# Are there subsets of $\mathbb{Q}$ with a finite number of upper-bounds?

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.

• You can have subsets of $\Bbb Q$ with no upper bounds. – Lord Shark the Unknown May 24 at 17:04
• 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? – Julien watson May 24 at 17:14
• upper-bounds of what? – user661541 May 24 at 17:22
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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).