\cup_i^nS_i$ bounded? If so, why?

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What have you tried? –  lhf Mar 13 '12 at 20:57
    
My idea is to show that $diam(M)<\infty$ this is easy if the distance between two subsets is finite. The problem is that I am not sure if this distance is finite. –  Godisemo Mar 13 '12 at 20:59
    
What is your definition of bounded? –  azarel Mar 13 '12 at 21:00
    
my definition of bounded is that $diam(M)<\infty$, that is $diam(S)=sup\{d(x,y); x,y\in S\}$. –  Godisemo Mar 13 '12 at 21:02
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3 Answers 3

up vote 7 down vote accepted

Since each $S_j$ is bounded, there exists a point $p_j$ such that $S_j\subset B(p_j,r_j)$. Now take $p=p_1$, $r=\max\{r_1,\ldots,r_n\}+\max_j\{d(p_1,p_j)\}$. If $x\in S_j$, then $$ d(x,p)\leq d(x,p_j)+d(p_j,p_1)\leq r_j+d(p_j,p_1)\leq r. $$ So $x\in B(p,r)$, and this shows that $S_j\subset B(p,r)$ for all $j$. Thus $$ \bigcup_j S_j\subset B(p,r). $$

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Do you mean to say 'there exists a point $p_j$ and there exists $r_j$ such that...'? I didn't see how you introduced the $r_j$'s. –  Robert Feb 2 '13 at 6:06
    
Also, does $p_j$ have to be in $S_j$? –  Robert Feb 2 '13 at 6:51
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To your first question, yes; to your second question, no. –  Martin Argerami Feb 2 '13 at 16:37
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In the comments you say that your definition of bounded $S$ is $$ \mathrm{diam}(S) < +\infty. $$

You want to prove that if $S$ is bounded and $x\in M$ is any point, then $$ \sup_{y\in S} d(x,y) < +\infty $$ (use triangular inequality) and then you can easily prove this last inequality for the finite union of bounded sets.

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Of course. A set is $S$ bounded iff for every $p$ in the space there is some $r_S$ such that $S \subset B(p,r_S)$. For finite unions we take the maximum of the $r_S$ in the union.

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So if I understand this correct, the distance between two bounded subsets cannot be infinitely large? –  Godisemo Mar 13 '12 at 21:07
    
That's not really the definition of bounded, but a provable property about boundedness (that there is an $r$ for every $p$.) –  Thomas Andrews Mar 13 '12 at 21:16
    
I know, that's why I wrote iff –  Henno Brandsma Mar 13 '12 at 21:48
    
@Godisemo indeed. –  Henno Brandsma Mar 13 '12 at 21:49
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