8
$\begingroup$

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?

  • 0
    my 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 3

12

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). $

  • 1
    To your first question, yes; to your second question, no.2013-02-02
6

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.

  • 0
    @Godisemo indeed.2012-03-13
0

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.