Prove that $B = \bigcup\{A_\alpha \mid \alpha \in[1,2]\}$ I am working this question:

Set $B = \{(x, y)\mid 1 \le x^2 + y^2 \le 4\}$, $A_\alpha = \{(x, y)\mid x^2 + y^2 = α^2\}$. Prove that $\bigcup\{A_\alpha\mid \alpha \in [1, 2]\} = B$. 

because this is a equality, I need to prove it into two cases.
Proof:
Case 1, $B\subseteq\bigcup\{A_\alpha \mid \alpha \in[1,2]\}$
let $(a,b)\in B$, since $B= \{ (x,y)\mid 1\leq x^2+y^2 \leq 4 \}$,
then $1\leq^2+b^2\leq 4$ $\Rightarrow a^2+b^2\geq1 $, $a^2+b^2\geq 4$
now let $\alpha\in\mathbb R^+$ such that $a^2+b^2=\alpha ^2$
then we have $\alpha ^2\geq 1$, and $\alpha ^2\geq 4$ $\Rightarrow 1\leq\alpha ^2 \leq 4\Rightarrow 1\leq\alpha  \leq 2$ 
therefore, $(a,b)\in\bigcup\{A_\alpha \mid \alpha \in[1,2]\}$ .
Case 2, $\bigcup\{A_\alpha \mid \alpha \in[1,2]\}\subseteq B$
let $(a,b)\in \bigcup\{A_\alpha \}$, then $(c,d)\in\bigcup A_\alpha\mid\alpha\in[1,2]$
since $A_\alpha =\{(x,y)\mid x^2+y^2=\alpha ^2\}$ and $\alpha\in[1,2]$
so, $1\leq \alpha\leq2 \Rightarrow 1\leq \alpha ^2\leq2^2 $
thus $ 1\leq c^2+d^2\leq 2^2\Rightarrow 1\leq c^2+d^2\leq 4 \Rightarrow (c.d)\in B$
therefore, $\bigcup\{A_\alpha \mid \alpha \in[1,2]\}\subseteq B$
follow those two cases, $B = \bigcup\{A_\alpha \mid \alpha \in[1,2]\}$ $\square$
I think my proof miss some details that I need to show. Can anyone give me a hit or show me how to write a better proof for this question?
Thanks you so much!
 A: The two set inclusions that you’re proving aren’t actually cases; they’re simply two things to be proved in order to get the desired conclusion. 

In a proof by cases you have a list of possibilities (cases), and you know that one of them is true, but you don’t know which one, so you have to treat each of them separately: if the first possibility holds, the result follows for one reason; if the second possibility holds, the result follows for another reason; and so on.

In the first part of your argument you have a few typos. You wanted to say that since $(a,b)\in B$, $1\le a^2+b^2\le 4$, and therefore $\color{red}a^2+b^2\ge 1$ and $a^2+b^2\color{red}{\le} 4$. You repeated the wrong inequality two lines down: you meant to say that we have $\alpha^2\ge 1$ and $\alpha^2\color{red}{\le}4$. The intended argument is correct, however.
The second part has a much bigger problem: you never explain where $(c,d)$ comes from, or what it has to do with the $(a,b)$ with which you started. You should be saying something like this:

Let $(a,b)\in\bigcup\{A_\alpha:\alpha\in[1,2]\}$. Then there is an $\alpha\in[1,2]$ such that $(a,b)\in A_\alpha$. Certainly $1^2\le\alpha^2\le 2^2$, and $(a,b)\in A_\alpha$ means that $a^2+b^2=\alpha^2$, so $1\le a^2+b^2\le 4$, and therefore $(a,b)\in B$. This shows that $\bigcup\{A_\alpha:\alpha\in[1,2]\}\subseteq B$ and completes the proof that $\bigcup\{A_\alpha:\alpha\in[1,2]\}=B$.

A: Just for fun, and to give a different viewpoint, here is a more 'logical' proof of
$$
\tag{0} \bigcup\{A_\alpha\mid \alpha \in [1, 2]\} \;=\; B
$$
with the given definitions of $\;A_\alpha\;$ and $\;B\;$, and with all variables being real numbers.

$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
The pairs $\;(x,y)\;$ in the left hand side are
$$\calc
(x,y) \in \bigcup\{A_\alpha\mid \alpha \in [1, 2]\}
\op\equiv\hint{definition of $\;\bigcup\;$}
\langle \exists \alpha : \alpha \in [1, 2] : (x,y) \in A_\alpha \rangle
\op\equiv\hints{rewrite using inequalities -- to better fit the definition of $\;B\;$;}\hint{definition of $\;A_\alpha\;$}
\langle \exists \alpha : 1 \le \alpha \le 2 : x^2 + y^2 = \alpha^2 \rangle
\endcalc$$
Combining this with the definition of $\;B\;$, which gives us the $\;(x,y)$ in the right hand side of $\ref 0$, we are asked to prove that
$$
\tag{1}
\langle \exists \alpha : 1 \le \alpha \le 2 : x^2 + y^2 = \alpha^2 \rangle
\;\equiv\;
1 \le x^2 + y^2 \le 4
$$
for all $\;x,y\;$.

All we have done until this point is to use the definitions to translate from the 'set level' to the 'logic level': we have removed all mention of sets, and now can focus on the arithmetical part.

From the shape of $\ref 1$, it is at least very likely that the more general
$$
\tag{2}
\langle \exists \alpha : 1 \le \alpha \le 2 : z = \alpha^2 \rangle
\;\equiv\;
1 \le z \le 4
$$
holds for all $\;z\;$: if we can prove $\ref 2$, then $\ref 1$ immediately follows by substituting $\;z:=x^2+y^2\;$.
Let's see if we can start with the left hand side of $\ref 2$, and simplify to reach the right hand side:
$$\calc
\langle \exists \alpha : 1 \le \alpha \le 2  : z = \alpha^2 \rangle
\op{\tag{*}\equiv}
  \hints{arithmetic: square $\;\alpha\;$ in the inequality}
  \hint{-- to make the left part more similar to the right}
\langle \exists \alpha : 1 \le \alpha^2 \le 4 \;\land\; \alpha \ge 0 : z = \alpha^2 \rangle
\op\equiv\hint{logic: substitute $\;z\;$ in left part}
\langle \exists \alpha : 1 \le z \le 4 \;\land\; \alpha \ge 0 : z = \alpha^2 \rangle
\op\equiv\hint{logic: extract part not using $\;\alpha\;$ out of $\;\exists \alpha\;$}
1 \le z \le 4 \;\land\; \langle \exists \alpha : \alpha \ge 0 : z = \alpha^2 \rangle
\op{\tag{**}\equiv}\hint{arithmetic: the non-negative numbers are the squares}
1 \le z \le 4 \;\land\; z \ge 0
\op\equiv\hint{arithmetic: simplify inequalities}
1 \le z \le 4
\endcalc$$
That proves $\ref 2$, which completes our proof.

Finally, note that the key steps were $\ref *$ and $\ref{**}$: the rest is just definitions, and basic logical and arithmetical manipulations.
