# 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!

• What is $A_{\alpha}$? – Uncountable Feb 28 '15 at 2:53
• @Uncountable Thanks for pointing out, edited – Simple Feb 28 '15 at 2:55
• In case $2$, why start with $(a,b)$ and go on with $(c,d)$? Other than that and a few typographical errors, it seems like the proof is correct. – Uncountable Feb 28 '15 at 3:00

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 $\bigcup\{A_\alpha:\alpha\in[1,2]\}=B$.

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.