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text: '[summary: \nA **complete lattice** is a [3rb poset] that is closed under arbitrary [3rc joins and meets]. A complete lattice, being closed under arbitrary joins and meets, is closed in particular under binary joins and meets. A complete lattice is thus a specific type of [46c lattice], and hence satisfies [3h4 associativity], [3jb commutativity], idempotence, and absorption of joins and meets. Complete lattices can be equivalently formulated as posets which are closed under arbitrary joins; it then follows that complete lattices are closed under arbitrary meets as well.\n\nBecause complete lattices are closed under all joins, a complete lattice $L$ must contain both $\\bigvee \\emptyset$ and $\\bigvee L$ as elements. Since $\\bigvee \\emptyset$ is a lower bound of $L$ and $\\bigvee L$ is an upper bound of $L$, complete lattices are bounded. \n]\n\nA **complete lattice** is a [3rb poset] that is closed under arbitrary [3rc joins and meets]. A complete lattice, being closed under arbitrary joins and meets, is closed in particular under binary joins and meets; a complete lattice is thus a specific type of [46c lattice], and hence satisfies [3h4 associativity], [3jb commutativity], idempotence, and absorption of joins and meets.\n\nComplete lattices can be equivalently formulated as posets which are closed under arbitrary joins; it then follows that complete lattices are closed under arbitrary meets as well.\n\n%%hidden(Proof):\nSuppose that $P$ is a poset which is closed under arbitrary joins. Let $A \\subseteq P$. Let $A^L$ be the set of lower bounds of $A$, i.e. the set $\\{ p \\in P \\mid \\forall a \\in A. p \\leq a \\}$. Since $P$ is closed under joins, we have the existence of $\\bigvee A^L$ in $P$. We will now show that $\\bigvee A^L$ is the meet of $A$.\n\nFirst, we show that $\\bigvee A^L$ is a lower bound of $A$. Let $a \\in A$. By the definition of $A^L$, $a$ is an upper bound of $A^L$. Because $\\bigvee A^L$ is less than or equal to any upper bound of $A^L$, we have $\\bigvee A^L \\leq a$. $\\bigvee A^L$ is therefore a lower bound of $A$.\n\nNow we will show that $\\bigvee A^L$ is greater than or equal to any lower bound of $A$. Let $p \\in P$ be a lower bound of $A$. Then $p \\in A^L$. Because $\\bigvee A^L$ is an upper bound of $A^L$, we have $p \\leq \\bigvee A^L$.\n%%\n\nComplete lattices are bounded\n========================\n\nAs a consequence of closure under arbitrary joins, a complete attice $L$ contains both $\\bigvee \\emptyset$ and $\\bigvee L$. The former is the least element of $L$ satisfying a vacuous set of constraints; every element of $L$ satisfies a vacuous set of constraints, so this is really the minimum element of $L$. The latter is an upper bound of all elements of $L$, and so it is a maximum. A lattice with both minimum and maximum elements is called bounded, and as this discussion has shown, all complete lattices are bounded.\n\nBasic examples\n=============\n\nFinite Lattices\n------------------------\n\nThe collection of all subsets of a finite lattice coincides with its collection of finite subsets. A finite lattice, being a finite poset that is closed under finite joins, is then necessarily closed under arbitrary joins. All finite lattices are therefore complete lattices.\n\nPowersets\n------------------\n\nFor any set $X$, consider the poset $\\langle \\mathcal P(X), \\subseteq \\rangle$ of $X$'s powerset ordered by inclusion.\nThis poset is a complete lattice in which for all $Y \\subset \\mathcal P(X)$, $\\bigvee Y = \\bigcup Y$.\n\nTo see that $\\bigvee Y = \\bigcup Y$, first note that because union contains all of its constituent sets, for all $A \\in Y$, $A \\subseteq \\bigcup Y$. This makes $\\bigcup Y$ an upper bound of $Y$. Now suppose that $B \\in \\mathcal P(X)$ is an upper bound of $Y$; i.e., for all $A \\in Y$, $A \\subseteq B$. Let $x \\in \\bigcup Y$. Then $x \\in A$ for some $A \\in Y$. Since $A \\subseteq B$, $x \\in B$. Hence, $\\bigcup Y \\subseteq B$, and so $\\bigcup Y$ is the least upper bound of $Y$.\n\n\nThe Knaster-Tarski fixpoint theorem\n=============================\n\nSuppose that we have a poset $X$ and a [5jg monotone function] $F : X \\to X$. An element $x \\in X$ is called $F$**-consistent** if $x \\leq F(x)$ and is called $F$**-closed** if $F(x) \\leq x$. A fixpoint of $F$ is then an element of $X$ which is both $F$-consistent and $F$-closed.\n\nLet $A \\subseteq X$ be the set of all fixpoints of $F$. We are often interested in the maximum and minimum elements of $A$, if indeed it has such elements. Most often it is the minimum element of $A$, denoted $\\mu F$ and called the **least fixpoint** of $F$, that holds our interest. In the deduction system example from [5lf], the least fixpoint of the deduction system $F$ is equal to the set of all judgments which can be proven without assumptions. Knowing $\\mu F$ may be first step toward testing a judgment's membership in $\\mu F$, thus determining whether or not it is provable. In less pedestrian scenarios, we may be interested in the set of all judgments which can be proven without assumption using *possibly infinite proof trees*; in these cases, it is the **greatest fixpoint** of $F$, denoted $\\nu F$, that we are interested in.\n\nNow that we've established the notions of the least and greatest fixpoints, let's try an exercise. Namely, I'd like you to think of a lattice $L$ and a monotone function $F : L \\to L$ such that neither $\\mu F$ nor $\\nu F$ exists.\n\n%%hidden(Show solution):\nLet $L = \\langle \\mathbb R, \\leq \\rangle$ and let $F$ be the identity function $F(x) = x$. $x \\leq y \\implies F(x) = x \\leq y = F(y)$, and so $F$ is monotone. The fixpoints of $F$ are all elements of $\\mathbb R$. Because $\\mathbb R$ does not have a maximum or minimum element, neither $\\mu F$ nor $\\nu F$ exist.\n%%\n\nIf that was too easy, here is a harder exercise: think of a complete lattice $L$ and monotone function $F : L \\to L$ for which neither $\\mu F$ nor $\\nu F$ exist.\n\n%%hidden(Show solution):\nThere are none. :p\n%%\n\nIn fact, every monotone function on a complete lattice has both least and greatest fixpoints. This is a consequence of the **Knaster-Tarski fixpoint theorem**.\n\n**Theorem (The Knaster-Tarski fixpoint theorem)**: Let $L$ be a complete lattice and $F : L \\to L$ a monotone function on $L$. Then $\\mu F$ exists and is equal to $\\bigwedge \\{x \\in L \\mid F(x) \\leq x\\}$. Dually, $\\nu F$ exists and is equal to $\\bigvee \\{x \\in L \\mid x \\leq F(x) \\}$.\n\n%%hidden(Proof):\nWe know that both $\\bigwedge \\{x \\in L \\mid F(x) \\leq x\\}$ and $\\bigvee \\{x \\in L \\mid F(x) \\leq x \\}$ exist due to the closure of complete lattices under meets and joins. We therefore only need to prove that $\\bigwedge \\{x \\in L \\mid F(x) \\leq x\\}$ is a fixpoint of $F$ that is less or equal to all other fixpoints of $F$. The rest follows from duality.\n\nLet $U = \\{x \\in L \\mid F(x) \\leq x\\}$ and $y = \\bigwedge U$. We seek to show that $F(y) = y$. Let $V$ be the set of fixpoints of $F$. Clearly, $V \\subseteq U$. Because $y \\leq u$ for all $u \\in U$, $y \\leq v$ for all $v \\in V$. In other words, $y$ is less than or equal to all fixpoints of $F$.\n\nFor $u \\in U$, $y \\leq u$, and so $F(y) \\leq F(u) \\leq u$. Since $F(y)$ is a lower bound of $U$, the definition of $y$ gives $F(y) \\leq y$. Hence, $y \\in U$. Using the monotonicity of $F$ on the inequality $F(y) \\leq y$ gives $F(F(y)) \\leq F(y)$, and so $F(y) \\in U$. By the definition of $y$, we then have $y \\leq F(y)$. Since we have established $y \\leq F(y)$ and $F(y) \\leq y$, we can conclude that $F(y) = y$.\n%%\n \n\nTODO: Prove the knaster tarski theorem and explain these images\n\nadd !'s in front of the following two lines\n[A Knaster-Tarski-style view of complete latticess](http://i.imgur.com/wKq74gC.png)\n[More Knaster-Tarski-style view of complete latticess](http://i.imgur.com/AYKyxlF.png)\n\n\n\n\n\n\n\n\n\n',
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