Maximal element of a linearly ordered set

First, this is the definition of Zorn's Lemma in my text. (Just in case it's not a standard way of defining it.)

Quote:

If A is a non-empty partially ordered set such that every chain in A (a sequence $\displaystyle a_1 \leq a_2 \leq ~..~\leq a_n$ where $\displaystyle a_i \in A$) has an upper bound in A then A contains a maximal element.

The problem is:

Quote:

Let $\displaystyle (A, \leq )$ be a linearly ordered set. The immediate successor of $\displaystyle a \in A$ (if it exists) is the least element in the set $\displaystyle \{ x \in A | a < x\}$. Prove that if A is well ordered by $\displaystyle \leq$, then at most one element of A has no immediate successor.

So. The proof.

If $\displaystyle (A, \leq)$ is a linear order then by Zorn's Lemma all chains (a, b, ..., z) in A have a maximal element. Since all elements of A are comparable in a linearly ordered set (and therefore all elements of A form a single chain) then there exists a single element z in A such that a < z for all $\displaystyle a \in A$. Thus z has no immediate successor. That is $\displaystyle \{ x | z < x \}$ is empty.

I think the proof looks good, but the trouble is I have what I think is a counter-example. Define the set A to be the natural numbers, N, linearly ordered by the usual definition of $\displaystyle \leq$ on the real numbers. I can find no (infinite) chain in N that has a maximal element.

My proof evidently missed something, but I can't tell what is missing.

-Dan