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\begin{document}

\head{Worksheet 7}{Week 7: 2024-10-08}

We've seen sets; that's about half of math. Now time for the other half: functions! As always,
work together, and ask lots of questions, both of each other, and of me! \vspace{5mm}

\begin{framed}
  \textbf{Karen Uhlenbeck Is First Woman to Win Abel Prize for Mathematics} by Kenneth Chang,
  March 19, 2019 (excerpt from
  \href{https://www.nytimes.com/2019/03/19/science/karen-uhlenbeck-abel-prize.html}{NY Times
    article}).

  For the first time, one of the top prizes in mathematics has been given to a woman.

  On Tuesday, the Norwegian Academy of Science and Letters announced it has awarded this year's
  Abel Prize---an award modeled on the Nobel Prizes---to Karen Uhlenbeck, an emeritus professor
  at the University of Texas at Austin. The award cites ``the fundamental impact of her work on
  analysis, geometry and mathematical physics.''

  One of Dr. Uhlenbeck’s advances in essence described the complex shapes of soap films not in a
  bubble bath but in abstract, high-dimensional curved spaces. In later work, she helped put a
  rigorous mathematical underpinning to techniques widely used by physicists in quantum field
  theory to describe fundamental interactions between particles and forces.

  In the process, she helped pioneer a field known as geometric analysis, and she developed
  techniques now commonly used by many mathematicians.

  ``She did things nobody thought about doing,'' said Sun-Yung Alice Chang, a mathematician at
  Princeton University who served on the five-member prize committee, ``and after she did, she
  laid the foundations of a branch of mathematics.''

  The Abel, named after the Norwegian mathematician Niels Henrik Abel, is set up more like the
  Nobels. Since 2003, it has been given out annually to highlight important advances in
  mathematics. The previous 19 laureates---in three years, the prize was split between two
  mathematicians---were men, including Andrew J. Wiles, who proved Fermat’s last theorem and is
  now at the University of Oxford; Peter D. Lax of New York University; and John F. Nash Jr.,
  whose life was portrayed in the movie ``A Beautiful Mind.''

  In 1990, she became the second woman to give one of the highlighted plenary talks at the
  International Congress of Mathematicians, a quadrennial meeting. At each congress, there are
  10 to 20 plenary talks, but for decades, all of the speakers had been men. (Emmy Noether, a
  prominent German mathematician, was the first woman to give a plenary talk, in 1932.)

  ``That was almost more unnerving'' than being the first woman to receive an Abel, Dr.
  Uhlenbeck said.

  Dr. Uhlenbeck said she recognized that she was a role model for women who followed her in
  mathematics.

  ``Looking back now I realize that I was very lucky,'' she said. ``I was in the forefront of a
  generation of women who actually could get real jobs in academia.''

  But she also noted: ``I certainly very much felt I was a woman throughout my career. That is,
  I never felt like one of the guys.''

  \noindent\makebox[\textwidth][c]{
  \begin{minipage}[t]{0.5\linewidth}
    \vspace{-2ex}
    \centering
    \includegraphics[width=6cm]{uhlenbeck}
  \end{minipage}}
\end{framed}

\begin{problem}
  We have already encountered the \textit{existential quantifier} \( \exists \), which can be
  used in a proposition like ``\( \exists x \in X \) such that \( P(X) \)''. A slight
  strengthening is the \textit{unique existential quantifier}: the proposition
  ``\( \exists !\, x \in X \) such that \( P(X) \)'' says two things: first that there exists an
  \( x \) satisfying \( P \), \textbf{and} that there is exactly one such \( x \). How can you
  express this meaning \textit{without} using the special \( \exists ! \) quantifier? What steps
  are required to prove such a proposition?
\end{problem}

\vspace{4cm}

\begin{problem}
  Let \( X \) and \( Y \) be sets. A \textit{function} \( f \) from \( X \) to \( Y \) (denoted
  \( f : X \to Y \)) is a specification of outputs \( f(x) \in Y \) for each input
  \( x \in X \), so that
  \[ \forall x \in X, \exists !\, y \in Y, y = f(x). \]
  The set \( X \) is called the \textit{domain} while the set \( Y \) is called the
  \textit{codomain}. A well-defined function must satisfy the following properties:
  \begin{enumerate}
  \item \textbf{Totality:} Every value \( x \in X \) is assigned a value \( f(x) \).
  \item \textbf{Existence:} For each \( x \in X \), the value \( f(x) \) should actually exist
    as an element of the codomain.
  \item \textbf{Uniqueness:} For each \( x \in X \), there should be only one value assigned as
    \( f(x) \).
  \end{enumerate}
  Using whatever notation/description you prefer, give three examples of ``non-functions'' that
  fail to satisfy each of the above criteria.
\end{problem}

\vspace{5cm}

\begin{problem}
  Prove that there exists a unique real number \( a \) for which the equation
  \( x^2 + a^2 = 0 \) has a real solution for \( x \)
\end{problem}

\vspace{4cm}

\begin{problem}
  For each of the following ``specifications'', determine if they are well-defined functions. If
  not, identify the property that fails.
  \begin{enumerate}
  \item \( f : \QQ \to \QQ \) given by \( f(x) = (x + 1)^{-1} \).
  \item \( g : \ZZ_{\geq 0} \to \QQ \) given by \( g(x) = (x + 1)^{-1} \).
  \item \( h : \NN \to \NN \) given by \( h(x) = (x + 1)^{-1} \).
  \end{enumerate}
  [Note: even though \( f \), \( g \), and \( h \) are all defined by identical formulae, they
  are not the same mathematical objects because their (co-)domains are different.]
\end{problem}

\vspace{4cm}

\begin{problem}
  Does the specification
  \[ f(x) =
    \begin{cases}
      x + 1 & \textnormal{if } x \geq 0, \\
      0 & \textnormal{if }x \leq 0,
    \end{cases}
  \]
  determine a well-defined function? If not, why not? Can this be remedied easily?
\end{problem}

\vspace{4cm}

\begin{problem}
  Give a condition on the sets \( X \) and \( Y \) that will ensure that the specification of a
  well-defined function \( i : X \cup Y \to \{ 0, 1 \} \) given by
  \[ i(x) =
    \begin{cases}
      1 & \textnormal{if } x \in X, \\
      0 & \textnormal{if } x \in Y.
    \end{cases}
  \]
\end{problem}

\vspace{3cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( U \subseteq X \). The \textit{image} of
  \( U \) under \( f \), denoted \( f(U) \), is the subset of \( Y \) given by
  \( f(U) = \{ f(x) \st x \in U \} \). Prove that we could equivalently define
  \( f(U) = \{ y \in Y \st \exists x \in X \textnormal{ so that } y = f(x) \} \).
\end{problem}

\vspace{3cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( V \subseteq Y \). The \textit{preimage} of
  \( V \) under \( f \), denoted \( f^{-1}(V) \), is the subset of \( X \) given by
  \( f^{-1}(V) = \{ x \in X \st f(x) \in V \} \).
  \begin{enumerate}
  \item Is it true that \( f(X) = Y \)?
  \item Is it true that \( f^{-1}(Y) = X \)? When does there exist a smaller set
    \( V \subseteq Y \) so that \( f^{-1}(V) = X \)?
  \end{enumerate}
\end{problem}

\vspace{3cm}

\begin{problem}
  Let \( f : \RR \to \RR \) be given by \( f(x) = \sqrt{1 + x^2} \). Is this a well-defined
  function? If so, describe its image \( f(X) \).
\end{problem}

\vspace{3cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( U, V \subseteq X \). Is it true that
  \( F(U) \cup f(V) = f(U \cup V) \)? (Provide proof.)
\end{problem}

\vspace{4cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( U, V \subseteq Y \). Is it true that
  \( f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) \)? (Provide proof.)
\end{problem}

\vspace{4cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( U, V \subseteq X \). Is it true that
  \( f(U) \cap f(V) = f(U \cap V) \)? (Provide proof.)
\end{problem}

\vspace{4cm}

\begin{problem}
  Let \( f : X \to Y \) be a function, and let \( U \subseteq X \) and \( V \subseteq Y \). Is
  it true that \( f^{-1}(f(U)) = U \)? If not, which (if any) containment \textit{is} true?

  Similarly, is it true that \( f(f^{-1}(V)) = V \)? Which containment is true in this case?
\end{problem}

\end{document}