Find the supremum and infimum of {x $\in$ [0,1]: x $\notin$ $\mathbb Q$}. Prove why your assertions are correct Ok I am lost from this question. Does that mean $x$ can only be $0$ or $1$? And it can't be any rational?

Remember that $[0,1]$ is an interval on the real line which contains all of the numbers between $0$ and $1$ including $0$ and $1$.
Now, $\{x \in [0,1] \mid x \not \in \Bbb Q \}$ is a subset of the interval $[0,1]$, but it is the subset we get by removing all of the rational points from $[0,1]$.
So, you have to think about all of the irrational numbers in $[0,1]$, and find the supremum and infimum of that set.
We know since $\Bbb R$ is complete that since the set in question is bounded above by $5$, for example, it has a least upper bound which is the supremum.  Do you know what the smallest upper bound is?  It is $1$, since $1$ is larger than every irrational in $[0,1]$ but it is smaller than every upper bound of the irrationals in $[0,1]$ (why?).
Now you have to find the infimum of the set of irrationals in $[0,1]$ on your own.  If you get stuck, let me know.
