I have encountered a question while I was practicing the topic 'upper and lower bound of partially ordered sets'.
Let $\mathbb{Q}$ be the set of rational numbers. Let
$$
B = \{ x \in \mathbb{Q} : x < \sqrt[3]{3} \}
$$
Prove that $\sup{B}$ and $\inf{B}$ does not exists.
Here is my proof to prove that $\sup{B}$ does not exists –
Let $r$ be any upper bound of $B$.
Then $r$ is a rational number hence $r\not=\sqrt[3]3$
But there always exists
$p \in \mathbb{Q}$ such that $\sqrt[3]3 < p < r$ as $\displaystyle\lim_{m\to\infty}(r-\frac1m)=r$, hence for all large enough $m$ we have $\displaystyle\sqrt[3]3<r-\frac1m<r$ and we can take $\displaystyle p=r-\frac1m$ for some $m$.
Hence $r$ is an upper bound of $B$ but not the least one proving that $\sup{B}$ does not exists.
$■$
How can we prove that $\sup{B}$ does not exists?
Can there exist a more formal proof ?
A detailed answer would be helpful.
Best Answer
Let $r$ be an irrational number,
$B = \{ x \in \mathbb{Q} : x < r \}$ and
${q_n}$ a sequence within $B$ that converges to $r$.
Let $b$ be, within $\mathbb{Q}$, the least upper bound of $B$.
If $b < r$, then exists $n$ in $N$ with $b < q_n$. So r < b.
Let $\left\{p_n\right\}$ be a sequence rationals $> r$ that converges to $r$.
Thus some $n$ in $N$ with $p_n < b$.
As $p_n$ is an upper bound of $B$, a contradiction ensues.
Thusforth one concludes within the rationals that $\sup B$ is fantasy.
How to generate the sequences.
Since Q is a dense subset of R, every not empty open set will contain a rational.
Whence for all n in N, there is a rational $q_n$ in (r-1/n, r).