here's a question I have for homework:
We have seen already the $\mathbb{R}$ is an archimedean field, meaning that $\mathbb{N}$ is not bounded in $\mathbb{R}$. Show that if $L \subset \mathbb{N}$ is infinite it is also not bounded in $\mathbb{R}$.
So, here's what I did and I just want to make sure I'm right:
Suppose by contradiction that $L$ is bounded. We know that in a complete ordered field every non-empty set that is bounded from above has a supremum, meaning: $$\forall x \in L:x \le sup(L)$$ But $(x+1)\in L$ and therefore we get $$\forall x \in L:x \le sup(L)-1$$ Contradicting the face that $sup(L)$ is the lowest upper bound. Proving that there's no lower bound is similar.
Here's my problem: I don't know the elements of $L$, only that $L$ is infinite. So even if $x+1$ is not in $L$, I know that ther'e another number that is in $L$ and is greater then $x$.
Two questions:
- Was I right about what I wrote?
- How do I formally define an element in $L$ that is greater then $x$, when I don't have the definition of $L$?
Thanks!