Supremum of a set of irrational numbers I need help with the following example:
Let S be the set of all irrationals in $[0,1]$. Show that $\sup(S) = 1$.
Is there some property that I should be referring to when proving problems like these? The set definition states that it is bounded between $[0,1]$ so is it possible to just say $\sup(S) = 1$ based off that knowledge or is that an insufficient proof?
Thanks in advance.

You need to show two things:
a. $s=1$ is indeed an upper bound of $S$ - this is obvious, as all the elements of $S$ are less than $1.
b. If $t<s=1$, then $t$ is not an upper bound of $S$. In other words, for every such $t<1$, you can find an element of $S$ which is larger than $t$. Take for example the sequence
$$
a_n=1-\frac{\sqrt{2}}{n+1}\in S, \quad n\in\mathbb N.
$$
Then $a_n\to 1$, and hence $t<a_n<1$, for sufficiently large $n$ - this comes from the definition of convergence of $a_n\to 0$, for $\varepsilon=1-t$. 
