0
$\begingroup$

Consider the poset $(\mathbb{Q}, \leq)$. Is there a subset of $\mathbb{Q}$ with a finite number of elements that are upper-bounds?

I tried to prove this as follows:

Suppose $K \subset \mathbb{Q}$, if $x \in K$ then $x+1 \in \mathbb{Q}$ and $x+2 \in \mathbb{Q}$ and so goes. This implies that for any subset of $\mathbb{Q}$ there is an infinite number of elements that are upper-bounds.

$\endgroup$
  • 1
    $\begingroup$ You can have subsets of $\Bbb Q$ with no upper bounds. $\endgroup$ – Lord Shark the Unknown May 24 at 17:04
  • $\begingroup$ i think i'm not getting the definition of upperbounds then, if k={1} isn't j={2, 3...} upperbound of k since k ⊂ Q and J ⊂ Q? $\endgroup$ – Julien watson May 24 at 17:14
  • $\begingroup$ upper-bounds of what? $\endgroup$ – user661541 May 24 at 17:22
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$ – dantopa May 25 at 2:26
0
$\begingroup$

Just take $\mathbb{Q}$. It has no upper bounds. More generally, you can take any subset of $\mathbb{Q}$ that is unbounded above (there are many).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.