Someone explain me, please. Examples 11.2.3b

Question

loser66 · Answer

attachment

loser66 · Answer

At: " However, if $\{G_{n_1},G_{n_2},......,G_{n_k}\}$ is any finite subcollection of G, and if we let m:= sup $\{n_1,n_2,....,n_k\}$, then 
$G_{n_1}\cup G_{n_2}\cup .....\cup G_{n_k}=G_m =(-1,m)$ "

loser66 · Answer

@ganeshie8 @Zarkon

loser66 · Answer

One more question: if they defined $G_n := (-1,n) $ like that, is it not that $\bigcup\limits_{n=1}^{\infty}G_n = G_n$ . Because I think:
$G_1= (-1,1)\G_2=(-1,2)\......\G_n= (-1,n)$
then, 
 $G_1\subset G_2\subset G_3\subset .....\subset G_n$
Hence $\bigcup\limits_{n=1}^{\infty} G_n = G_n$

ganeshie8 · Answer

heyy

ganeshie8 · Answer

$G_n$ is a "single" set/interval

$\{G_n\}$ is a "family/collection" of sets/intervals

ganeshie8 · Answer

For both `b` and `c` examples, notice that the supremum of indices( `m`) is in the set but it will not be in the finite subcover.

loser66 · Answer

from another part:
if $G_n:= (1/n, n)$
then if $\{G_n\} = \bigcup\limits_{n=1}^{\infty} G_n$ is a open cover of set A. It is a union of $G_n$ and each of them can't be a subcover. I mean this set is not decomposed to smaller one. 
To the example above, if $G_n$ is defined as shown, then $\{G_n\}=\{G_1, G_2,.....,G_n\}$This is an open cover of the given set, but $G_n$ (just only ) itself is an opencover of the set.  That shows this open cover has a finite subcover on the set A. Hence it is compact. 
Am I right?