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.
 A: 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.
A: 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 $\varepsilon=1-t$. 
A: Observe that $\sup (S) \leq 1$ since $1$ is an upper bound for $S$. If $\sup (S) < 1$. Let $a = \sup (S)$, then $a < 1$. Choose an irrational number $q$ such that $a < q < 1$, then $a$ is no longer an upper bound of $S$. Thus $a = 1$.
