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?


Does that mean x can only be 0 or 1? And it can't be any rational?

No, it means $x$ is any irrational between $0$ and $1$.
$\{x\in [0;1]: x\notin \Bbb Q\}$ is: the set of numbers from the real interval $0$ to $1$ inclusive, such that these numbers are also not in the rationals.
Now, find the supremum of this set.   First, what is your definition of a supremum?

A set $E ⊂  R$  is bounded above iff $∃  b ∈  R$  s.t. $a \leq b, ∀  a ∈  E$ ($b$ is an upper bound for $E$) 

No, that is not the definition of supremum.  That is the definition of upper bound.  A supremum is a special kind of upper bound.  It's the least such.
A value $s\in R$ is the supremum of $E$ if $\;\nexists b\in R: \forall a\in E: ((a\leq b) \wedge (b< s)\wedge (a\leq s))$
Now, can you find such an upper bound for your set?  Can you show that this is the supremum.
