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.

Note that $1$ is an upper bound for the set $S.$ We must show that is it the least upper bound.
(It means we have to show that any thing less than $1$ cannot be an upper bound.)
Take any $\epsilon>0.$
Then we can find a large natural number $n>1$ such that $\dfrac{1}{n}<2\epsilon.$
This gives us $$1-2\epsilon<1-\dfrac{1}{n}<1$$ and $$\dfrac{1}{\sqrt{5}n}<\dfrac{1}{2n}<\epsilon$$
Therefore $$1-\epsilon<1-\dfrac{1}{n}-\dfrac{1}{\sqrt{5}n}<1.$$
Note that $$1-\dfrac{1}{n}-\dfrac{1}{\sqrt{5}n}$$ is an irrational.  Hence $1$ is the least upper bound.
