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\title{Bounds and bijections}
\date{Due January 31, 2022}

\begin{document}
\ \vspace{-.25in}

\maketitle

\vspace{-.2in}

\lem{[Triangle inequality]
For all $x, y \in \RR$, $\lvert x + y \rvert \leq \lvert x \rvert + \lvert y \rvert$.
}

\dfn{
A function $f: D \to \RR$ is \textbf{bounded} if there is $M \in \RR$ such that $\lvert f(x) \rvert \leq M$ for all $x \in D$.
% (or, equivalently, $-M \leq f(x) \leq M$ for all $x \in D$).
}

\vspace{-.15in}

\prob{
Use the triangle inequality to show that the function $f : [-3, 5] \to \RR$ defined by $f(x) = x^2 - 6x + 1$ is bounded.
}

\vspace{1.75in}

\dfn{
Let $f : D \to \RR$ be a function. 
Then $f(D) := \{ f(x): x \in D\}$ and \[
\sup_{x \in D} f(x) := \sup f(D) \text{ and }
\inf_{x \in D} f(x) := \inf f(D). 
\]
}

\prob{
Find $\inf_{x \in D} f(x)$ and $\sup_{x \in D} f(x)$ for the function in the previous problem.
}

\vfill

\prob{
Does your answer to the previous problem change if we change the domain to $(-3,5)$ but otherwise keep the function the same?
}

\vspace{1in}

\newpage

\prop{
Let $D$ be a nonempty set and $f:D \to \RR$ and $g: D \to \RR$ be bounded functions such that $\forall x \in D,\ f(x) \leq g(x)$.
Then  \[ \sup_{x \in D} f(x) \leq \sup_{x \in D} g(x) \text{ and } \inf_{x \in D} f(x) \leq \inf_{x \in D} g(x).\]
}

\prob{
Prove that it is possible for $f$ and $g$ to satisfy the conditions of the previous proposition but still have $\inf_{x \in D} g(x) \leq \sup_{x \in D} f(x)$.
(This means producing an example of two bounded functions $f$ and $g$ such that $\forall x \in D,\ f(x) \leq g(x)$ and $\inf_{x \in D} g(x) \leq \sup_{x \in D} f(x)$).
}

\vfill

\dfn{
Sets $A$ and $B$ have the same cardinality (notation: $\lvert A \rvert = \lvert B \rvert$) if there is a bijection $f: A \to B$.
}

\prob{
Prove that the real intervals $(0,1)$ and $(0, \infty)$ have the same cardinality by producing a bijection.
}

\vfill


\chal{
Prove that $(0,1)$ and $\RR$ have the same cardinality by producing a bijection.
(The composition of two bijections is again a bijection so it might be helpful to work in two steps: e.g.\@ $(0,1) \to (-1,1) \to \RR$)
}

\end{document}