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\begin{center}{\bf Homework 2, due Thursday, September 6, 2012}\end{center}
Let $f:X\to Y$ be a mapping of sets. We say $f$ is one to one (or injective) if
whenever $x_1$ and $x_2$ are in $X$ with $f(x_1)=f(x_2)$ then $x_1=x_2$.
Another way to say it is: $f$ is injective if $x_1\neq x_2$ always implies $f(x_1)\neq f(x_2)$.
We say $f$ is onto (or surjective) if for each $y\in Y$ there is an $x\in X$ with $f(x)=y$.
Another way to say it is: $f$ is surjective if $f(X)=Y$.
And we say $f$ is bijective if $f$ is injective and surjective.
Remember: the goal in writing proofs is not only to be right,
but to be understood. Never turn in an assignment without proofreading it.
(As Truman Capote said in reference to the Beat writers,
``That isn't writing at all, it's typing.'')
And when you proofread your proofs, check not only for correctness
but ask yourself: is it clear? Is it unambiguous? Could I say the same thing
more simply? Could I shorten it and still be clear? Would a few extra words be helpful?
Also remember that proof writing is a form of writing and thus requires sentences.
Proofs with no words are hard to read. A specific general rule is never to
begin a sentence with a symbol. So every proof should have at least one sentence
and thus at least one word!
Each problem is worth 20 points. Solutions will be graded for correctness, clarity and style.
\begin{itemize}
\item[(1)] Let $X$ be a set with $|X|>1$. Show that the indiscrete topology is not a metric
topology on $X$. (I.e., show that if $d$ is a metric on $X$, then some union of open discs
is not an open set in the indiscrete topology.)
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\item[(2)] Let $f:X\to Y$ be a map of sets.
Prove that $f:X\to Y$ is injective if and only if
$f(A)\cap f(B)\subseteq f(A\cap B)$ for all subsets $A,B\subseteq X$.
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\item[(3)] Let $f:X\to Y$ be a map of sets. Show that $f$ is surjective if and only if
$C\subseteq f(f^{-1}(C))$ for all subsets $C\subseteq Y$.
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\item[(4)] Let $X$ be the set of all continuous functions $f:[0,1]\to {\bf R}$.
Given $f,g\in X$, define $d(f,g)=\int_0^1|f(t)-g(t)|\,dt$. Prove that $d$ is a metric on $X$.
(You may assume usual facts from calculus.)
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\item[(5)] Let $X$ be a nonempty set and let $d:X\times X\to {\bf R}$ be a function
such that $d(x,y)=0$ if and only if $x=y$ and such that
$d(x,y)+d(z,y)\geq d(x,z)$ for all $x,y,z\in X$.
Show that $d(x,y)\geq 0$ and $d(x,y)=d(y,x)$ for all $x,y\in X$. Conclude that $d$ is a metric.
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\end{itemize}
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