I have a question regarding whether two sets are equal or not.
Basically I have a equivalence relation given by $x \sim y \in Q$ where $Q$ is the set of rationals.
I know that based on the above, there is a partition of the real numbers $R$ given by say [0],[$z_1$],[$z_2$],[$z_3$],.... where [0] is the set that contains all the rational number, [$z_1$] is a set that contains only irrational numbers and the numbers in the [$z_1$] set is all different from the number in the [$z_2$] set by an irrational number,.... so each [$z_i$] contains uncountably number of irrational numbers.
My question is regarding the following:
I was reading a proof that says
[0,1] $\subseteq \bigcup_{q \in Q \cap [-1,1]} (q+E) \subseteq [-1,2]$
where E is a set that contains just one element from each of the equivalence class defined above. So one element from [0], one from [$z_1$],one from [$z_2$],...
My question is how does the set [0,1] being contained in the union of (q+E) defined above? because I know that say if we pick a rational in [-1,1], say we pick 0.3, and say a rational 0.7 from the set [0], then I have 0.3 + 0.7 = 1 , so yes 1 is indeed contained in the set $\bigcup_{q \in Q \cap [-1,1]} (q+E) $, but how about 0? because I already use 0.7 from the set [0], is it still possible to have 0 in the set [0,1] to be in the set $\bigcup_{q \in Q \cap [-1,1]} (q+E) $?
because in the set E, you can only have one rational number, and already use that (in this case I use 0.7) to obtain 1, i.e. 0.3+0.7 =1, then how can we obtain 0 from the equation of x +y = 0 , if y can no longer be rational, and x is a rational in the set Q $\cap [-1,1]$ ?
My question is does the set [0,1] actually a subset of the set $\bigcup_{q \in Q \cap [-1,1]} (q+E) $? or there is a mistake in the notes?
this is the notes I was reading:
https://www.maths.tcd.ie/~richardt/MA2224/MA2224-ch2.pdf
the very last page of the pdf above.
It's clear that you are a little confused about something, but reading through your question, I'm not quite sure exactly what it is your confused about. I'll just try to explain as clearly as I can.
Ok, firstly, the answer. Yes, $$[0,1]\subseteq \bigcup_{q\in\Bbb{Q}\cap[-1,1]} (q+E).$$
Why?
Well, if $x\in [0,1]$, then $x\sim e$ for some $e\in E$, since $E$ contains one element in every equivalence class. Moreover, since $e\in [0,1]$ as well, then $-1\le -e \le 0$, and $0\le x\le 1$, so $-1\le x-e\le 1$. Thus $x-e$ is a rational (since $x\sim e$) in $[-1,1]$, so $x\in (x-e)+E$. Hence every element of $[0,1]$ is in the union $\bigcup_{q\in\Bbb{Q}\cap[-1,1]} (q+E)$. Thus, by definition $$[0,1]\subseteq \bigcup_{q\in\Bbb{Q}\cap[-1,1]} (q+E).$$