Determine $\inf\{x \mid x \in [0, \infty) \setminus \mathbb{Q} \}$. 
Determine $\inf\{x \mid x \in [0, \infty) \setminus \mathbb{Q} \}$.

I was convinced this would be $0$, but $0$ isn't irrational number. This set is the irrationals right? In order to find the infimum I need to get that if $\inf\{x \mid x \in [0, \infty) \setminus \mathbb{Q} \} = a$, then for any $\varepsilon >0$ there exists $x_\varepsilon$ such that $x_\varepsilon < a + \epsilon$. So if I add any positive real number to $a$ then it isn't a lower bound anymore. How should I approach this?
 A: The infimum of a set is just its greatest lower bound.   $0$ is clearly a lower bound of the set.  If you had any other lower bound, you can show it must be negative (Because any positive number isn't a lower bound, you can find a smaller positive irrational).   Thus 0 is the infimum.
We use inf and sup specifically in place of maximum and minimum to allow for the cases where there isn't a smallest/biggest point in the set but it is still bounded
A: Let us mark the set of positive irrationals $A:=\{x|x \in [0,\infty) \setminus \Bbb Q \}$. Then $ \forall x \in A: 0 \leq x$, thus $0$ is a lower bound of $A$. Let now an $\epsilon >0$; from density of the irrationals in $\Bbb R$ there exists an irrational $r \in (0,\epsilon)$. But $r$ is a positive irrational, so $r \in A$. By the characterization of infimum we have prooved that $\inf A=0$.
A: Hint. For all $n\in\mathbb{N}\backslash\{0\}$ we have
$$
1-\sqrt[\,n\,]{2}\in \{x \, | \, x\in [0,\infty)\backslash\mathbb{Q}\}
$$
and
$$
\lim_{n\to\infty}( 1- \sqrt[\, n \, ]{2})=0
$$
