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?
Is a finite union of bounded sets bounded in any metrical space?
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0my definition of bounded is that diam(M)<\infty, that is $diam(S)=sup\{d(x,y); x,y\in S\}$. – 2012-03-13
3 Answers
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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1To your first question, yes; to your second question, no. – 2013-02-02
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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0@Godisemo indeed. – 2012-03-13
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.