Foundations of Advanced Mathematics

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We follow [Bir48] and [LP98] in part. As a preliminary, we prove the following two facts: (1) $x\bigsqcup x=x$ for all $x\in A\text{;}$ and (2) $x\bigsqcup y=x$ if and only if $x\sqcap y=y\text{,}$ for all $x,y\in A\text{.}$ Let $x,y,z\in A\text{.}$ Using both parts of Property (c), we see that $x\bigsqcup x=x\bigsqcup (x\sqcap (x\bigsqcup x))=x\text{,}$ which proves Fact (1). Suppose that $x\bigsqcup y=x\text{.}$ Then by Properties (a) and (c) we see that $x\sqcap y=(x\bigsqcup y)\sqcap y=y\sqcap (y\bigsqcup x)=y\text{,}$ which proves one of the implications in Fact (2); a similar argument proves the other implication, and we omit the details.

We now show that $(A,\preccurlyeq )$ is a poset. Because $x\bigsqcup x=x$ by Fact (1), it follows from the definition of $\preccurlyeq$ that $x\preccurlyeq x\text{.}$ Hence $\preccurlyeq$ is reflexive. Now suppose that $x\preccurlyeq y$ and $y\preccurlyeq z\text{.}$ Then $x\bigsqcup y=x$ and $y\bigsqcup z=y\text{.}$ By Property (b) we see that $x\bigsqcup z=(x\bigsqcup y)\bigsqcup z=x\bigsqcup (y\bigsqcup z)=x\bigsqcup y=x\text{.}$ It follows that $x\preccurlyeq z\text{.}$ Therefore $\preccurlyeq$ is transitive. Next, suppose that $x\preccurlyeq y$ and $y\preccurlyeq x\text{.}$ Then $x\bigsqcup y=x$ and $y\bigsqcup x=y\text{.}$ It follows from Property (a) that $x=y\text{.}$ Therefore $\preccurlyeq$ is antisymmetric. We conclude that $(A,\preccurlyeq )$ is a poset.

Finally, we show that $\bigsqcup$ and $\sqcap$ are the meet and join of $(A,\preccurlyeq )\text{,}$ respectively. It will then follow from this fact that meet and join always exist for any two elements of $A\text{,}$ and hence $(A,\preccurlyeq )$ is a lattice. We start with $\bigsqcup \text{.}$ Using Property (b) and Fact (1) we see that $(x\bigsqcup y)\bigsqcup y=x\bigsqcup (y\bigsqcup y)=x\bigsqcup y\text{.}$ Hence $x\bigsqcup y\preccurlyeq y\text{.}$ Because $x\bigsqcup y=y\bigsqcup x$ by Property (a), a similar argument shows that $x\bigsqcup y\preccurlyeq x\text{.}$ Therefore $x\bigsqcup y$ is a lower bound of $\{x,y\}\text{.}$ Now suppose that $z\in A$ is a lower bound of $\{x,y\}\text{.}$ Then $z\preccurlyeq x$ and $z\preccurlyeq y\text{,}$ and therefore $z\bigsqcup x=z$ and $z\bigsqcup y=z\text{.}$ By Property (b) we see that $z\bigsqcup (x\bigsqcup y)=(z\bigsqcup x)\bigsqcup y=z\bigsqcup y=z\text{.}$ Hence $z\preccurlyeq (x\bigsqcup y)\text{.}$ It follows that $x\bigsqcup y$ is the greatest lower bound of $\{x,y\}\text{,}$ which means that $x\bigsqcup y$ is the meet of $x$ and $y\text{.}$

We now turn to $\sqcap \text{.}$ By Property (c) we know that $x\bigsqcup (x\sqcap y)=x\text{.}$ Hence $x\preccurlyeq x\sqcap y\text{.}$ Because $x\sqcap y=y\sqcap x$ by Property (a), a similar argument shows that $y\preccurlyeq x\sqcap y\text{.}$ Hence $x\sqcap y$ is an upper bound of $\{x,y\}\text{.}$ Now suppose that $w\in A$ is an upper bound of $\{x,y\}\text{.}$ Then $x\preccurlyeq w$ and $y\preccurlyeq w\text{,}$ and therefore $x\bigsqcup w=x$ and $y\bigsqcup w=y\text{.}$ By Fact (2) we deduce that $x\sqcap w=w$ and $y\sqcap w=w\text{.}$ Property (b) then implies that $(x\sqcap y)\sqcap w=x\sqcap (y\sqcap w)=x\sqcap w=w\text{.}$ Hence $(x\sqcap y)\bigsqcup w=x\sqcap y$ by Fact (2). Therefore $x\sqcap y\preccurlyeq w\text{.}$ It follows that $x\sqcap y$ is the least upper bound of $\{x,y\}\text{,}$ which means that $x\sqcap y$ is the join of $xandy\text{.}$

Let $C=\{x\in L|x\preccurlyeq f(x)\}\text{.}$ Observe that $L$ is non-empty because it is a poset, and all posets are assumed to be non-empty. Let $m$ be the greatest lower bound of $L\text{,}$ which exists by hypothesis. Then $m$ is a lower bound of $L\text{,}$ and therefore $m\preccurlyeq x$ for all $x\in L\text{.}$ In particular, we see that $m\preccurlyeq f(m)\text{.}$ It follows that $m\in C\text{,}$ and so $C$ is non-empty.

Let $a$ be the least upper bound of $C\text{.}$ Let $x\in C\text{.}$ Then $a$ is an upper bound of $C\text{,}$ and therefore $x\preccurlyeq a\text{.}$ Using the definition of $C$ and the fact that $f$ is an order homomorphism, we deduce that $x\preccurlyeq f(x)\preccurlyeq f(a)\text{.}$ It follows that $f(a)$ is an upper bound for $C\text{.}$ Because $a$ is the least upper bound of $C\text{,}$ we deduce that $a\preccurlyeq f(a)\text{.}$ Because $f$ is an order homomorphism, it follows that $f(a)\preccurlyeq f(f(a))\text{.}$ Hence $f(a)\in C\text{,}$ and therefore $f(a)\hookrightarrow a\text{,}$ because $a$ is an upper bound of $C\text{.}$ By antisymmetry, we deduce that $f(a)=a\text{.}$