Open Subsets in a Metric Space .... Stromberg, Theorem 3.6 ... ....

[Math Amateur]

I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.6 on page 94 ... ... Theorem 3.6 and its proof read as follows:
View attachment 9114
In the above proof by Stromberg we read the following:

" ... ... Letting $$r = \text{min} \{ r_1, r_2, \ ... \ ... \ r_n \}$$ we see that $$B_r (a) \subset U_j \text{ for each } j = 1,2, \ ... \ ... n$$ ... ... "Although it seems plausible ... I do not see ... rigorously speaking, why the above statement is true ...

Can someone demonstrate rigorously that letting $$r = \text{min} \{ r_1, r_2, \ ... \ ... \ r_n \}$$ ...

... implies that $$B_r (a) \subset U_j$$ for each $$j = 1,2, \ ... \ ... n$$ ... ... Surely it is possible that $$B_r (a)$$ lies partly outside some $$U_j$$ ... ...

Help will be appreciated ...

Peter

 

[HallsofIvy]

The Theorem makes reference to "Definition 3.3" but you don't give us that definition. Also you are asking about "B_r(a)" but don't tell us what that means. Since the theorem is about "open sets", I strongly suspect, but can't be sure, that "Definition 3.3" is the definition of "open set" and that "B_r(a)" is the "open ball" centered at a with radius r.

If that is correct then B_r(a) is \{ p| d(p, a)< r\}, the set of all points whose distance from point a (the center of the ball) is less than r (the radius of the ball). From that it follows immediately that if r_1< r_2 then B_{r_1}(a)\subset B_{r_2}(a)- the smaller radius ball is inside the larger radius ball.

 

[Olinguito]

Can someone demonstrate rigorously that letting $$r = \text{min} \{ r_1, r_2, \ ... \ ... \ r_n \}$$ ...

... implies that $$B_r (a) \subset U_j$$ for each $$j = 1,2, \ ... \ ... n$$ ... ...

$r$ is the minimum of all the $r_j$ and so $r\leqslant r_j$ for all $j$; hence $B_r(a)\subseteq B_{r_j}(a)\subset U_j$ for all $j=1,\ldots,n$.

 

[Math Amateur]

$r$ is the minimum of all the $r_j$ and so $r\leqslant r_j$ for all $j$; hence $B_r(a)\subseteq B_{r_j}(a)\subset U_j$ for all $j=1,\ldots,n$.

Thanks for the help Olinguito ...

Peter

- - - Updated - - -

The Theorem makes reference to "Definition 3.3" but you don't give us that definition. Also you are asking about "B_r(a)" but don't tell us what that means. Since the theorem is about "open sets", I strongly suspect, but can't be sure, that "Definition 3.3" is the definition of "open set" and that "B_r(a)" is the "open ball" centered at a with radius r.

If that is correct then B_r(a) is \{ p| d(p, a)< r\}, the set of all points whose distance from point a (the center of the ball) is less than r (the radius of the ball). From that it follows immediately that if r_1< r_2 then B_{r_1}(a)\subset B_{r_2}(a)- the smaller radius ball is inside the larger radius ball.

Thanks for the help HallsofIvy ...

Peter