Prove that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is contained in $[3,5]$. 
Prove that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is contained in $[3,5]$.

By contradiction: Assume that $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$ is not contained in $[3,5]$. So, there exists $a$ such that $a$ is an element of $\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2]$, but $a$ is not an element of $[3,5]$, so $a < 3$ or $a > 5$. Using $a < 3$, we know that there is a rational number $b > 0$ such that $a = 3-b$.
I'm not sure where to go from here to get a contradiction.  
 A: It's easier to just prove it directly. Choose any $a \in \bigcap_{x \in \mathbb R} [3 - x^2, 5 + x^2]$. We want to show that $a \in [3, 5]$. But since $a$ belongs to every interval of the form $[3 - x^2, 5 + x^2]$, we know in particular that it must belong to the interval obtained by letting $x = 0$. So $a \in [3, 5]$, as desired.
A: So if $a < 3$, prove that there is an $x$ so that $a < 3-x^2$.  That would prove that $a \not \in \cap [3-x^2, 5+x^2]$.
If $a > 5$, prove that there is an $x$ so that $5+x^2 < a$.  That would prove that $a \not \in \cap [3-x^2, 5+x^2]$
To prove the first.  If $a < 3$, then $3- a > 0$.  Have you proven that square positive square roots exist for every positive real number yet?  If so just let $x <\sqrt{3 -a}$ so $3 - x^2 > 3 - (\sqrt{3 -a })^2 = a$.  But if you haven't proven that let $0 < x < \min(3-a, 1)$.  Since $x < 1$, $x^2 < x$.  so $x^2 < 3-a$ and $3 - x^2 > 3-(3-a) = a$.
A: Since $0 \in \mathbb{R}$ we have
$$\bigcap_{x\in\mathbb R}[3 - x^2, 5 + x^2] \subseteq [3-0^2, 5+0^2] = [3,5]$$

It is more interesting to prove that $$\bigcap_{x\in\mathbb R\setminus \{0\}}[3 - x^2, 5 + x^2] \subseteq [3,5]$$
