# Equality on Partial and Total Orders

If I have a partial order, is the following conclusion valid? $$a \geq c$$ $$b \geq c$$

Then, $$a = b$$

Does the result change if the partial order were to become a total order? Any and all explanations would be helpful. Cheers.

Follow-up: This seems quite obvious but I'm curious nonetheless. First, call S a finite subset of the naturals. Now define a and b with the same definition:
$$a, b \geq n$$ for all n in S. Does it follow that a = b by definition? From a method standpoint, is it enough to draw an equality between two elements by showing that those elements are defined in the same way?

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The only case where that is enough information is when $c$ is maximal. For, if $c$ were not maximal, then there exists $a>c$ satisfying $a\geq c$ and $c\geq c$, but we hypothesized $a\ne c$. – peoplepower Oct 3 '12 at 3:36

Nope, you don't have this conclusion at all. The only axioms you get on an arbitrary partial order are reflexivity, antisymmetry, and transitivity, and only antisymmetry permits you to deduce an equality. So you need to get $a\leq b$ and $b\leq a$. But you can't deduce anything at all about the relationship between $a$ and $b$ from the given information.
If you think of literally any nontrivial (some different elements are comparable) partial (or total) order on a set of more than one element, you'll immediately see this doesn't hold. Take $a\neq b$ and $a\geq b$. If your inference worked we'd have $a=b$ from $a\geq b$ and $b\geq b$, a contradiction.
For a concrete counterexample, on the real line $1\geq 0$ and $2\geq 0$, but $1\neq 2$.
To infer equality from defining two elements in the same way you have two things to check: that some element exists satisfying your definition, and that any element satisfying your definition is unique. The maximum of a total order, for instance, is unique whenever it exists, while if I "define" $x$ as a natural number greater than 0, $x$ could be any of infinitely many different numbers. Look at your proposed definition: $a,b \geq n$ for every $n$. Does there exist a natural number at least as big as every natural number? No, so your definition in fact specifies no element at all. – Kevin Carlson Oct 3 '12 at 3:52