To find supremum of this

For $s=1$ to be supremum of the set $Q=(0,1)cap mathbb{Q}$ it need to be an upper bound for the set, i.e., $1geq x$ for all $xin Q$, and for given $nin mathbb{N}$ should exist $r_nin Q$ satisfying $1-1/n<r_n$. Note that in every interval of real numbers there exists rational numbers. Why not the supremum is a number greater than one? Because if $s=sup Q>1$ you can always find a real number $rin(1,infty)$ that $r<s$. So, $sup Q=1$.

Characteristic for the supremum $s$ of a set $X$ are the following properties:

• $xleq s$ for every $xin X$ (i.e. $s$ is an upper bound of $X$)
• if $r<s$ then some $xin X$ exists with $r<x$ (i.e. every $r<s$ is not an upper bound of $X$).

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If $Bsubseteq A$ then you can immediately conclude that $sup A$ is an upper bound of $B$ or equivalently that: $$sup Bleqsup A$$

Now you can start examining which of the following statements is true:

• $sup B=sup A$
• $sup B<sup A$

The second is true if and only if an element $ain A$ exists with $b<a$ for every $bin B$.

Investigate this for the sets $A=(0,1)$ and $B=(0,1)capmathbb Q$.

For irrationals investigate this for the sets $A=(0,1)$ and $B=(0,1)cap mathbb Q^{complement}$.

Both cases use same technics. Firstly consider $A = (0, 1) cap mathbb{Q}$.

1. It's obvious, that $forall x in A, x<1$
2. Let's take $forall epsilon > 0, epsilon < 1$ and consider $1-epsilon < 1$ . As we know for any real numbers $a<b$, there exists rational between them. So exists $x in A$ and $x>1-epsilon$ which finish proof.

For irrational case we use analogical proof.

$(0,1)cap Q$ is the set of rational numbers between 0 and 1 (not including 0 and 1). We can find both rational numbers and irrational numbers arbitrarily close to 1 so the supremum, in both cases, is 1 (the answer doesn't change). 1 is not an irrational number, but the supremum of a set does not have to be in that set.