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\begin{center}
{\Large Intro to Topology -- HW 4.}
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\noindent
\begin{enumerate}



\item 
Show that if $U$ is closed in $X$ and $V$ is closed in $Y$ then
$U\times V$ is closed in $X \times Y$. 

(It is assumed that $X$ has topology $\tau_X$, $Y$ has topology $\tau_Y$ and $X\times Y$ has the product topology.)

\item Let $(X,\tau_X)$, $(Y,\tau_Y)$ be topological spaces and equip the product $X\times Y$ with the product topology. In class we discussed
that the projections maps $proj_x:X\times Y\to X$, $proj_x((x,y))=x$ and
$proj_y:X\times Y\to Y$, $proj_x((x,y))=y$ are continuous. Now prove that they are also open maps (that is: they take open sets to open sets).


\item Decide if the following statements are true or false
(i.e. for all sets $A,B$ and in any topology):

$$cl(A\cap B)=cl(A)\cap cl(B)$$


$$cl(A\cup B)=cl(A)\cup cl(B)$$

If false, give a counterexample to the statement. 

If true, prove it {\bf two ways}:

first ("proof 1") by element chasing - that is, show that if an element $x$ is in the left hand side (LHS) set,
 then it is in the right hand side set (RHS) and vice-versa, and also that
$x\in cl(D)$ for some set $D$ if and only if $\forall U\in\tau$ such that $x\in U$ we have $D\cap U\not=\emptyset$.

Then ("proof 2") by using "properties of closure":

\noindent $\forall A\subset X$ we have

(1) $cl(A)$ is a closed set in $X$

(2) $A\subset cl(A)$ 

(3) $A$ is closed if and only if $cl(A)=A$

(4)  if $W$ is a closed set for which
$A\subset W\subset cl(A)$ then $W=cl(A)$


At each step of proof 2, state which of properties 1-4 you are using.

\item 
 In $\R$ with the standard (or usual) topology consider

$$A=(0,1)\cup (1,2)\cup\{(2,3)\cap\Q\}\cup\{4\}$$

i.) What are $cl(A)$, $int(A)$, $\partial A$?

ii.) Check that
 $cl(int(A))=cl(int(cl(int(A)))$. i.e. work out both sides.

iii.) 
Show that, in fact,  $cl(int(A))=cl(int(cl(int(A)))$ in general. (i.e. this is true in all
topologies, for any $A$).


\item [EXTRA CREDIT] 
Pythagorean triples are natural numbers $a, b, c \in\ N$ satisfying the equation 
$a^2 + b^2 = c^2$. The Pythagorean
triple that is probably most commonly known is 3, 4, 5.

In this problem you can work out formulae to generate such triples, starting a rational number, using stereographic projection.

a.) Work out a formula for stereographic projection for $n=1$,
that is $\phi: S^1\setminus\{(0,1)\}\longrightarrow \R=\{(x,0)\,|\, x\in\R\}$, where
$S^1=\{(x,y)\,|\, x^2+y^2=1\}$ and its inverse, to show that there is a
one-to-one correspondence between those
 points  
 of the unit circle take away the North pole  whose both coordinates are rational
 and rational points of the x-axis (ie points of the form $(s,0)$
 with $s\in\Q$).

b.) Consider now $a, b, c \in\Z$ (not $\N$), not all zero.

Show that Pythagorean triples determine points of the unit circle  whose both coordinates are rational. How?

Find an inverse procedure, ie formulae which given a "rational point" of $S^1$, 
provide a Pythagorean tripple.


c.) If $k$ is not zero and $m, k\in\Z$, then ${m\over k}\in\Q$.
Use formulae you worked out in part a) and b) to
find formulae which for a pair of non-zero integers $m, k\in\Z$ provide a Pythagorean triple $a, b, c \in\ Z$. (These  formulae were first already known to Euclid).

d.) Is it possible to get all Pythagorean triples this way?




\end{enumerate}

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