I am struggling with this proof:
If for a fixed $\epsilon$ we choose for each j a cube $Q_j$$\subset$$S_j$ and such that $$\left| S_j\right| \leq (1+e) \left| Q_j\right|$$
Then I have a struggle with the following:
My assumption is(please let me know if I am wrong!)
If $Q_j$$\subset$$S_j$, then $\left| Q_j\right|<=\left|S_j\right|$
Now if that assumption is correct:
then choose $a=\left|S_j\right|$
and $b=\left|Q_j\right|$
Now if $$a \leq (1+e) b$$ for all e>0
then $$a \leq b$$
proof: assume the opposite: $$b <a$$ then a-b>0, so we can choose $\epsilon=a-b$
it the follows that $$a \leq b$$ and we end up with a contradiction.
In other words:
$$\left|S_j\right|<=\left| Q_j\right|$$
Therefore,
$$\left| S_j\right| \leq (1+e) \left| Q_j\right|$$
holds for a particular e and not for all if $Q_j$$\subset$$S_j$
But this is then used later on in the proof, where they conclude with since e is arbitrary, but I have just shown that the equation does not hold for all e, and hence e is not arbitrary.
What am I doing wrong?
The cube $S_j$ depends on $\epsilon$: after fixing $\epsilon>0$, you choose $S_j$ such that $Q_j\subseteq S_j$ and $|S_j|\leq(1+\epsilon)|Q_j|$. So we have a different $S_j$ for each choice of $\epsilon$, and we do not conclude that $|S_j|=|Q_j|$ for any single one of these choices.
This does not change the fact that $\epsilon$ is arbitrary, since we fix an arbitrary $\epsilon$ first before we choose $S_j$. For any particular $\epsilon$, it is possible to choose such an $S_j$.