auto-update(nvim): 2025-02-06 00:26:56

content/Math/pstat.md

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+---
+id: pstat
+aliases: []
+tags: []
+---
+
+

documents/by-course/math-8/pset-4/main.typ

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+#import "@youwen/zen:0.1.0": *
+
+#show: zen.with(
+  title: "Homework 4",
+  author: "Youwen Wu",
+)
+
+#set heading(numbering: none)
+
+#set par(first-line-indent: 0pt, spacing: 1em)
+
+#let nonzero = $ZZ_(!=0)$
+
+Problems:
+
+2.2: \#1dfh, 2bdf, 5, 6d, 7q, 8h, 9b, 10e, 11b, 13a, 15b, 16
+
+2.3: \#1bhLmn, 2bhLmn, 3, 12, 16d, 18bd
+
+= 2.2
+
+*1d.*
+
+$A - (B - C) = {1,3,5,7,9}$
+
+*1f.*
+
+$A union (C sect D) = {1,2,3,5,7,8,9}$
+
+*1h.*
+
+${1,5,7}$
+
+*2b.*
+
+$[2,8)$
+
+*2d.*
+
+$[3,6]$
+
+*2f.*
+
+$(6,8)$
+
+*5.*
+
+Only $C$ and $D$ are disjoint.
+
+*6d.*
+
+$A = {1, 2}, B = {3, 4}, C = {2, 3}$
+
+*7q.*
+
+Claim: if $A subset.eq B$, then $A union C subset.eq B union C$.
+
+#proof[
+  Suppose $A subset.eq B$. Therefore $forall a in A, a in B$. This implies that
+  $forall a in A, a in A union C$, and $forall a in A, a in B union C$.
+  Note that $forall c in C$, both $c in A union C$ and $c in B union C$. Since
+  $forall x in A union C$, either $x in A$ or $x in C$, and either way $x in B$
+  and $x in C$, then $A union B subset.eq A union C$.
+]
+
+*8h.*
+
+Claim: $(A union B)^c = A^c sect B^c$.
+
+#proof[
+  For all $x$ in $(A union B)^c$, $x$ is not in $A union B$. Therefore $x$ is
+  not in $A$ and $x$ is not in $B$. In other words $(A union B)^c$ is comprised of all $x$ not in $A$ and not in $B$, which are all the elements in both $A^c$ and $B^c$, which is $A^c sect B^c$.
+]
+
+*9b.*
+
+Claim: If $A subset.eq B union C$ and $A sect B = emptyset$, then $A subset.eq C$.
+
+#proof[
+  Suppose $A subset.eq B union C$. Then $forall a in A$, either $a in B$ or $a
+  in C$. However $a$ is never in $B$, as $A$ and $B$ are disjoint and $a in B$
+  would be a contradiction of this fact. So in fact the only case is $a in C$.
+  Therefore we have $forall a in A, a in C$, which is the definition of $A
+  subset.eq C$.
+]
+
+*10e.*
+
+Claim: if $A union B subset.eq C union D$, $A sect B = emptyset$, and $C subset.eq A$, then $B subset.eq D$.
+
+#proof[
+  Suppose $C subset.eq A$. Then $forall c in C, c in A$. Since $A$ and $B$ are
+  disjoint, $forall b in B, b in.not C$. Because we assume $A union B subset.eq
+  C union D$, $forall x in A union B, x in C union D$. Then this implies
+  $forall a in A$ and $forall b in B$, either $a in C$, or $b in C$, or $a in
+  D$, or $b in D$. However we established earlier that $forall b in B, b in.not
+  C$. This means that $forall b in B, b in D$.
+  Therefore $B subset.eq D$.
+]
+
+*11b.*
+
+Claim: if $A sect C subset.eq B sect C$, then $A subset.eq B$.
+
+Counterexample: $A = {1,2,3}, B = {2,3,4}, C = {2,3,4}$.
+
+*13a.*
+
+(I wrote a Vim macro to compute these because it was so annoying).
+$
+  A times B = \
+  {(1,a), (1, e), (1, k), (1, n), (1, r), (3,a), (3, e), (3, k), (3, n), (3, r), (5,a), (5, e), (5, k), (5, n), (5, r)} \
+  B times A = \
+  {(a,1), (e, 1), (k, 1), (n, 1), (r, 1), (a,3), (e, 3), (k, 3), (n, 3), (r, 3), (a,5), (e, 5), (k, 5), (n, 5), (r, 5)} \
+$
+
+*15b.*
+
+Claim: $A times emptyset = emptyset$.
+
+#proof[
+  Let there be a set $X$ such that $A times emptyset = X$. Then any element in
+  $X$ is an ordered pair $(a,b)$ such that $a in A$ and $b in emptyset$.
+  But $b in emptyset$ is a contradiction no ordered pair can exist. Therefore
+  $X$ contains no elements and $X = emptyset$, so $A times emptyset =
+  emptyset$.
+]
+
+*16a.*
+
+$
+  A = {1}, B = {2}, C = {3}, D = {4}
+$
+
+*16b.*
+
+$
+  A = {1}, B = {2}, C = {2}
+$
+
+*16c.*
+
+$
+  A = {1}, B = {2}, C = {3}
+$
+
+= 2.3
+
+*1b.*
+
+$
+  union.big_(A in cal(A)) A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} \
+  sect.big_(A in cal(A)) A = emptyset
+$
+
+*1h.*
+
+$
+  union.big_(A in cal(A)) A = [-pi, infinity) \
+  sect.big_(A in cal(A)) A = [-pi, 0]
+$
+
+*1L.*
+
+$
+  union.big_(L in cal(l)) L = (-infinity, infinity) \
+  sect.big_(L in cal(l)) L = emptyset
+$
+
+*1m.*
+
+$
+  union.big_(A in cal(A)) A = (-infinity, infinity) - ZZ \
+  sect.big_(A in cal(A)) A = emptyset
+$
+
+*1n.*
+
+$
+  union.big_(D in cal(D)) D = (-infinity, 1) \
+  sect.big_(D in cal(D)) D = (-1, 0]
+$
+
+*2b.*
+
+No.
+
+*2h.*
+
+No.
+
+*2L.*
+
+Yes.
+
+*2m.*
+
+Yes.
+
+*2n.*
+
+No.
+
+*3a.*
+
+Claim: let $cal(A)$ be a family of sets, for every set B in the family
+$cal(A)$, $B subset.eq union.big _(A in cal(A)) A$
+
+#proof[
+  By the definition of the union of sets, $forall b in B$, $b in union.big _(A
+  in cal(A)) A$. Therefore $B subset.eq union.big _(A in cal(A))$.
+]
+
+*3b.*
+
+Claim: let $cal(A)$ be a nonempty family of sets and $B$ be a set. Then if $A
+subset.eq B$ for all $A in cal(A)$, then $union.big _(A in cal(A)) A subset.eq
+B$.
+
+#proof[
+  Suppose $A subset.eq B$ for all $A in cal(A)$. Then $forall a in A, a in B$
+  for all $A in cal(A)$. Then note that $forall  in union.big _(A in cal(A))
+  A$, there exists some $A$ with $x in A$. This implies that any element of
+  $union.big _(A in cal(A)) A$ is an element of some $A in cal(A)$. By our
+  earlier assumption this implies that any element in $union.big _(A in cal(A))
+  A$ is an element of $B$. This is the definition of $union.big _(A in cal(A))
+  A subset.eq B$.
+]
+
+*12a.*
+
+$
+  cal(A) = {{1, x} : x in {2, 3, dots.c, 20}}
+$
+
+*12b.*
+
+$
+  cal(B) = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}}
+$
+
+*12c.*
+
+$
+  cal(l) = {{x} : x in {1, 2, 3, dots.c, 20}}
+$
+
+*16d.*
+
+Claim: $sect.big _(i = 1) ^infinity A_i subset.eq sect.big _(i=k) ^m A_i$.
+
+#proof[
+  Note that
+
+  $
+    sect.big_(i = 1)^infinity A_i = (sect.big_(i = 1)^(k-1) A_i) sect (sect.big_(i = k)^m A_i) sect (sect.big_(i = m + 1)^infinity A_i)
+  $
+
+  This implies that the set $sect.big _(i=1) ^infinity A_i$ contains only
+  elements belonging to (by the definition of the set intersection) $sect.big
+  _(i=k) ^m A_i$. Therefore $sect.big _(i=1) ^infinity A_i$ is indeed a subset
+  of $sect.big _(i=k) ^m A_i$.
+]
+
+*18b.*
+
+$
+  {(-infinity ,infinity), (-infinity, 1]}
+$
+
+*18d.*
+
+${{1, 2, 3}, {1, 2}, {1}, emptyset}$

documents/by-course/math-8/pset-4/package.nix

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+{
+  pkgs,
+  typstPackagesCache,
+  typixLib,
+  cleanTypstSource,
+  flakeSelf,
+  ...
+}:
+let
+  src = cleanTypstSource ./.;
+  commonArgs = {
+    typstSource = "main.typ";
+
+    fontPaths = [
+      # Add paths to fonts here
+      # "${pkgs.roboto}/share/fonts/truetype"
+    ];
+
+    virtualPaths = [
+      # Add paths that must be locally accessible to typst here
+      # {
+      #   dest = "icons";
+      #   src = "${inputs.font-awesome}/svgs/regular";
+      # }
+    ];
+
+    XDG_CACHE_HOME = typstPackagesCache;
+    SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified;
+  };
+
+in
+typixLib.buildTypstProject (
+  commonArgs
+  // {
+    inherit src;
+  }
+)