I'm reading about Zorn's Lemma:
Let S be a partially ordered set whose chains all have an upper bound. Then S has a maximal element.
Is it true that for a chain to lack an upper bound it must be inifinte (necessary condition)? I'm thinking of the analogy with sequences that converge towords some limit that is not in the same space, here "the space" would be the one containing comparable elements.
Yes, a finite chain must have an upper bound, because it is either empty [and assuming your partial order is not empty] so every element is an upper bound, or it has a maximal element since every finite partial order has a maximal element.
But let me try and root out a much worse problem here. If a sequence is convergent, then its limit is always in the space. If the limit is not in the space, then the sequence is not convergent. Period.
Yes, you were probably trying to say something like $x_n=n$ for the real numbers, which would be a sequence converging to a point "outside the real numbers" (in this case $+\infty$). But this is not quite that, because the sequence $q_n=\sum_{k<n}\frac1{k!}$ is an increasing sequence of rational numbers, and while it is not convergent in $\Bbb Q$, it still has many upper bounds (e.g. $3$).
The problem with the notion of "converging to an element outside the space" is that one is very free to introduce an element $\star$ and claim that all the sequeneces converge to $\star$, even those that were already converging somewhere. And similarly, given any partial order $P$, we can always add a new element $\star$ and declare that all the members of $P$ are smaller than $\star$, making it the maximum.
So what can you do?
Well, if nothing else, forget your intuition and work with the definition. It's a fairly straight forward definition: a subset of $S$ is bounded if it has an upper bound (not necessarily strict!), an element which is greater-than-or-equal-to all the members of the subset.
A subset, therefore, is unbounded, if no such element exists. It doesn't have to be increasing. Take a set $X$ with at least two elements, and look at the partial order obtained by removing $X$ from $\mathcal P(X)$ with $\subseteq$. Then the collection of all the singletons is unbounded, there is no single member of $\mathcal P(X)$ which contains all the singletons except $X$ itself—which is not a member of our partial order. Here we didn't even
But if you focus your attention only to chains, then indeed the chains are increasing. But not necessarily forming a sequence. Their ordering can be fairly complex, fairly dense, and fairly incomprehensible to our imagination. Which is why we work with the definitions. So a chain is unbounded, well, if no single element is larger than (or equal to) all members of the chain.