Prove that $\bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]=[3,5]$. Where have I gone wrong?

Question

Prove that $\bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]=[3,5]$.

Assume $[a,b]\in\bigcap\limits_{x\in\Bbb R}[3-x^2,5+x^2]$. Note that $\vert x\vert\ge 0$. If $x=0$, then $[a,b]=[3,5]$. If $\vert x\vert\gt 0$, then $a\lt 3$ and $b\gt 5$. This shows that $[3,5]$ must be contained within $[a,b]$. Therefore it must follow that $\bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]=[3,5]$.

$\blacksquare$

Given sets $A$ and $B$, I know that the general approach to show that $A=B$ is to demonstrate that $A\subseteq B\wedge B\subseteq A$. However, I'm a bit confused how that would work in this case. Is this a valid proof? If not, where have I gone wrong?

Answer 1

For set equality, you should start with an element of one set to show that it is in the other set and vice versa. The elements in both sets are real numbers. So what you need to show is

• if $s \in \bigcap_{x \in \Bbb R}[3-x^2,5+x^2]$, then $s \in [3,5]$, and
• if $t \in [3,5]$, then $t \in \bigcap_{x \in \Bbb R}[3-x^2,5+x^2]$.

You have the right idea for the second containment. But you need to work on the first containment. Essentially speaking, you need to show if $3-x^2 \leq s \leq 5+x^2$ for every real $x$, then $3 \leq s \leq 5$.

Can you proceed from here?

Answer 2

Firstly, you should have written $[a,\,b]\color{limegreen}{\subseteq}\bigcap_{x\in\Bbb R}[3-x^2,\,5+x^2]$. Secondly, while each $[3-x^2,\,5+x^2]$ has $[3,\,5]$ as a subset, you need to emphasize no proper superset thereof is a subset of the $x=0$ case; indeed, the $x=0$ case is a subset of all cases, so is the intersection. Thirdly, if you want to use extensionality as you discuss, you should say$$\begin{align}y\in\bigcap_{x\in\Bbb R}[3-x^2,\,5+x^2]&\iff\forall x\in\Bbb R(3-x^2\le y\le 5+x^2)\\&\iff3=\max_{x\in\Bbb R}(3-x^2)\le y\le\min_{x\in\Bbb R}(5+x^2)=5\\&\iff y\in[3,\,5].\end{align}$$

Answer 3

Let $A=\bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]$ and $B=[3,5]$.

$B \subseteq A$:

Since $[3,5]\subseteq [3-x^2, 5+x^2]$ for each $x$ we have $[3,5]\subseteq \bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]$ so $B \subseteq A$.

$A \subseteq B$:

Now take any $y\in\bigcap\limits_{x\in\Bbb R}[3-x^2, 5+x^2]$ and suppose $y\notin[3,5]$.

• If $y<3$ then exists $x_0$ such that $3-x_0^2>y$ so $y\notin [3-x_0^2, 5+x_0^2]$ and thus $y\notin A$. A contradiction.

• If $y>5$ then exists $x_0$ such that $5+x_0^2>y$ so $y\notin [3-x_0^2, 5+x_0^2]$ and thus again $y\notin A$. A contradiction.

  So $y\in B$ and thus $A\subseteq B$ and we are done.