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text: '\nTry these exercises to test your poset knowledge.\n\n# Corporate Ladder\n\nImagine a company with five distinct roles: CEO, marketing manager, marketer, IT manager, and IT worker. At this company, if the CEO gives an order, everyone else must follow it. If an order comes from an IT manager, only IT workers are obligated to follow it. Similarly, if an order comes from a marketing manager, only marketers are obligated to follow it. Nobody is obligated to follow orders from marketers or IT workers.\n\nDo the workers at this company form a poset under the "obligated to follow orders from" relation?\n\n%%%hidden(Show solution):\nWhile not technically a poset due to its lack of reflexivity, it is pretty close. It is actually a strict ordering, whose underlying partial order could be obtained by making the reasonable assumption that each worker will follow her own orders.\n%%%\n\n# Bag Inclusion\n\nWe can define a notion of [power_set power sets] for [3jk bags] as follows. Let $X$ be a set, then we use $\\mathcal{M}(X)$ to denote the set of all bags containing elements of $X$. Let $A \\in \\mathcal{M}(X)$. The multiplicity function of $A$, $1_A : X \\rightarrow \\mathbb N$ maps each element of $X$ to the number of times that element occurs in $A$. We can use multiplicity functions to define a inclusion relation $\\subseteq$ for bags. For $A, B \\in \\mathcal M(X)$, we write $A \\subseteq B$ whenever for all $x \\in X$, $1_A(x) \\leq 1_B(x)$.\n\nDoes $\\mathcal{M}(X)$ form a poset under the bag inclusion relation $\\subseteq$? If so, prove it. Otherwise, show that it does not satisfy one of the three poset properties. \n\n# Duality\n\nGive the dual of the following proposition. \n\nFor all posets $P$ and all $p, q \\in P$, $q \\prec p$ implies that $\\{ r \\in P~|~r \\leq p \\}$ is a superset of $\\{ r \\in P~|~r \\leq q\\}$.\n\n%%hidden(Show solution):\nFor all posets $P$ and all $p, q \\in P$, $q \\succ p$ implies that $\\{ r \\in P~|~r \\geq p \\}$ is a superset of $\\{ r \\in P~|~r \\geq q\\}$ (where $q \\succ p$ means $p \\prec q$).\n%%\n\n# Hasse diagrams\n\nLet $X = \\{ x, y, z \\}$. Draw a Hasse diagram for the poset $\\langle \\mathcal P(X), \\subseteq \\rangle$ of the power set of $X$ ordered by inclusion.\n\n%%hidden(Show solution):\n\n\n%%%comment:\ndot source:\n\ndigraph G {\n node [width = 0.1, height = 0.1]\n edge [arrowhead = "none"]\n e [label = "{}"]\n x [label = "{x}"]\n y [label = "{y}"]\n z [label = "{z}"]\n xy [label = "{x,y}"]\n xz [label = "{x,z}"]\n yz [label = "{y,z}"]\n xyz [label = "{x,y,z}"]\n\n rankdir = BT;\n e -> x\n e -> y\n e -> z\n x -> xy\n x -> xz\n y -> xy\n y -> yz\n z -> xz\n z -> yz\n xy -> xyz\n xz -> xyz\n yz -> xyz\n}\n%%%\n\n%%\n\n#Hasse diagrams (encore)\n\nIs it possible to draw a Hasse diagram for any poset?\n\n%%hidden(Show solution):\nNote that our description of Hasse diagrams made use of the covers relation $\\prec$. The covers relation, however, is not adequate to describe the structure of many posets. Consider the poset $\\langle \\mathbb R, \\leq \\rangle$ of the real numbers ordered by the standard comparison $\\leq$. We have $0 < 1$, but how would we convey that with a Hasse diagram? The problem is that $0$ has no covers, even though it is not a maximal element in $\\mathbb R$. In fact, for any $x \\in \\mathbb R$ such that $x > 0$, we can find a $y \\in \\mathbb R$ such that $0 < y < x$. This "infinite density" of $\\mathbb R$ makes it impossible to depict using a Hasse diagram.\n%%\n\n\n\n\n',
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