Find the supremum $(0,1)\setminus \mathbb{Q}$. My intuition tells me that $ A = (0,1)\setminus\mathbb{Q}$  has no supremum. But I'm not sure if the following arguments are sufficient or if my intuition is misguided...
We have that 1 is an upper bound of A, since:$$1 > a  \quad \forall a \in A$$
then because $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a $q$ in $\mathbb{Q}$ such that: 
$$a < q < 1 \quad \forall a \in A$$
hence we can find an $r$ in $\mathbb{Q}$ such that 
$$a < r < q \quad \forall a \in A$$
and the process repeats indefinitely, hence we will never find a least upper bound for the set. I assume I can use something similar to prove that $A$ has no infimum...
Any help is appreciated!
 A: The set $A$ is non-empty and bounded above by $1$. It therefore has a supremum $\sup A \le 1$ in $\mathbb{R}$. You can also verify that 
$$1-\varepsilon < \sup A,$$ for all $\varepsilon > 0$. Therefore
$$
\sup A = 1.
$$
A: I suspect your problem is this: you think that the supremum of a set $S$ must belong to $S$. This is not true! If you keep that in mind, I think you will easily see what the supremum must be.
By the way, the supremum is different from the maximum: the maximum of a set $S$ does have to belong to $S$. So not every bounded set has a maximum, although every bounded non-empty set has a supremum.
A: 
We have that 1 is an upper bound of A, since:$$1 > a  \quad \forall a \in A$$

Yes, this is right.

then because $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a $q$ in $\mathbb{Q}$ such that: 
  $$a < q < 1 \quad \forall a \in A$$

Not so fast.  Yes, $\mathbb{Q}$ is dense in $\mathbb{R}$, but you have your quantifiers mixed up.  It is true that for all $a \in A$ there exists $q \in \mathbb{Q}$ such that $a < q < 1$.  What you wrote is that there exists $q \in \mathbb{Q}$ such that for all $a \in A$, $a < q < 1$.  Do you see the difference?  The second (false) statement asserts that there is an upper bound for $A$ less than $1$.  The first (true) statement asserts that no $a\in A$ is maximal.
I think what you mean to say is that for all $x\in\mathbb{R}$ with $x<1$, there exists $a\in A$ such that $x < a < 1$.  
