\chapter{Bases, Subspaces, Products: Creating New Spaces}

\begin{thm}
\label{thm:basis.given.top}  
Let $(X,{\mathcal T})$ be a topological space
and $\mathcal B$ be a collection of subsets of $X$.  Then $\mathcal B$ is a 
basis for $\mathcal T$ if and only if:
\be
\item ${\mathcal B}\subset {\mathcal T}$, and
%\item $\emptyset \in {\mathcal B}$,
\item for each set $U$ in $\mathcal T$ and point $p$ in $U$ there is a set
$V$ in $\mathcal B$ such that $p\in V\subset U$.
\ee
\end{thm}

\oneperpage

\begin{exercise} 
\begin{enumerate}   
\item Let ${\mathcal B}_1 = \{ (a,b) \subset \R \mid a, b
\in \Q\}$. Show that ${\mathcal B}_1$ is a basis for the \emph{standard topology} on
$\R$. 
\item Let ${\mathcal B}_2 = \{ (a,b)\cup(c,d) \subset \R \mid a<b<c<d \text{ are distinct irrational numbers}\}$.  Show that  ${\mathcal B}_2$ is also a basis for the \emph{standard topology} on $\R$.  
\end{enumerate}
\end{exercise}

\oneperpage

\begin{thm} 
\label{thm:basis.some.top}   
Suppose $X$ is a set and $\mathcal B$ is a collection
of subsets of $X$.  Then $\mathcal B$ is a basis for some topology on $X$
if and only if:
\be
%\item $\emptyset \in {\mathcal B}$,
%\item for each point $p$ in $X$ there is a set $U$ in $\mathcal B$ with $p\in U$, and 
\item each point of $X$ is in some element of $\mathcal B$, and 
\item  if $U$ and $V$ are sets in $\mathcal B$ and $p$ is a point
        in $U\cap V$, there is a set $W$ in $\mathcal B$ such that
        $p\in W \subset (U\cap V)$.
\ee
\end{thm}

\oneperpage

\begin{exercise}
Show that the basis proposed above for the lower limit topology is in fact a basis.
    
\end{exercise}

\oneperpage

\begin{thm}
\label{std.in.lowerlim} 
Every open set in $\R_\mathrm{std}$ is an open set in $\RLL$, but not vice versa.
\end{thm}

\oneperpage

\begin{exercise}
Give an example of two topologies on $\R$ such that neither is finer than the other, that is, the two topologies are not comparable.
\end{exercise}

\oneperpage

\begin{exercise}
Check that the collection of sets that we specify as a basis in the Double Headed Snake actually forms the basis for a topology.
\end{exercise}

\oneperpage

\begin{exercise}
In the Double Headed Snake, show that every point is a closed set; however, it is impossible to find disjoint open sets $U$ and $V$ such that $0' \in U$ and $0'' \in V$. 
\end{exercise}

\oneperpage

\begin{exercise}
\begin{enumerate}
\item In the topological space  $\R_\mathrm{har}$, what is the closure of the set $H=\{1/n\}_{n\in\N}$?
\item In the topological space  $\R_\mathrm{har}$, what is the closure of the set $H^{-}=\{-1/n\}_{n\in\N}$?
\item Is it possible to find disjoint open sets $U$ and $V$ in $\R_\mathrm{har}$ such that $0 \in U$ and $H \subset V$?
\end{enumerate}
\end{exercise}

\oneperpage

\begin{exercise}
\begin{enumerate}
\item In $\Hbub$, what is the closure of the set of rational points on the $x$-axis?
\item In $\Hbub$, which subsets of the $x$-axis are closed sets?
\item In $\Hbub$, let $A$ be a countable set on the $x$-axis and let $z$ be a point on the $x$-axis not in $A$. Then there exist disjoint open sets $U$ and $V$ such that $A \subset U$ and $z \in V$.  (Do you need the countability hypothesis on $A$?)
% (Francis: what do you think about this one? Also what do you think about asking whether they could separate the rationals from the irrationals? 
% Mike: These are hard, but I like them.  It was a problem in the separation chapter but it could be put here.
\item In $\Hbub$, let $A$ and $B$ be countable sets on the $x$-axis such that $A$ and $B$ are disjoint. Then there exist disjoint open sets $U$ and $V$ such that $A \subset U$ and $B \subset V$. 
\item In $\Hbub$, let $A$ be the rational numbers and let $B$ be the irrational numbers.  
Do there exist disjoint open sets $U$ and $V$ such that $A \subset U$ and $B \subset V$?
\end{enumerate}
\end{exercise}

\oneperpage

\begin{exercise}
Check that the arithmetic progressions form a basis for a topology on $\Z$.
\end{exercise}

\oneperpage

\begin{thm}
There are infinitely many primes.
  \begin{inst}
  The proof is due to F\"urstenburg.
  \end{inst}
\end{thm}

\oneperpage

\begin{exercise}   
A basis for a topology is also a subbasis for that topology.
%  
\end{exercise}

\oneperpage

\begin{thm}    
Let $(X,{\mathcal T})$ be a topological space and let
$\mathcal S$ be a collection of subsets of $X$.  Then $\mathcal S$ is a subbasis
for $\mathcal T$ if and only if 
\be
\item $\mathcal S \subset \mathcal T$, and
%\item there is a finite collection $\{ V_i\}_{i=1}^n\subset\mathcal S$ such that $\bigcap_{i=1}^n V_i = \emptyset$, 
\item for each set $U$ in $\mathcal T$ and point $p$ in $U$ there is a
finite collection $\{ V_i\}_{i=1}^n$ of elements of $\mathcal S$ such that
$$
p\in {\textstyle\bigcap\limits_{i=1}^n}
 V_i \subset U\ .
$$
\ee
\end{thm}

\oneperpage

\begin{exercise}    
Let $\mathcal S$ be the following collection of subsets of $\R$:
$\{ x\mid x<a$ for some $a\in \R\}$ and $\{ x\mid a<x$ for some $a\in \R\}$. 
Then $\mathcal S$ is a subbasis for $\R$ with the standard topology.
\end{exercise}

\oneperpage

\begin{thm} \label{subbasis.thm}   
Suppose $X$ is a set and $\mathcal S$ is a collection of subsets of $X$.  
Then $\mathcal S$ is a subbasis for some topology on $X$ if and
only if
%\begin{enumerate}
%\item 
every point of $X$ is in some element of $\mathcal S$.
%and \item there are sets $\{ U_i\}_{i=1}^n$ in $\mathcal S$ such that  $\cap_{i=1}^n U_i = \emptyset$.
%\end{enumerate}
\end{thm}

\oneperpage

\begin{exercise}
Let $\mathcal S$ be the following collection of subsets of $\R$:
$\{ x\mid x<a$ for some $a\in \R\}$ and $\{ x\mid a \leq x$ for some $a\in \R\}$. 
For what topology on $\R$ is $\mathcal S$ a subbasis?
\end{exercise}

\oneperpage

\begin{exercise}
Let $X$ be a set totally ordered by $<$.  Let $\mathcal S$ be the collection of sets of the following forms
\[\{ x\in X\ | \ x <a\} \quad \text{ or } \quad \{ x\in X\ | \  a <x \}\]
for $a \in X$.  Then $\mathcal S$ forms a subbasis for the order topology on $X$.
\end{exercise}

\oneperpage

\begin{exercise}    
Verify that the order topology on $\R$ with the usual $<$ order is the standard topology on $\R$.
\end{exercise}

\oneperpage

\begin{exercise} 
In the lexicographically ordered square find the closures of the following subsets:
\begin{eqnarray*}
A & = & \left\{ \left(\frac{1}{n},0\right)\  |\  n\in \N \right\}.\\
B & = & \left\{ \left(1-\frac{1}{n},\frac{1}{2}\right)\  | \ n\in \N \right\}. \\
C & = & \left\{\left(x,0\right)\ |\ 0<x<1\right\} .\\
D & = & \left\{\left(x,\frac{1}{2}\right)\ |\ 0<x<1\right\} .\\
E & = & \left\{\left(\frac{1}{2},y\right)\ |\ 0<y<1\right\} .
\end{eqnarray*}
\end{exercise}

\oneperpage

\begin{exercise}
Assume that $\N$ has the usual order.  
Let $\N^\omega$ denote the Cartesian product of a countable number of copies of the space $\N$.  It can be endowed with the dictionary order in a natural way.
Show that $\N^\omega$ with the dictionary order topology is uncountable, is not well-ordered, and any set that does not have a least element does have a limit point.
\end{exercise}

\oneperpage

\begin{thm}   
Consider the topological space $\omega_1$ consisting of all ordinals less than $\omega_1$, the first uncountable ordinal, with the order topology.  Let $A$ be an infinite set of ordinals in $\omega_1$.  
Then there is an ordinal $\beta<\omega_1$ that is a limit point of $A$.
\end{thm}

\oneperpage

\begin{thm}   
Let $A$  and $B$ be unbounded closed sets in the topological space $\omega_1$. 
Then $A \cap B \neq \emptyset$.
\end{thm}

\oneperpage

\begin{thm}
Let $(X,\mathcal{T})$ be a topological space and $Y\subset X$.
Then the collection of sets $\mathcal{T}_Y$ is in fact a topology on $Y$.
\end{thm}

\oneperpage

\begin{exercise}
Consider $Y=[0,1)$ as a subspace of $\R_{\mathrm{std}}$.  In $Y$, is the set $[1/2,1)$ open, closed, neither, or both?
\end{exercise}

\oneperpage

\begin{exercise} 
Consider a subspace $Y$ of the topological space $X$.  Is every subset $U\subset Y$ 
that is open in $Y$ also open in $X$?
%If so, prove it.  If not, give an example of a set $Y$ and a subset $U$ 
%that is open in $Y$
%but not open in $X$.  Find conditions on $Y$ such there are no such
%subsets.
\end{exercise}

\oneperpage

\begin{thm}
Let $(Y, \mathcal T_Y)$ be a subspace of $(X,\mathcal T)$. 
A subset $C \subset Y$ is closed in $(Y, \mathcal T_Y)$ 
if and only if there is a 
set $D \subset X$, closed in $(X,\mathcal T)$, such that $C=D \cap Y$.
\end{thm}

\oneperpage

\begin{cor}
Let $(Y, \mathcal T_Y)$ be a subspace of $(X,\mathcal T)$. 
A subset $C \subset Y$ is closed in $(Y, \mathcal T_Y)$
if and only if
$\cl_X(C)\cap Y=C$.
\end{cor}

\oneperpage

\begin{thm}
Let $(X,\mathcal{T})$ be a topological space, and
$(Y, T_Y)$ be a subspace.
If $\mathcal{B}$ is a basis for $\mathcal{T}$,
then $\mathcal{B}_Y=\{B\cap Y | B\in\mathcal{B}\}$
is a basis for $\mathcal T_Y$.
\end{thm}

\oneperpage

\begin{exercise}
Consider the following subspaces of the lexicographically ordered square:
\be
\item $D  =  \left\{\left(x,\frac{1}{2}\right)\ |\ 0<x<1\right\}$.
\item $E  = \left\{\left(\frac{1}{2},y\right)\ |\ 0<y<1\right\}$.
\item $F  = \left\{\left(x,1\right)\ |\ 0<x<1\right\}$.
\ee
As sets they are all lines.  
Describe their relative topologies, especially noting any connections to topologies you have seen already.
\end{exercise}

\oneperpage

\begin{exercise}
Verify that the collection of basic open sets above satisfies the conditions of Theorem \ref{thm:basis.some.top}, thus confirming that this collection is the basis for a topology.
%Verify that the product topology on $X\times Y$ is actually a topology.
\end{exercise}

\oneperpage

\begin{exercise}
Draw examples of basic and arbitrary open sets in 
$\R^2=\R \times \R$ using the standard topology on $\R$.
Find 
(i) an open set in $\R \times \R$ that is not the product of open sets, and 
(ii) a closed set in $\R \times \R$ that is not the product of closed sets.
\end{exercise}

\oneperpage

\begin{exercise}
Is the product of closed sets closed?
\end{exercise}

\oneperpage

\begin{exercise}
Show that the product topology on $X\times Y$ is the same as the topology generated by the subbasis of inverse images of open sets under the projection functions, that is the subbasis is 
$\{ \pi_{X}^{-1}(U) \mid \text{ $U$ open in $X$}\} \cup 
\{ \pi_{Y}^{-1}(V) \mid \text{ $V$ open in $Y$}\}$.
\end{exercise}

\oneperpage

\begin{exercise}
Using the standard topology on $\R$, is the product topology on $\R \times \R$ the same as
the standard topology on $\R^2$?
\end{exercise}

\oneperpage

\begin{exercise}
\label{thm:finite-gate}
A {\em basis} for the 
{\em product topology} on $\Pi_{\alpha \in \lambda} X_\alpha$ 
is the collection of all sets of the form 
$\Pi_{\alpha\in\lambda} U_\alpha$ 
where $U_\alpha$ is open in $X_\alpha$ for each
$\alpha$ and $U_\alpha = X_\alpha$ for all but finitely many $\alpha$.
\end{exercise}

\oneperpage

\begin{exercise}
\label{basis-for-2^X}
Let $\Tau$ be the topology on $2^X$ with basis generated by the subbasis $\mathcal{S}$.
\begin{enumerate}
\item Every basic open set in $2^{X}$ is both open and closed.
\item Show that if a collection of subbasic open sets of $2^X$ has the property that every point of $2^X$ lies in at least one of those subbasic open sets, then there are two subbasic open sets in that collection such that every point of $2^X$ lies in one of those two subbasic sets.
\item Show that if a collection of basic open sets of $2^X$ has the property that every point of $2^X$ lies in at least one of those basic open sets, then there are a finite number of basic open sets in that collection such that every point of $2^X$ lies in one of those basic sets.
\end{enumerate}
\end{exercise}

\oneperpage

\begin{exercise}
 In the product space $2^\R$, what is the closure of the set $Z$ consisting of all elements of $2^\R$ that are $0$ on every rational coordinate, but may be $0$ or $1$ on any irrational coordinate?
 Equivalently, thinking of $2^\R$ as subsets of $\R$, what is the closure of the set $Z$ consisting of all subsets of $\R$ that do not contain any rational?
  \end{exercise}

\oneperpage

\begin{exercise}
\label{exercise:2^R}
Find a subset $A$ of $2^{\R}$ and a limit point $x$ of $A$
% that is not in $A$
such that no sequence in $A$ converges to $x$.  For an even greater challenge, determine whether you can find such an example if $A$ is countable.
\end{exercise}

\oneperpage

\begin{exercise}
\label{exercise:R^omega}
Let $\R^\omega$ be the countable product of copies of $\R$.  
So every point in $\R^\omega$ is a sequence $(x_1, x_2, x_3,...)$.
Let $A \subset \R^\omega$ be the set consisting of all points 
with only positive coordinates.  Show that in the product topology, ${\bf 0}=(0,0,0,...)$ is a limit
point of the set $A$, and there is a sequence of points in $A$
converging to ${\bf 0}$.  Then show that in the box topology, ${\bf 0}=(0,0,0,...)$ is a limit
point of the set $A$, but there is no sequence of points in $A$ converging to ${\bf 0}$.
\end{exercise}

\oneperpage

\begin{exercise}
Show that the set $2^\N$ in the box topology is a discrete space, whereas 
the set $2^\N$ in the product topology has no isolated points.
\end{exercise}

\oneperpage

