# Proving the a set $L \subset \mathbb{N}$ is not bounded in $\mathbb{R}$

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:

1. Was I right about what I wrote?
2. How do I formally define an element in $L$ that is greater then $x$, when I don't have the definition of $L$?

Thanks!

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$\Bbb Q$ is an ordered field in which sets bounded above do not necessarily have suprema. What you’re using there is the fact that the reals are completely ordered. –  Brian M. Scott Mar 19 '12 at 17:42
Changed it to complete ordered field –  yotamoo Mar 19 '12 at 17:56
You cannot automatically claim that $\sup (L)$ (if it exists) is an element of $L$ (you would have to make a further argument that this would be true). (As for your second question, recall that the set of natural numbers has a very nice property.) –  Arthur Fischer Mar 19 '12 at 18:02
I don't think I claimed anywhere the $sup(L) \in L$, did I? –  yotamoo Mar 19 '12 at 18:48
@yotamoo: My apologies, I seem to have misread/read too quickly. Upon a closer inspection (where's my brain today?) no, you don't assume that $\sup (L) \in L$. You do assume that $x \in L \rightarrow x+1 \in L$, which would imply that (assuming $L \neq \emptyset$) $L$ is a final segment of $\mathbb{N}$. –  Arthur Fischer Mar 19 '12 at 18:56

Hint: Suppose that $L\subseteq \mathbb{N}$ is bounded in $\mathbb{R}$. Then, there exists $y\in \mathbb{R}$ such that $x\leq y$ for all $x\in L$. Let $n=\lceil y \rceil$ (where $\lceil \cdot \rceil$ denotes the ceiling function). Then $L\subseteq \{1,2,\cdots,n\}$ (why?), and hence $L$ is finite.