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Exercise: Let $L$ be a lattice in which every chain has an upper bound. Prove that $L$ has a unique maximal element.

I said:

p1) Suppose $L$ has two maximal elements. (Proof by contradiction is where I'm headed)

p2) Since $L$ is a lattice, then every two elements in the set have a least upper bound (lub) and a greatest lower bound (glb).

p3) Suppose there exists a $2$-element lattice $\{x_\alpha, x_\beta\}\in L$ such that $x_\alpha <x_\beta$ or $x_\beta<x_\alpha$ or they're equal. In this way we have also defined the requirement for a chain.

p4) Since $L$ has two maximal elements, $x_\alpha$ has to equal $x_\beta$.

p5) But that's impossible, because there would be no glb/lub defined in that lattice.

p6) The same can be said for three element subsets - if $x_\gamma\le x_\alpha$ and $x_\gamma\le x_\beta$, $x_\alpha$ and $x_\beta$ can't be equal - or else there is no glb/lub between $x_\alpha$ and $x_\beta$. If $x_\gamma$ is greater than both $x_\alpha$ and $x_\beta$, then $x_\gamma$ is maximal, contradictory to the premises.

Q) Thus $L$ has a unique maximal element.

I would gladly consider some criticism of my proof. Thanks!

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Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. – Zev Chonoles Jul 9 '13 at 0:54
You're making this much too complicated. Sometimes the best thing is to take a break and start over from scratch in order to find an approach that is simpler and easier to find than the one you have attempted. – dfeuer Jul 9 '13 at 1:17
up vote 2 down vote accepted

The biggest problem is that you’ve not shown that $L$ has any maximal element at all: you’ve only tried to show the other half of the result, i.e., that $L$ cannot have more than one maximal element. I realize that you may have omitted the existence part on the grounds that it’s a trivial application of Zorn’s lemma, so I’ll concentrate on the argument that you did present.

Your proof of uniqueness is horribly overcomplicated and not very clearly organized. The first point is fine, but you’ll make matters much clearer if you name the supposed maximal elements right at the start:

Suppose that $x$ and $y$ are maximal elements of $L$.

Your second point is premature, and your third is irrelevant. Your fourth is unclear, because you never said that $x_\alpha$ and $x_\beta$ are the maximal elements of the first point; you also haven’t clearly stated any argument to justify the conclusion that $x_\alpha=x_\beta$, though I think that you probably have the right general idea in mind. Now is actually the time to bring in your second point, but don’t clutter things up with the irrelevant part of it:

$L$ is a lattice, so $x$ and $y$ have a least upper bound $z$. Then $x\le z$, and since $x$ is maximal in $L$, it follows that $x=z$. Similarly, $y\le z$, and $y$ is maximal in $L$, so $y=z$. Thus, $x=y$, and $L$ has exactly one maximal element.

Note that it need not be cast as a proof by contradiction: the foregoing argument shows directly that any two maximal elements of $L$ are actually the same element.

I’ll comment briefly on the remaining points of your argument. The fifth is simply wrong: if $x_\alpha=x_\beta$, the least upper bound of $\{x_\alpha,x_\beta\}$ is simply $x_\alpha$. The sixth is partly wrong (for much the same reason) and wholly unnecessary.

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Why can't the lub also be x(beta)? – MathApprentice Jul 9 '13 at 1:49
@MathApprentice: It obviously is $x_\beta$: $x_\alpha$ and $x_\beta$ are the same element of $L$! – Brian M. Scott Jul 9 '13 at 1:50
Thanks @Brian. You were a great help. Here is my new proof: p1) Suppose L has two maximal elements x(alpha) and x(beta). p2) Then both x(alpha) and x(beta) have upper bound z. p3) Since xalpha and xbeta are maximal, xalpha <= z and xbeta <= z only hold for the equivalence relation =, so xalpha = xbeta = z. Q) Since xalpha and xbeta are equal, L has one maximal element. – MathApprentice Jul 9 '13 at 2:04

By Zorn's lemma, $L$ has a maximal element.

Let $x$ and $y$ be maximal elements of $L$.

Then as you noted, $\{x,y\}$ has a least upper bound $m\in L$.

By the definition of an upper bound, $x \le m$ and $y\le m$.

By the definition of a maximal element, $x = m$ and $y=m$.

Thus $x=y$.

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Yes, but you don't make any contradictions in this proof. For example, you don't elaborate on x and y's relation. I'm also trying to make it as detailed as possible since I have to present this to a class. My proof is not too different from yours. – MathApprentice Jul 9 '13 at 1:11
There is no need for a contradiction per se. There is a maximal element and any two maximal elements are equal. That's the same as saying there is a unique maximal element. – dfeuer Jul 9 '13 at 1:13
But that's a singleton, and lattices can't be singletons. So that means that there can't be two maximal elements, or two minimal elemenets - so there can only be a unique max/min element. – MathApprentice Jul 9 '13 at 1:17
What on Earth are you talking about? – dfeuer Jul 9 '13 at 1:18
A lattice is a poset in which every two elements have a lub/glb. – MathApprentice Jul 9 '13 at 1:19

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