MA0301/exercise7/main.tex

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\documentclass[12pt]{article}
\usepackage{ntnu}
\usepackage{ntnu-math}
\author{Øystein Tveit}
\title{MA0301 Exercise 7}
\usepackage{amsthm}
\usepackage{mathabx}
\begin{document}
\ntnuTitle{}
\break{}
\begin{excs}
\exc{}
\begin{subexcs}
\subexc{}
By the definitions given in the exercise, we can define the PoS (partition of S) as
\[ \forall y \in T \{ x \in S \mid f(x) = y\} \]
In order for PoS to be a partition of a set, the following conditions have to hold: \\
The PoS can not contain the empty set
This holds because $f$ is a surjective function which by its definition $\forall y \in T \exists x \in S [f(x)=y]$ needs every $y$ in $T$ to have an $x$ in $S$. There are no $y$s without and $x$ and therefore no empty sets. \\
The union of all the subsets in $S$ has to be equal to $S$
This holds because $f$ is a function. Every $x$ of the domain needs to have a $y$ in the range, and because the union of the blocks of the PoS contains every $x$ for which there exists a $y$, that would mean it covers the whole domain. \\
No pair of sets in the PoS contains any common elements
This holds because $f$ is a function. No $f(x)$ can have multiple values, and therefore there will not be any $x$s in several blocks of the partition.
\subexc{}
If $f$ were to only be a function, the PoS would not fulfill the first condition in part a, and therefore it would not be a proper partition of a set.
\subexc{}
If $f$ was bijective, every block in the PoS would have a cardinality of $1$, meaning that every block would only contain one element of $S$
\setsubexc{4}
\subexc{}
A block of $f^{-1}[\N]$ representing a natural number $n$ would be $\{x \mid n \leq x < n + 1\}$
\end{subexcs}
\exc{}
\begin{figure}[H]
\begin{mgraphbox}[width=5cm]
\center
\begin{tikzpicture}[scale=1]
\tikzset{every node/.style={shape=circle,draw,inner sep=2pt}}
\node (2) at (0,0) {$2$};
\node (3) at (1,0) {$3$};
\node (4) at (-1,1) {$4$};
\node (16) at (0,2) {$16$};
\draw (2) -- (4) -- (16);
\end{tikzpicture}
\end{mgraphbox}
\caption{Hasse diagram of $R$ on $X$}
\end{figure}
\begin{center}
Minimal elements $= \{3, 3\}$ \\
Maximal elements $= \{3, 16\}$
\end{center}
\exc{}
Reflexive:
\[ (a,a), (b,b), (c,c), (d,d) \]
Antisymmetric:
There are no cases where $(x,y) \wedge (y,x)$
Transitive:
Because $(c,a) \wedge (a,d)$, $(c,d)$ is also included.
\begin{figure}[H]
\begin{mgraphbox}[width=5cm]
\center
\begin{tikzpicture}[scale=1]
\tikzset{every node/.style={shape=circle,draw,inner sep=2pt}}
\node (c) at (0,0) {$c$};
\node (a) at (-1,1) {$a$};
\node (b) at (1,1) {$b$};
\node (d) at (0,2) {$d$};
\draw (c) -- (a) -- (d);
\draw (c) -- (b);
\end{tikzpicture}
\end{mgraphbox}
\caption{Hasse diagram of $P$}
\end{figure}
\exc{}
The function is injective, because it is a linear polynomial. However, it is not bijective, because all integers of the form $3x$ or $3x-1$ as the input of $f^{-1}$ does not result in an integer
\exc{}
In order for $f$ to be an injective function, it has to hold that $f(x) = f(y) => x = y$.
Suppose $f(x) = f(y)$ for $x,y \in \Z$
Case i) both are even
\[ -2x=-2y => x = y \]
Case ii) both are odd
\[ 2x-1=2y-1 => x = y \]
Therefore $f$ is injective\\
In order for $f$ to be surjective, it has to hold that $\forall n \in \N \exists x \in \Z [f(x)=n]$
Case i) $n$ is even
The $x$ in this case has to be in the form of
\[n = -2x \Leftrightarrow x = -\frac{n}{2}\]
which would be an integer, because $n$ is even and therefore divisible by $2$
Since $x \leq 0$
\[ f(x) = f\left(\frac-{n}{2}\right) = -2\left(-\frac{n}{2}\right) = n\]
Case ii) $n$ is odd
The $x$ in this case has to be in the form of
\[n = 2x-1 \Leftrightarrow x = -\frac{n+1}{2}\]
which would be an integer, because $n$ is odd and therefore $n+1$ is divisible by $2$
Since $x > 0$
\[ f(x) = f\left(\frac{n+1}{2}\right) = 2\left(\frac{n+1}{2}\right) - 1 = n\]
Therefore $f$ is surjective
From here, I will create the inverse function piece by piece
Piece 1) $x \leq 0$
\begin{align*}
y &= -2x \\
-y &= 2x \\
\frac{-y}{2} &= x \\
x &= \frac{-y}{2} \\
\end{align*}
Piece 2) $x > 0$
\begin{align*}
y &= 2x - 1 \\
y + 1 &= 2x \\
\frac{y + 1}{2} &= x \\
x &= \frac{y + 1}{2} \\
\end{align*}
hence
\[ f^{-1} =
\begin{cases}
\frac{-y}{2} & \text{for } n \leq 0 \\
\frac{y + 1}{2} & \text{for } n > 0 \\
\end{cases} \]
\exc{}
\begin{subexcs}
\subexc{}
Because $(f \circ g)$ is surjective, we know that
\[\forall c \in C \exists a \in A [(f \circ g)(a) = c]\]
therefore
\[\forall f(a) = b \in B [g(b) = c]\]
therefore $g$ is surjective
\subexc{}
Because an injective function is one to one, we know that if their output is equal, their inputs must also be equal. Therefore
\begin{align*}
(f \circ g)(x) &= (f \circ g)(y) \\
f(g(x)) &= f(g(y)) \\
g(x) &= g(y) \\
x &= y \\
\end{align*}
\[ ((f \circ g)(x) = (f \circ g)(y) \Leftrightarrow x = y) \Rightarrow (f\text{ is injective} \wedge g\text{ is injective} \Leftrightarrow f \circ g\text{ is injective}) \]
\end{subexcs}
\exc{}
\begin{figure}[H]
\begin{mgraphbox}[width=15.8cm]
\center
\begin{tikzpicture}[scale=2]
% Domains
\filldraw[fill=blue!20, draw=blue!60] (0,0) circle (1cm);
\filldraw[fill=blue!20, draw=blue!60] (2.5,0) circle (1cm);
\filldraw[fill=blue!20, draw=blue!60] (5,0) circle (1cm);
% Ranges
\filldraw[fill=red!20, draw=red!60] (0,0) circle (0.5cm);
\filldraw[fill=red!20, draw=red!60] (2.5,0) circle (1cm);
\filldraw[fill=red!20, draw=red!60] (5,0) circle (0.5cm);
% Labels
\node at (0, 1.2) {$A$};
\node at (2.5, 1.2) {$B$};
\node at (5, 1.2) {$C$};
\node at (1.25, 0.8) {\tiny$f(a)$};
\node at (3.75, 0.8) {\tiny$g(b)$ or $h(b)$};
\node at (0, 0.7) {\tiny Preimage};
\node at (0, 0) {\tiny Domain};
\node at (2.5, 0) {\tiny Range of $f$};
% Nodes for arrows
\node (a1) at (0, 0.3) {};
\node (b1) at (2.5, 0.3) {};
\node (c1) at (5, 0.3) {};
% Arrows
\draw[->] (a1) to [out=30,in=150] (b1);
\draw[->] (b1) to [out=30,in=150] (c1);
\end{tikzpicture}
\end{mgraphbox}
\caption{Diagram of $g \circ f: A \to C$ in the case where $f$ is surjective}
\end{figure}
Because the range of $f$ covers the whole preimage of $g$ or $h$ when it is surjective, it means that if $g \circ f = h \circ f$ then $g = h$
\begin{figure}[H]
\begin{mgraphbox}[width=15.8cm]
\center
\begin{tikzpicture}[scale=2]
% Domains
\filldraw[fill=blue!20, draw=blue!60] (0,0) circle (1cm);
\filldraw[fill=blue!20, draw=blue!60] (2.5,0) circle (1cm);
\filldraw[fill=blue!20, draw=blue!60] (5,0) circle (1cm);
% Ranges
\filldraw[fill=red!20, draw=red!60] (0,0) circle (0.5cm);
\filldraw[fill=red!20, draw=red!60] (2.5,0) circle (0.8cm);
\filldraw[fill=red!20, draw=red!60] (5,0) circle (0.5cm);
% Labels
\node at (0, 1.2) {$A$};
\node at (2.5, 1.2) {$B$};
\node at (5, 1.2) {$C$};
\node at (1.25, 0.8) {\tiny$f(a)$};
\node at (3.75, 0.8) {\tiny$g(b)$ or $h(b)$};
\node at (0, 0.7) {\tiny Preimage};
\node at (0, 0) {\tiny Domain};
\node at (2.5, 0) {\tiny Range of $f$};
\node at (3.7, -1) {\tiny$g(b_1)$};
\node at (3.7, -1.35) {\tiny$h(b_1)$};
% Nodes for arrows
\node (a1) at (0, 0.3) {};
\node (b1) at (2.5, 0.3) {};
\node (b2) at (2.5, -0.85) {};
\node (c1) at (5, 0.3) {};
\node (c2) at (5.7, -0.5) {};
\node (c3) at (4.5, -0.6) {};
% Arrows
\draw[->] (a1) to [out=30,in=150] (b1);
\draw[->] (b1) to [out=30,in=150] (c1);
\draw[->] (b2) to [out=-30,in=-140] (c2);
\draw[->] (b2) to [out=-30,in=-130] (c3);
\end{tikzpicture}
\end{mgraphbox}
\caption{Diagram of $g \circ f: A \to C$ in the case where $f$ is not surjective}
\end{figure}
Because the range of $f(a)$ restricts the domain of $g(b)$ or $h(b)$, as long as they map to the same elements within their restricted domain, $g \circ f = h \circ f$. \\
However, since $f(a)$ is not surjective, it doesn't imply that $g$ and $h$ can not differ outside of their domain. In order for $g \circ f = h \circ f$ to imply that $g = h$, $f$ has to be surjective. \\
Therefore \[ f(a)\text{ is surjective} \Leftrightarrow (g \circ f = h \circ f \Rightarrow g = h) \]
\exc{}
\begin{align*}
f^{-1}(B_1 \cap B_2) &= \{a \mid a \in f^{-1}(B_1 \cap B_2)\} \\
&= \{a \mid f(a) \in B_1 \cap B_2\} \\
&= \{a \mid f(a) \in B_1 \wedge f(a) \in B_2\} \\
&= \{a \mid a \in f^{-1}(B_1) \wedge a \in f^{-1}(B_2)\} \\
&= \{a \mid a \in f^{-1}(B_1) \cap f^{-1}(B_2)\} \\
&= f^{-1}(B_1) \cap f^{-1}(B_2)
\end{align*}
\begin{align*}
f^{-1}(\overline{B_1}) &= \{a \mid a \in f^{-1}(\overline{B_1})\} \\
&= \{a \mid f(a) \in \overline{B_1}\} \\
&= \{a \mid f(a) \notin B_1\} \\
&= \{a \mid a \notin f^{-1}(B_1)\} \\
&= \{a \mid a \in \overline{f^{-1}(B_1)}\} \\
&= \overline{f^{-1}(B_1)} \\
\end{align*}
\end{excs}
\end{document}