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\begin{center}\textbf{Problem Set 2 \hspace{.7in} Set Theory \& Topology \hspace{.7in} Due 1/29/16} \end{center}

\noindent You are expected to do the following problems to a high standard (i.e., at least well enough to be published in a textbook) for full credit.

\begin{enumerate}

\item If $A$ and $B$ are sets and $A=B$, then $A-B = \varnothing$.
\item If $A$ and $B$ are sets and $A-B = \varnothing$, then  $A=B$.
\item If both $A$ and $B$ are the empty set, then $A \times B = \varnothing$.
\item If $A$ and $B$ are sets and $A \times B = \varnothing$, then $A = \varnothing$ and $B = \varnothing$.
\item If $A, B$, and $C$ are sets, then $A-(B-C)=(A-B)-C$.
\item If $A$ and $B$ are subsets of $X$ and $A \cap B \neq \varnothing$, then $B \nsubseteq X - A$.
\item Let $\{A_\alpha : \alpha \in \Lambda \}$ be an indexed collection of sets.  If $\bigcap \{ A_\alpha : \alpha \in \Lambda \} = \varnothing$, then for any distinct $\alpha$ and $\beta$ in $\Lambda$, $A_\alpha \cap A_\beta = \varnothing$.
\item Let $\{A_\alpha : \alpha \in \Lambda \}$ be an indexed collection of sets.  If for any distinct $\alpha$ and $\beta$ in $\Lambda$ $A_\alpha \cap A_\beta = \varnothing$, then $\bigcap \{ A_\alpha : \alpha \in \Lambda \} = \varnothing$.
\item If $f:X \to Y$ is a function and $f(x_1)=f(x_2)$, then $x_1=x_2$.
\item If $f:X \to Y$ is a one-to-one function and $f(x_1)=f(x_2)$, then $x_1=x_2$.
\item If $f:X \to Y$ is a function and $V$ is a nonempty subset of $Y$, then $f^{-1}(V)$ is a nonempty subset of $X$.
\item If $f:X \to Y$ is an onto function and $V$ is a nonempty subset of $Y$, then $f^{-1}(V)$ is a nonempty subset of $X$.
\item The inverse of the inverse of a one-to-one onto function is the original function.
\item Let $f:X \to Y$ be a function and let $A$ and $B$ be subsets of $Y$.  If $f^{-1}(A)=f^{-1}(B)$, then $A=B$.
\item If $f:X \to Y$ is a function, then $f(X)=Y$.
\item If $f:X \to Y$ is onto, then $f(X)=Y$.
\item Inverse images of sets are only defined for one-to-one functions.
\item If $f:X \to Y$ is a function, then $f^{-1}(Y)=X$.
\item If $f:X \to Y$ is a function and $U$ and $V$ are subsets of $X$, then $f(U \cap V) = f(U) \cap f(V)$.
\item If $f:X \to Y$ is a function and $U$ and $V$ are subsets of $X$, then $f(U \cap V) \subseteq f(U) \cap f(V)$.

\end{enumerate}

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