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\begin{center}
POINT SET TOPOLOGY
\end{center}
\begin{description}
\item[Definition 1]
A topological structure on a set $X$ is a family
${\cal{O}}\subset{\cal{P}}(X)$ called open sets and satisfying
\begin{description}
\item[($O_1$)]
${\cal{O}}$ is closed for arbitrary unions
\item[($O_2$)]
${\cal{O}}$ is closed for finite intersections.
\end{description}
\item[Definition 2]
A set with a topological structure is a topological space
$(X,{\cal{O}})$
$$\cup_{\emptyset}=\cup_{i\in\emptyset}E_i=\{x:x\in E_i\textrm{for
some }i\in\emptyset\}=\emptyset$$
so $\emptyset$ is always open by $(O_1)$
$$\cap_{\emptyset}=\cap_{i\in\emptyset}E_i=\{x:x\in E_i\textrm{for
all }i\in\emptyset\}=X$$
so $X$ is always open by $(O_2)$.
\item[Examples]
\begin{description}
\item[(i)]
${\cal{O}}={\cal{P}}(X)$ the discrete topology.
\item[(ii)]
${\cal{O}}\{\emptyset,X\}$ the indiscrete of trivial topology.
These coincide when $X$ has one point.
\item[(iii)]
${\cal{Q}}$=the rational line.
${\cal{O}}$=set of unions of open rational intervals
\end{description}
\item[Definition 3]
Topological spaces $X$ and $X'$ are homomorphic if there is an
isomorphism of their topological structures i.e. if there is a
bijection (1-1 onto map) of $X$ and $X'$ which generates a
bijection of ${\cal{O}}$ and $\cal{O}$.
e.g. If $X$ and $X$ are discrete spaces a bijection is a
homomorphism. (see also Kelley p102 H).
\item[Definition 4]
A base for a topological structure is a family
${\cal{B}}\subset{\cal{O}}$ such that every $o\in{\cal{O}}$ can be
expressed as a union of sets of ${\cal{B}}$
\item[Examples]
\begin{description}
\item[(i)]
for the discrete topological structure $\{x\}_{x\in X}$ is a base.
\item[(ii)]
for the indiscrete topological structure $\{\emptyset, X\}$ is a
base.
\item[(iii)]
For ${\cal{Q}}$, topologised as before, the set of bounded open
intervals is a base.
\item[(iv)]
Let $X=\{0,1,2\}$
Let ${\cal{B}}=\{(0,1),(1,2),(0,12)\}$. Is this a base for some
topology on $X$? i.e. Do unions of members of $X$ satisfy $(O_2)$?
$(0,1)\cap(1,2)=(1)$- which is not a union of members of
${\cal{B}}$, so ${\cal{B}}$ is not a base for any topology on $X$.
\end{description}
\item[Theorem 1]
A necessary and sufficient condition for ${\cal{B}}$ to be a base
for a topology on $X=\cup_{o\in{\cal{B}}}o$ is that for each $O'$
and $O''\in{\cal{B}}$ and each $x\in O'\cap O''\exists
O\in{\cal{B}}$ such that $x\in O\subset O'\cap O''$.
\item[Proof]
Necessary: If ${\cal{B}}$ is a base for ${\cal{O}},\ O'\cap
O''\in{\cal{O}}$ and if $x\in O'\cap O''$, since $O'\cap O''$ is a
union of sets of ${\cal{B}}\ \exists O\in{\cal{B}}$ such that
$x\in O\subset O'\cap O''$.
Sufficient: let ${\cal{O}}$ be the family of unions of sets of
${\cal{B}}$.
$(O_1)$ is clearly satisfied.
$(O_2)\ (\cup A_i)\cap(\cup B_j)=\cup(A_i\cap B_j)$ so that it is
sufficient to prove that the intersection of two sets of
${\cal{B}}$ is a union of sets of $\cal{B}$.
Let $x\in O'\cap O''$. Then $\exists Ox\in\cal{B}$ such that $x\in
O\ x\subset O'\cap O''$ so that $O'\cap O''=\cup_{x\in O'\cap O''}
Ox$.
\item[Theorem 2]
If ${\cal{S}}$ is a non-empty family of sets the family
${\cal{B}}$ of their finite intersections is a base for a topology
on $\cup_{\cal{S}}$
\item[Proof]
Immediate verification of $O_1$ and $O_2$. The topology generated
in this way is the smallest topology including all the sets of
${\cal{S}}$.
\item[Definition 4]
A family ${\cal{S}}$ is a sub base for a topology if the set of
finite intersections is a base for the topology.
e.g. $\{a\infty\}_{a\in Q}$ and $\{(-\infty\ a)\}_{a\in Q}$ are
sub bases for ${\cal{Q}}$
\item[Definition 5]
If a topology has a countable base it satisfies the second axiom
of countability.
\item[Definition 6]
In a topological space a neighbourhood of a set $A$ is a set which
contains an open set containing $A$. A neighbourhood of a point
$X$ is a neighbourhood of $\{x\}$.
\item[Theorem 3]
A necessary and sufficient condition that a set be open is that it
contains (is) a neighbourhood of each of its points.
\item[Proof]
Necessary: Definition of a neighbourhood
Sufficient: Let $O_A=\cup$ open subsets of $A$. $O_A$ is open
$(O_1)$ and $O_A\subset A$.
If $x\in A\ A\supset$ a neighbourhood of $x\supset$ open set $\ni
x$ therefore $x\in O_A$ therefore $A\subset O_A$.
Let $V(x)$ denote the family of neighbourhoods of $x$. Then $V(x)$
has the following properties.
\begin{description}
\item[($V_1$)]
Every subset of $X$ which contains a member of $V(x)$ is a member
of $V(x)$
\item[($V_2$)]
$V(x)$ is closed for finite intersections
\item[($V_3$)]
$x$ belongs to every member of $V(x)$.
\item[($V_4$)]
If $v\in V(x)\exists W\in V(x)$ such that $v\in V(y)$ for all
$y\in W$.
(Take $W$ to be an open set $\ni x$ and $\subset V$.)
\end{description}
\item[Theorem 4]
If for each point $x$ of $X$ there is given a family $V(x)$ of
subsets of $X$, satisfying $V_{1-4}$ then $\exists$ a unique
topology on $X$ for which the sets of neighbourhoods of each point
$x$ are precisely the given $V(x)$. (Hausdorff).
\item[Proof]
If such a topology exists theorem 3 shows that the open sets must
be the O such that for each $x\in O\ O\in V(x)$ and so there is at
most one such topology. Consider the set $O$ so defined.
$(O_1)$ Suppose $x\in \cup O$. Then $x\in $some $O'\ O'\in V(x)\
O'\subset |cup O$ so $\cup O\in V(x)$.
$(O_2)$ Suppose $x\in\cap_Fo\Rightarrow x\in$ each $O$. each $O\in
V(x)$ therefore $\cap_FO\in V(x)$ by $V_2$.
Now consider the system $U(x)$ of neighbourhoods of $x$ defined by
this topology.
\begin{description}
\item[(i)]
$U(x)\subset V(x)$. Let $U\in U(x)$. Then $U\supset O\ni x$. But
$O\in V(x)$ so by $V_1\ U\in V(x)$.
\item[(ii)]
$V(x)\subset U(x)$. Let $V\in V(x)$. It is sufficient to prove
that $\exists O\in{\cal{O}}$ such that $x\in O\subset V$.
Let $O=\{y:V\in V(y)\}\ x\in O$ since $V\in V(x)$.
$O\subset V$ since $y\in V$ for all $y\in O$ by $V_3$.
To prove that $O\in{\cal{O}}$ it is sufficient to prove that $V\in
V(y)$ for all $y\in O$.
If $t\in O\ V\in V(y)$ by definition of $O$ therefore $\exists
W\in V(y)$ such that $V\in V(z)$ for all $z\in W$ by $V-4$.
Therefore $W\subset O$ (definition of $O$)
Therefore $O\in V(y)$ by $V_1$
e.g. $R_1:V(x)$= sets which contain interval $(a,b)\ a0\right\}$ etc.
\item[(iv)]
In $R_2$ the squares, discs, ellipses et. centre 0 form bases for
the filter of neighbourhoods of 0.
\item[(v)]
In $R_1\ \{x,|x|>n\}$ is a filter base.
\end{description}
\item[Definition 22]
A fundamental system of neighbourhoods of a point in a topological
space is a base for the filter of neighbourhoods of the point.
\item[Definition 23]
A space satisfies the first axiom of countability if every point
has a countable fundamental system of neighbourhoods.
2nd axiom $\Rightarrow$ 1st axiom
1st axiom $\not\Rightarrow$ 2nd axiom.
e.g. $X$ uncountable, with discrete topology. $\{x\}$ is a base
for the neighbourhoods of $x$ and is countable.
\item[Theorem 19]
Let ${\cal{F}}$ be a filter on $X\ A\subset X$. A necessary and
sufficient condition that $\ds{\cal{F}}_A=\{F\cap
A\}_{F\in{\cal{F}}}$ be a filter on $A$ is that $F\cap
A\neq\emptyset\forall F\in {\cal{F}}$.
If $A\in{\cal{F}}\ {\cal{F}}_A$ is a filter on $A$.
\item[Definition 24]
${\cal{F}}_A$ is a filter induced on $A$ by ${\cal{F}}$
e.g. $X$ a topological space $A\subset X$ $V_A(x)$ is a filter on
$A\Leftrightarrow x\subset \overline{A}$
Let $f:X\to Y$ and let ${\cal{F}}$ be a filter on $X$. Then in
general $\ds\{f(F)\}_{F\in{\cal{F}}}$ is not a filter, for $F_2$
breaks down. But if ${\cal{B}}$ is a filter base for a filter on
$X$ then $\ds\{f(B)\}_{B\in{\cal{B}}}$ is a base for a filter i=on
$Y$.
\item[Definition 25]
$X$ a topological space. ${\cal{F}}$ a filter on $X$. $x$ is a
limit point of ${\cal{F}}$ if ${\cal{F}}\supset V(x)$ we say
${\cal{F}}$ converges to $X$.
$x$ is a limit of a filter base ${\cal{B}}$ if the filter with
base ${\cal{B}}$ converges to $x$
e.g.
\begin{description}
\item[(i)]
In $R_1$, the filter with base
$\ds\left\{\left(-\frac{1}{n},\frac{1}{n}\right)\right\}$
converges to 0, but that with base $\ds\{\{x:|x|>n\}\}$ does not
converge.
\item[(ii)]
\begin{eqnarray*}
X&=&\{x,y,z\}\\ {\cal{O}}&=&\{\emptyset,\{x,y\},\{z\} X\}\\
{\cal{B}}&=&\{\{x,y\} X\}\\ V(x)&=&\{\{x,y\} X\}
\end{eqnarray*}
Therefore ${\cal{B}}$ converges to $x$, $V(y)=\{(x,y)\ X\}$
therefore ${\cal{B}}$ converges to $y$.
\end{description}
\item[Definition 26]
$X$ a topological space.
${\cal{B}}$ is a filter base on $X$. $x$ is adherent to
${\cal{B}}$ if it is adherent to every set of ${\cal{B}}$.
$x$ is adherent to ${\cal{B}}\Leftrightarrow V\cap
B\neq\emptyset\forall V\in V(x),\ B\in{\cal{B}}$.
Every limit point of a filter is adherent to the filter.
The set of point adherent to a filter is
$\ds\cap_{B\in{\cal{B}}}\overline{B}$ and is closed.
\item[Definition 27]
Let $X$ be a set, $Y$ a topological space. Let ${\cal{P}}:X\to Y$.
Let ${\cal{F}}$ be a filter on $X$.
$y\in Y$ is a limit of ${\cal{F}}$ along ${\cal{F}}$ if $y$ is
adherent to ${\cal{P}}({\cal{F}})$
\item[Examples]
\begin{description}
\item[(i)]
Fr\'{e}chet ${\cal{F}}$ (of sections of $Z$) \ $\alpha: Z\to R\ \
n\mapsto a_n$.
A set of the filter is a set $\supset\{m:m\geq n\}=F\ \
\alpha(F_\ni\{a_m:m\geq n\}$.
$a$ is a limit of $\alpha$ along ${\cal{F}}\Leftrightarrow a$ is
adherent to every set of ${\cal{F}}$ i.e. every
$(a-\varepsilon,a+\varepsilon)$ meets every set of ${\cal{F}}$
i.e. $(a-\varepsilon,a+\varepsilon)\cap\{a_m:m\geq
n\}\neq\emptyset$ for all $n$.
\item[(ii)]
$X$ a topological space, $Y$ a topological space. ${\cal{p}}:X\to
Y$. $V(a)$ is the filter of neighbourhoods of $a\in X$.
In this case we write $\ds y=\lim_{x\to a}{\cal{P}}(x)$ instead of
$\ds\lim_{\cal{F}}{\cal{P}}$.
\end{description}
\item[Theorem 20]
$X,Y$ topological spaces.
${\cal{P}}:X\to Y$ continuous at $\ds a\in
Y\Leftrightarrow\lim_{x\to a}{\cal{P}}(x)={\cal{P}}(a)$.
\item[Proof]
$\ds\lim_{x\to a}{\cal{P}}(x)={\cal{P}}(a)\Leftrightarrow$ given
$u\in U({\cal{P}}(a))\exists V\in V(a)$ such that
${\cal{P}}(V)\subset U$.
let $X,Y$ be topological spaces. Let $A\subset X$ and let
$a\in\overline{A}$. Let $f:A\to Y$. Let
$\ds{\cal{F}}=V_A(a)=\{A\cap V\}_{v\in V(a)}$.
We write $\ds\lim_{x\to a,x\in A}f(x)$ instead of
$\ds\lim_{\cal{F}}f$.
\item[Definition 28]
$\ds\lim_{x\to a,\ x\in A}f(x)$ is a limit of $f$ at $a$ relative
to $A$.
\item[Theorem 21]
Let $X$ be a set and let $\{Y_i\}_{i\in I}$ be a family of
topological spaces. Let $f_i:X\to Y_i$.
A filter ${\cal{F}}$ on $X$ converges to $a\in X$ in the initial
topology ${\cal{O}}$ on $X\Leftrightarrow$ the filter base
$f_i({\cal{F}})$ converges to $f_i(a)\ \forall i\in I$.
\item[Proof]
Necessary condition: The $f_i$ are continuous by Theorem 20.
Sufficient condition: Let $w\in V(a)$. By definition of
${\cal{O}}\exists \cap_J A_i\subset W$ where $a\in
A_i=f^{-1}(O_i),\ O_i$ open in $Y_i$ and $J$ is a finite set.
Since $f_i({\cal{F}})$ converges to $f_i(a)$ in $Y_i$ $O_i\in
f_i({\cal{F}})$ and $f^{-1}(O_i)\in{\cal{F}}$ so that $\ds\cap_J
f^{-1}(O_i)\in{\cal{F}}$ i.e. $V(a)\subset F$.
\item[Corollary]
A filter ${\cal{F}}$ on a product space $\ds X=\prod_I X_i$
converges to $x\Leftrightarrow$ the filter base $pr_i({\cal{F}})$
converges to $x_i\ \forall i\in I$.
\end{description}
\end{document}