# Understanding Zorn's lemma.

A lot of authors assume Zorn's lemma. I am told it is not an obvious mathematical fact, but I am having problems understanding why that is.

Zorn's lemma states that if every chain in a partially ordered set $P$ has an upper bound in $P$, then it has a maximal element.

Both the maximal element and the upper bound of every chain have to belong to $P$.

Say any one chain of $P$ has an upper bound in $P$. Won't that be the maximal element by default. Is there some fine difference between maximal element and upper bound which I am unable to grasp?

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"Then "" it"" has a maximal element"...""it"" being here the partially ordered set P , so no: the maximal element is not necessarily in the chain. –  DonAntonio Feb 16 at 11:10
uhmm, that's not exactly Zorn's Lemma. Zorn's Lemma states that (assumming the statement about the chains) the partially ordered set has at least one maximal element, not the chain. ('it has a maximal element is rather ambigously formulated'). –  Thomas Feb 16 at 11:11
Why would the upper bound of the chain not necessarily be the maximal element of the partially ordered set? Say $a\in P$ is the upper bound of a chain. If we could find an element greater than $a$ in $P$, clearly $a$ would not be the upper bound of that chain. –  Ayush Khaitan Feb 16 at 11:13
I might be confused. Say we have a chain $b_1\leq b_2\leq b_3$. If $b_4\in P$ such that $b_3\leq b_4$, would we be required to add $b_4$ to the chain, or would that not be compulsory? As in, are we not required to keep on adding elements to a chain as long as we keep getting bigger upper bounds? –  Ayush Khaitan Feb 16 at 11:14
Ayush, that is not what an upper bound means. An upper bound of a set $S$ is any element $x$ which is greater than every element of $s$; that is it satisifies $x\geq s$ for all $s\in S$. So any element larger than $x$ is also an upper bound of $S$, but that does not mean they "discredit" the fact that $x$ is an upper bound. Or to phrase it as an answer to your other question: no, it is not compulsory to make chains as large as possible. –  Eric Stucky Feb 16 at 11:16
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As others have pointed out, your statement of Zorn's lemma is ambiguous, and the one reasonable interpretation in English is not what is wanted (and in fact makes the statement false). This might be the source of your confusion. However, to address your questions at face value:

A maximal element of $P$ is an element such that no other is greater than it. That is, $x$ is maximal $p\not\geq x$ for any $p\in P$. Of course, because $P$ is not necessarily totally ordered, this is weaker than the notion of a maximum element, which demands that every other element is less than it.

Upper bounds require a bit more subtlety to define: Given a partially ordered set $P$ and a subset $S\subseteq P$, we say that $x$ is an upper bound of $S$ if every element of $S$ is less than $x$. Therefore, an upper bound is more similar to a maximum element than a maximal element. But it is importantly different because the upper bound does not need to belong to $S$ itself, just to $P$.

(A classic example for when this distinction is important is that open sets on the real line never have maximum elements, but they could have upper bounds.)

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Now, to your question, the assumption just says that a chain has an upper bound, which is an element of the partial order which is larger than the members of the chain. A maximal element, however, is an element which no point is strictly above it. Consider the interval $[-1,1]$ in the real numbers, the set of negative numbers has an upper bound, $0$. But this is not a maximal element.