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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 $(a,b)\ a&lt;x&lt;b$.

\end{description}

\item[Definition 7]
A metric for a set $X$ is a function $\rho$ form $X\times X$ to
$R^+$ (non-negative reals) such that

\begin{description}

\item[$M_1$]
$\rho(x,y)=\rho(y,x)$ for all $x,y$

\item[$M_2$]
$\rho(x,z)\leq\rho(x,y)+\rho(y,z)$ for all $x,y,z$

\item[$M_3$]
$\rho(x,x)=0$ for all $x$.

\end{description}

This is sometimes called a pseudo metric and for a metric we have

($M_3')\ \ \rho(x,y)\geq0,\ =0\Leftrightarrow x=y$.

\item[Definition 8]
The open $r$- Ball about $x\ =\{y:\rho(x,y)&lt;r\}$ and is denoted by
$B(r,x)$

The closed $r$-ball around $x\ =\{y:\rho(x,y)\leq R\}$ and is
denoted by $\overline{B}(r,x)$.

$V(x)=\{$ sets which contain one of $B\left(\frac{1}{n},x\right)\
n=1,2,\ldots\}$ satisfies $V_{1-4}$ and so defines a topology on
$X$. This topology is the metric topology defined on $X$ by
$\rho$.

The development of topology from the neighbourhood point of ve=iew
is due to H. Weyl and Hausdorff. That from the open sets aspect is
due to Alexandroff and Hopf.

\item[Definition 9]
The closed sets $G$ of a topological space are the complements of
the open sets.

\begin{description}

\item[($G_1$)]
$G$ is closed for arbitrary intersections

\item[($G_2$)]
$G$ is closed for finite unions. $\emptyset$ and $X$ are both
closed and open.

\end{description}

Clearly given a family $G$ satisfying $G_1$ and $G_2$ the family
of complements is a topology for which the closed sets are the
sets of $G$.

\item[Definition 10]
a point $x$ is an interior point of a set $A$ if $A$ is a
neighbourhood of $x$.

The set of interior points of $A$ is the interior $A^0$ of $A$.

An exterior point of $A$ is an interior point of $cA$ (i.e.
$\exists$ a neighbourhood of $x$ which does not meet $A$, or $x$
is isolated, separated from $A$.)

\item[Theorem 5]

\begin{description}

\item[(i)]
The interior of a set is open and

\item[(ii)]
is the largest open subset.

\item[(iii)]
A necessary and sufficient condition for a set to be open is that
it coincides with its interior.

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
$x\in A^{0}\Rightarrow \exists$ open $O$ such that $x\in O\subset
A$.

$y\in O\Rightarrow y\in A^0$ therefore $x\in O\subset A^0$
therefore $A^0$ is a neighbourhood of each of its points and so it
is open.

\item[(ii)]
Let $O\subset A\Rightarrow O\subset A^o$ therefore $\cup_{O\subset
A^0}O\subset A^o$, but $A^0$ is such an $O$ therefore
$\cup_{O\subset A}O=A^0$.

\item[(iii)]
Sufficient condition from (i).

Necessary condition from (ii).

$\overbrace{A\cap B}^0=A^0\cap B^0$ but $\overbrace{A\cup B}^0\neq
A^0\cup B^0$

e.g. $X=R_1$ with metric topology, $A$=rationals, $B$=irrationals.

$A^0=\emptyset\ B^0=\emptyset$ therefore $A^0\cup B^0=\emptyset$.
But $A\cup B=R_1$ and so $\overbrace{A\cup B}^0=R_1$.

However $A^0\cup B^0\subset \overbrace{A\cup B}^0$ always.

\end{description}

\item[Definition 11]
A point $x$ is adherent to a set $A$ if every neighbourhood of $x$
meets $A$.

The set of points adherent to a set $A$ is called the adherence
(closure) $\overline{A}$ of $A$.

An adherent point of $A$ is an isolated point of $A$ if there is a
neighbourhood of $A$ which contains no point of $A$ other than
$x$; otherwise it is a point of accumulation (limit point) of $A$.

\item[Examples]

\begin{description}

\item[(i)]
$A={\cal{Q}}\subset{\cal{R}}\ \overline{A}=R$

\item[(ii)]
In a discrete space no set has accumulation points, every point is
isolated.

\item[(iii)]
In an indiscrete space every non-empty set has $X$ as its
adherence.

\end{description}

\item[Theorem 5]

\begin{description}

\item[(i)]
The adherence of a set is closed and

\item[(ii)]
is the smallest closed set containing the given set

\item[(iii)]
$A$ is closed $\Leftrightarrow A=\overline{A}\Leftrightarrow
A\supset$ its accumulation points.

\begin{eqnarray*}
c\overline{A}&amp;=&amp;\overbrace{cA}^0\\ cA^0&amp;=&amp;\overline{cA}
\end{eqnarray*}

\end{description}

\item[Corollary]
$\overline{\overline{A}}=\overline{A}$

$\overline{A}\cup\overline{B}=\overline{A\cup B}$

\item[Definition 12]
The set of accumulation points of $A$ is its derived set $A'$.

A perfect set is a closed set without isolated points.

Suppose we map ${\cal{P}}(X)\to{\cal{P}}(X)$ where
$A\to\overline{A}$ then

\begin{description}

\item[($C_1$)]
$\overline{\emptyset}=\emptyset$

\item[($C_2$)]
$A\subset\overline{A}$

\item[($C_3$)]
$\overline{\overline{A}}\subset\overline{A}$ (with $C_2\Rightarrow
\overline{\overline{A}}=\overline{A}$)

\item[($C_4$)]
$\overline{A\cup B}=\overline{A}\cup\overline{B}$.

\end{description}

\item[Theorem 6]
If we are given an operation mapping ${\cal{P}}(X)$ into
${\cal{P}}(X)$ which has the properties $C_{1-4}$ then the set of
complements of the sets $G$ such that $G=\overline{G}$ is a
topology on $X$ for which $\overline{AS}$ is the closure of $A$
for all $A\subset X$ (Kunatowski).

\item[Definition 13]
The frontier or boundary of a set $A$ is
$\overline{A}\cap\overline{cA}$ and is the set of points adherent
to $A$ and to $cA$, or is the set interior to neither $A$ nor
$cA$, or is the set neither interior or exterior to $A$. It is a
closed set.

A set is closed $\Leftrightarrow$ it contains its boundary.

A set is open $\Leftrightarrow$ it is disjoint from its boundary.

\item[Definition 14]
A set is dense in $X$ if $\overline{A}=X$.

$A$ is dense in itself if all its points are accumulation point,
i.e. $A\subset A'$.

A set is nowhere dense if $c\overline{A}$ is dense i.e.
$\overline{A}^0=\emptyset$.

\item[Induced topologies]
Given $(X{\cal{O}})$ and $Y\subset X$ , can we use ${\cal{O}}$ to
get a topology for $Y$.

\item[Theorem 7]
$\{Y\cap O\}_{O\in{\cal{O}}}$ is a topology $O_Y$ on $Y$ called
the topology induced on $Y$ by $(x{\cal{O}})$

\item[Proof]

$(O_1)\ \cup Y\cap O=Y_n\cup O$

$(O_2)\cap_F Y_nO=Y_n\cap_FO$

If $Z\subset Y\subset X$ then ${\cal{O}}_Y$ and ${\cal{O}}$ induce
the same topology on $Z$.

A (sub) base ${\cal{B}}$ for ${\cal{O}}$ induces a (sub) base
${\cal{B}}_Y$ for ${\cal{O}}_Y$.

\lq\lq $O$ open in (relative to )$Y$'' means\lq\lq $O$ open in
$(YO_Y)$.

A necessary and sufficient condition that every set open in $Y$ be
open in $X$ is that $Y$ be open in $X$.

\item[Theorem 8]

\begin{description}

\item[(i)]
$G\subset Y$ is closed in $Y\Leftrightarrow G=Y\cap G'$ where $G'$
is closed in $X$.

\item[(ii)]
$V_Y(x)=\{Y\cap V\}_{v\in V(x)}$

\item[(iii)]
If $z\subset Y\subset X$ then
$\overline{Z}_{mY}=Y_n\overline{Z}_{mx}$

\end{description}

\item[Proof]

\begin{description}

\item[(i)]
$G$ is closed in $Y$

$\Leftrightarrow Y\cap cG$ is open in $Y$

$\leftrightarrow Y\cap cG=Y\cap O; O$ open in $X$

$\Leftrightarrow Y\cap G=Y\cap cO$ (take comp. in $Y$)

$\Leftrightarrow G=Y\cap cO$

take $G'=cO$.

\item[(ii)]
$Y\supset U\in V_Y(x)$

$\Leftrightarrow U\supset O\ni x$, $O$  open in $Y$

$\Leftrightarrow U\supset O'\cap Y\ni x\ O'$ open in $X$

$\Leftrightarrow U=V\cap Y$ where $V\supset O'\ni x$

i.e. $U=Y\cap V$ where $V\in V(x)$.

\item[(iii)]
$\overline{Z}_{mY}=\bigcap$ closed sets in $Y$ which $\supset Z$

$=\bigcap(Y_\cap$ closed sets $\supset Z)$

$=Y\cap \bigcap$ closed sets $\supset Z$

$=Y\cap \overline{Z}$

\end{description}

\item[Continuous functions]

\item[Definition 15]
A map $F$ of a topological space $X$ into a topological space $Y$
is continuous at $x_0\in X$ if given a neighbourhood $V$ of
$f(x_0)$ in $Y\ \exists $ a neighbourhood $U$ of $x_+0$ in $X$
such that $f(U)\subset V$.

$f$ is continuous at $x_0$ if for every neighbourhood $V$ of
$f(x_0)\ f^{-1}(V)$ is a neighbourhood of $x_0$.

\item[Theorem 9]
If $f:X\to Y$ is continuous at $X$ and $x\in \overline{A}$ then
$f(x)\in \overline{f(A)}$

\item[Proof]
Let $V\in V(f(x0)$ then $f^{-1}(V)\in V(x)$ therefore
$f^{-1}(V)\cap A\neq\emptyset$. Therefore $f(f^{-1}(V))\cap
f(A)\neq f(\emptyset)=\emptyset$.

$f\circ f^{-1}$ is the identity therefore $V\cap
f(A)\neq\emptyset$

\item[Theorem 10]
if $f:X\to Y$ is continuous at $x_0$ and $g:Y\to Z$ is continuous
at $f(x_0)$ Then $g\circ f$ is continuous at $x_0$.

\item[Proof]
Let $W\in V(g\circ f(x_0))$ then $g^{-1}(W)\in V(f(x_0))$

$f^{-1}\circ g^{-1}(W)\in V(x_0)\ \ (g\circ f)^{-1}(W)\in V(x_0)$

\item[Definition 16]
A map $f:X\to $ is continuous (on $X$) if it is continuous at each
point of $X$.

\item[Theorem 11]
Let $f:X\to Y$. Then the following properties are equivalent.

\begin{description}

\item[(i)]
$f$ continuous on $X$

\item[(ii)]
$f(\overline{A})\subset\overline{f(A)}$ for all $A\in
{\cal{P}}(X)$

\item[(iii)]
the inverse image of a closed set is closed

\item[(iv)]
the inverse image of an open set is open.

\end{description}

\item[Proof]
(i)$\Rightarrow$ (ii) by theorem 9.

(ii)$\Rightarrow$ (iii) Let $G'$ be closed in $Y$ and
$f^{-1}(G')=G$.

$f(\overline{G})\subset\overline{f(G)}$ by (ii) $\subset
\overline{G'}=G'$.

Therefore $\overline{G}\subset f^{-1}(G')=G\subset \overline{G}$
therefore $\overline{G}=G$ therefore $f^{-1}(G')$ is closed.

(iii)$\Rightarrow$ (iv)

$cf^{-1}(A)=f^{-1}(cA)\ A\subset Y$

(iv)$\Rightarrow$ (i) Let $x\in X v\in V(f(x))$.

$\exists x\in O\subset V\ O$ open in $Y$. $f^{-1}(O)$ is open in
$Y$ and contains $x$ so is a neighbourhood of $x$ in $X
f(f^{-1}(O))\subset V$

\item[Note]
The image of an open set under a continuous map is not necessarily
open.

e.g. $f:R\to R\ x\mapsto\frac{1}{1+x^2}$

$f(R)=(0\ 1]$ not open.

\item[Comparison of Topologies]

\item[Definition 17]
If ${\cal{O}}_1$ and ${\cal{O}}_2$ are topologies on a set $X,\
{\cal{O}}_1$ is finer than ${\cal{O}}_2({\cal{O}}_2$ is coarser
then ${\cal{O}})$ if ${\cal{O}}_1\supset{\cal{O}}_2$ (strictly
finer if not equal).

[Topologies on $X$ are not necessarily complete]

e.e. $R_1:{\cal{O}}_1$:usual topology\ \ ${\cal{O}}_2$: open sets
are open sets of ${\cal{O}}_1$ which contains $O$ and $\{o\}$.

These topologies are not comparable.

\item[Theorem 12]
If ${\cal{O}}_1$ and ${\cal{O}}_2$ are topologies on $X$, the
following properties are equivalent:

\begin{description}

\item[(i)]
${\cal{O}}_1\supset{\cal{O}}_2$

\item[(ii)]
${\cal{O}}_2$ - closed sets are ${\cal{O}}_1$ closed

\item[(iii)]
$V_1(x)\supset V_2(x)$ for all $x\in X$

\item[(iv)]
The identity map $(X\ {\cal{O}})\to(X{\cal{O}}_2)$ is continuous.

\item[(v)]
$\overline{A}^{(1)}\subset\overline{A}^{(2)}$ for all $A\subset X$

\end{description}

We also have the following qualitative results:

the discrete topology on a set is the finest topology on the set,
and the indiscrete is the coarsest.

The finer the topology, the more open sets, closed sets,
neighbourhoods of a point, the smaller the adherence, the larger
the interior of a set, the fewer the dense sets.

If we refine the topology of $X$ we get more continuous functions.
If we refine the topology of $Y$ we get fewer continuous
functions.

\item[Final Topologies]

\item[Theorem 13]
Let $X$ be a set. Let $\ds(Y_i,{\cal{O}}_i=\{O_{ij}\}_{j\in
J_i})_{i\in I}$ be a family of topological spaces.

Let $f_i:Y_i\to X$. Let ${\cal{O}}=\{O\subset X:f_i^{-1}(o)$ open
in $Y_i$ for all $i\in I\}$.

Then ${\cal{O}}$ is a topology on $X$ and is the finest for which
the $f_i$ are continuous.

If $g:X\to Z$ is a map into a topological space $Z,\ g$ is
continuous from $(X{\cal{O}})\to Z\Leftrightarrow g\circ f_i$ are
all continuous.

\item[Proof]
${\cal{O}}$ is non-empty: $f^{-1}(X)=Y_i$

$(O_1)$ Let $f_i^{-1}(O_k)\in{\cal{O}}_1,\ O_k\in{\cal{O}}$. Then
$f_i^{-1}(\cup_k O_k)=\cup_K f_i^{-1}(O_k)\in{\cal{O}}_1$ for all
$i$.

$(O_2)$ Let $f_i^{-1}(O-k)\in{\cal{O}}_i,\ O_k\in{\cal{O}}$. Then
$f_i^{-1}(\cap_F O_k)=\cap_F f_i^{-1}(O_k)\in{\cal{O}}_i$ for all
$i$.

A necessary and sufficient condition for $f_i$ to be continuous is
that $f_i^{-1}(O)$ be open in $Y_i$ for all $i$. A finer topology
than ${\cal{O}}$ will not satisfy this.

Now let $f_i:Y_i\to X\ g:X\to Z\ g\circ f_i:Y_i\to Z$.

The necessary condition is obvious.

Sufficient condition:

$g:X\to Y$ continuous $\Leftrightarrow g^{-1}(O)$ open in $X$ for
all $O$ open in $Z$.

$g\circ f_i$ continuous $\Rightarrow {g\circ f_i}^{-1}(O)$ open in
$Y_i$ for all $i$

$\Rightarrow f_i^{-1}\circ g^{-1}(o)$ open in $Y_i$ for all $i$

$\Rightarrow g^{-1}(O)$ is open in $X$

We define ${\cal{O}}$ to be the final topology for $X$, the maps
$f_i$ and spaces $Y_i$.

\item[Examples]

\begin{description}

\item[(i)]
$X$ a topological space. R is an equivalence relation on $X$.

$\phi:X\to \frac{X}{r}=Y\ x\mapsto \dot{x}$ (~class)

The finest topology on $Y$ such that $\phi$ is continuous is the
quotient topology of that of $X$ by the relation $R$.

$f:\frac{X}{R}\to Z$ is continuous $\Leftrightarrow
f\circ\phi:X\to Z$ is continuous.

e.g. $X+R_2$

$\underline{R}:(x_1\ y_1)\sim(x_2\ y_2)\Leftrightarrow x_1=x_2$.

Then $\frac{X}{\underline{R}}=R$ (isomorphically).

\item[(ii)]
$X$ a set. $(X{\cal{O}}_i)$ a family of topological spaces.

$\phi_i:(X,{\cal{O}}_i\to X\ x\mapsto x$

The final topology on $X$ is the finest topology coarser than all
the ${\cal{O}}_i$.

${\cal{O}}$ is called the lower bound of the ${\cal{O}}_i$.
${\cal{O}}=\cap_i{\cal{O}}_i$

\end{description}

\item[Initial Topologies]

\item[Theorem 14]
Let $X$ be a set. Let $(Y_i{\cal{O}}_i)_{i\in I}$ be a family of
topological spaces. Let $f_i:X\to Y_i$. Let $f_i:X\to Y_i$. Let
$S=\{f_i^{-1}(O_{ij})\}_{i\in I, j\in J_i}$.

Then $S$ is a sub-base for a topology $\cal{J}$ on $X$, the
coarsest topology for which all the $f_i$ are continuous.

If $Z$ is a topological space $g:Z\to X$ is continuous
$\Leftrightarrow f_i\circ g$ continuous for all $i\in I$

\item[Proof]
$S$ is non-empty, for $\emptyset \in S$.

$\cup_{s\in S} S=X(-f_i^{-1}(Y_i))$ then use theorem 2.

A necessary and sufficient condition for $f_i$ to be continuous is
the $f_i^{-1}(O_{ij})$ be open in ${\cal{J}}$ for all $ij$.

the rest of the proof is similar to theorem 13.

We define ${\cal{J}}$ as the initial topology for $X$, the maps
$f_i$ and spaces $Y_i$.

\item[Examples]

\begin{description}

\item[(i)]
$X$ a set $(Y{\cal{O}})$ a topological space $f:X\to Y$.

the initial topology here is called the inverse image of
${\cal{O}}$.

\item[(ii)]
If $X\subset Y\ f:X\to Y\ x\mapsto x$ is the canonical injection.

$f^{-1}(A)=A\cap X$ and the open sets of the initial topology are
the intersections with $X$ of the open sets of $Y$- we have the
induced topology.

\item[(iii)]
$(X{\cal{O}}_i)\ \phi_1:X\to X\ x\mapsto x$.

The initial topology is the coarsest topology finer than all the
${\cal{O}}_i$

\item[(iv)]
$(X_i{\cal{O}}_i)_{i\in I}\ X=\prod_{i\in I}X_i$

$\phi_i=\textrm{proj}_i:X\to X_i\ \{x_i\}_{i\in I}\mapsto x_i$

the initial topology is the coarsest for which all the projections
are continuous and is called the product topology of the
${\cal{O}}_i$. $(X{\cal{J}})$ is the topological product of the
$(X_i,{\cal{O}}_1)$. The $(X_i,{\cal{O}}_i)$ are the Factor
spaces. The open sets of the product topology have as base the
finite intersections of sets $\textrm{proj}_i^{-1}(O_{ij})$ where
$O_{ij}$ is open in $(X_i\ {\cal{O}}_i)$.

proj$_i^{-1}(O_{ij}=\prod_{i\in I}A_i$ where $A_u=X_u\ i\neq j,\
A_j=O_{ij}$ and the base consists of sets $\prod_{i\in I} A_i$
where $A_i=X_i$ except for a finite set of $i$, where $A_i$ is
open in $X_i$. These are called elementary sets.

$g:Z\to\prod X_i$ is continuous $\Leftrightarrow $ proj$_i\circ g$
is continuous for all $i$ i.e. all the co-ordinates are
continuous.

\end{description}

\item[Limit Processes]
Consider the following limit processes:

\begin{description}

\item[(i)]
$\ds\lim_{n\to\infty}a_n$

\item[(ii)]
$\ds\lim_{x\to a}f(x)$

\item[(iii)]
$\ds R\int_a^bf(x)\,dx=\lim\sum(x_{i+1}-x_i)f(\xi_i)$

\end{description}

What have these in common.

\begin{description}

\item[A.]
A set with some order properties

\begin{description}

\item[(i)]
$Z\to R\ n\mapsto a_n$

\item[(ii)]
$V(a)\to R\ v\mapsto f(v)$

\item[(iii)]
Nets on $(a\ b)\ N\to I(N,f)$

\end{description}

\item[B.]
A map of the ordered set into a topological space.

\item[C.]
We consider the set of images as we proceed along the order.

We now unify all these.

\end{description}

\item[The Theory of Filters]
(H.CARTAN 1937)

\item[Definition 18]
A filter on a set $X$ is a family ${\cal{F}}\subset {\cal{P}}(X)$
such that

\begin{description}

\item[($F_1$)]
$A\in {\cal{F}}$ and $B\supset A\Rightarrow B\in {\cal{F}}$

\item[($F_2$)]
$\cal{F}$ is closed for finite intersections

\item[($F_3$)]
$\emptyset\not\in {\cal{F}}$.

\end{description}

$F_3\Rightarrow$ every finite $A^n$ in $F$ is non-empty.

$F_2\Rightarrow X\in {\cal{F}}$ i.e. ${\cal{F}}$ is non-empty.

\item[Examples]

\begin{description}

\item[(i)]
$X\neq\emptyset\ \{x\}$ is a filter on $X$.

\item[(ii)]
$\emptyset\neq A\subset X:\{Y:Y\supset A\}$ is a filter on $X$.

\item[(iii)]
$X$ an infinite set.

The complements of the finite subsets form a filter on $X$.

\item[(iv)]
If $X=Z$ (the positive integers) the filter of (iii) is called the
Fr\'{e}chet filter.

\item[(v)]
$X$ a topological space. The set of neighbourhoods of
$\emptyset\neq A\subset X$ is a filter. $A=\{x\}$ gives the
neighbourhoods of $x$.

\end{description}

\item[Comparison of Filters]

\item[Definition 19]
If $\cal{F}$ and ${\cal{F}}'$ are filters on a set $X$ and
${\cal{F}}\subset{\cal{F}}'$ we say ${\cal{F}}$ is coarser than
${\cal{F}}',\ {\cal{F}}'$ is finer than ${\cal{F}}$.

Filters are not neccessarily comparable e.g. the filters of
neighbourhoods of distinct points in a metric space.

If $\ds\{{\cal{F}}_n\}_{i\in I}$ is a family of filters on $X$
then $\ds{\cal{F}}=\bigcap_{i\in I}{\cal{F}}_i$ is a filter.

\item[Definition 20]
The intersection of the ${\cal{F}}_i$ is the finest filter coarser
than all the ${\cal{F}}_i$ and is the lower bound of the
${\cal{F}}_i$.

\item[Theorem 15]
Let $\cal{S}$ be a system of sets in $X$ . A necessary and
sufficient condition that $\exists$ a filter on $X$ containing
${\cal{S}}$ is that the finite intersections of members of
${\cal{S}}$ be non-empty.

\item[Proof]
Necessary: Immediate from $F_2$

Sufficient: Consider the family ${\cal{F}}$ of sets which contain
a member of ${\cal{S}}'$, the set of finite intersections on $
{\cal{S}}$.

${\cal{F}}$ satisfies $F_1,F_2,F_3$.

Any filter $\supset{\cal{S}}$ is finer than ${\cal{F}}$.
${\cal{F}}$ is the coarsest filter $\supset{\cal{S}}$.

${\cal{S}}$ is called a system of generators of ${\cal{F}}$.

\item[Corollary 1]
${\cal{F}}$ is a filter on $X,\ A\subset X$. A necessary and
sufficient condition that $\exists{\cal{F}}'\supset{\cal{F}}$ such
that $A\in{\cal{F}}'$ is that $F\cap A\neq\emptyset\forall
F\in{\cal{F}}$.

\item[Corollary 2]
A set $\Phi$ of filter on (non-empty) $X$ has an upper bound in
the set of all filters on $X\Leftrightarrow$ for every finite
sequence.

$\ds\{{\cal{F}}_i\}_{i=1,2,\ldots,n}\subset\Phi$ and every
$A_i\in{\cal{F}}_i\ i=1,2,\ldots n\cap A_i\neq\emptyset$.

\item[Filter Bases]
If ${\cal{S}}$ is a system of generators for $\cal{F,\ F}$ is not,
in general, the set of subsets of $X$ which contain an element of
${\cal{S}}$

\item[Theorem 16]
Given ${\cal{B}}\subset {\cal{P}}(X)$, a necessary condition that
the family of subsets of $X$ which contain an element of
${\cal{B}}$ be a filter is that ${\cal{B}}$ have the properties

\begin{description}

\item[($B_1$)]
The intersection of 2 sets of ${\cal{B}}$ contains a set of
${\cal{B}}$.

\item[($B_2$)]
${\cal{B}}$ is not empty; $\emptyset\not\in{\cal{B}}$.

\end{description}

\item[Definition 21]
A system ${\cal{B}}\subset{\cal{P}}(X)$ satisfying $B_1,B_2$ is
called a base for the filter it generates by Theorem 16.

2 filter bases are equivalent if they generate the same filter.

\item[Theorem 17]
${\cal{B}}\subset{\cal{F}}$ is a base for
${\cal{F}}\Leftrightarrow$ each set of ${\cal{F}}$ contains a set
of ${\cal{B}}$.

\item[Theorem 18]
A necessary and sufficient condition that ${\cal{F}}'$ with base
${\cal{B}}'$ be finer than ${\cal{F}}$ with base ${\cal{B}}$ id
${\cal{B}}'\subset{\cal{B}}$.

\item[Examples]

\begin{description}

\item[(i)]
Let $X$ be a non-empty partially ordered set $(\leq)$ in which
each pair of elements has an upper bound. The sections $\{x:x\geq
a\}$ of $X$ form a filter base. The filter it defines is called
the filter of sections of $X$.

\item[(ii)]
$X$=set of nets on $[a,b]\ N=(a=x_0&lt;x_1&lt;\ldots&lt;x_n=b)\ N_1\leq
N_2$ if $N_1\subset N_2$.

An upper bound for $N_1$ and $N_2$ is $N_1\cup N_2$.

\item[(iii)]
The filter of neighbourhoods of $O$ in $R_1$ has as bases
$\ds\left\{\left(-\frac{1}{n},\frac{1}{n}\right)\right\},\\\left\{
\left[-\frac{1}{n},\frac{1}{n}\right)\right\}\left\{(-a,b)\
ab&gt;0\right\}$ 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 $R_1\ \{x,|x|&gt;n\}$ 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 $\ds\{\{x:|x|&gt;n\}\}$ does not
converge.

\item[(ii)]
\begin{eqnarray*}
X&amp;=&amp;\{x,y,z\}\\ {\cal{O}}&amp;=&amp;\{\emptyset,\{x,y\},\{z\} X\}\\
{\cal{B}}&amp;=&amp;\{\{x,y\} X\}\\ V(x)&amp;=&amp;\{\{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}
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