To find supremum of this

Question

How do I find supremum of set ?

$ (0,1) \cap\Bbb{Q}$ , where $\Bbb Q$ is set of al rationals. How will answer change if rationals are replaced by irrationals? I know supremum of $(0,1)$ is $1$. How do I do?

Thank you.

Answer 1

$(0,1)\cap Q$ 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.

Answer 2

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.

Answer 3

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

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

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If $B\subseteq A$ then you can immediately conclude that $\sup A$ is an upper bound of $B$ or equivalently that: $$\sup B\leq\sup A$$

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 $a\in A$ exists with $b<a$ for every $b\in B$.

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

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

Answer 4

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., $1\geq x$ for all $x\in Q$, and for given $n\in \mathbb{N}$ should exist $r_n\in Q$ 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 $r\in(1,\infty)$ that $r<s$. So, $\sup Q=1$.