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HW 2 (p 83: 4, 8)

4a.) Since $\emptyset, X \in {\cal T}_\alpha$ for all $\alpha$,
$\emptyset, X \in \cap {\cal T}_\alpha$

Suppose $U_\beta \in \cap{\cal T}_\alpha$ for all $\beta \in B$.  Then
$U_\beta \in {\cal T}_\alpha$ for all $\alpha, \beta$.  Since ${\cal
T}_\alpha$ is a topology, $\cup_{\beta \in B} U_\beta \in {\cal
T}_\alpha$ for all $\alpha$.  Thus $\cup_{\beta \in B} U_\beta \in
\cap{\cal T}_\alpha$

Suppose $U_i \in \cap{\cal T}_\alpha$ for $i = 1, ..., n$.  Then $U_i
\in {\cal T}_\alpha$ for all $\alpha, i = 1, ..., n$.  Since ${\cal
T}_\alpha$ is a topology, $\cap_{i=1}^n U_i \in {\cal T}_\alpha$ for
all $\alpha$.  Thus $\cap_{i= 1}^n U_i \in \cap{\cal T}_\alpha$

Let ${\cal T}_1 = \{\emptyset, \{a, b\}, \{a, b, c\}\}$ and ${\cal T}_2
= \{\emptyset, \{b, c\}, \{a, b, c\}\}$.  Then ${\cal T}_1 \cup {\cal
T}_2 = \{\emptyset, \{a, b\}, \{b, c\}, \{a, b, c\}\}$.  $\{a, b\} \cap
\{b, c\} = \{b\} \not\in {\cal T}_1 \cup {\cal T}_2$.  Thus, ${\cal
T}_1 \cup {\cal T}_2$ is not a topology.

4b.)  Lemma:  $\cap{\cal T}_\alpha$ is the unique largest topology
contained in all the ${\cal T}_\alpha$.

$\cap{\cal T}_\alpha$ is a topology contained in all the ${\cal
T}_\alpha$. Suppose ${\cal T}$ is a topology contained in all the
${\cal T}_\alpha$.  Then ${\cal T} \subset {\cal T}_\alpha$ for all
$\alpha$ implies ${\cal T} \subset \cap {\cal T}_\alpha$.  Therefore
$\cap{\cal T}_\alpha$ is larger than or equal to all other topologies
contained in all the ${\cal T}_\alpha$ and thus $\cap{\cal T}_\alpha$
is
the unique
largest topology contained in all the ${\cal T}_\alpha$.

Lemma:  $\cup{\cal T}_\alpha$ is a subbasis for the unique smallest
topology containing all the ${\cal T}_\alpha$.

Since $X \in {\cal T}_\alpha$, $\cup_{U_\beta \in \cup{\cal T}_\alpha}
U_\beta = X$. Thus,  $\cup{\cal T}_\alpha$ is a subbasis.

Let ${\cal T}$ be the topology generated by the subbasis $\cup{\cal
T}_\alpha$. Suppose that ${\cal T}'$ is a topology containing all the
${\cal T}_\alpha$.  Then $\cup{\cal T}_\alpha \subset {\cal T}'$.  If
$U \in {\cal T}$, then $U = \cup_{\beta \in B} (\cap_{i=1}^n U_{i,
\beta}$) where $U_{i,  
\beta} \in \cup {\cal T}_\alpha \subset {\cal T}'$.  Hence $U \in {\cal
T}'$, and thus ${\cal T}
\subset {\cal
T}'$.  Therefore ${\cal T}$ is smaller than all or equal to other
topologies
containing all the ${\cal T}_\alpha$ and thus ${\cal T}$ is the unique
smallest topology containing all the ${\cal T}_\alpha$.

4c.)  The largest topology contained in ${\cal T}_1$ and ${\cal T}_2$ =
${\cal T}_1 \cap {\cal T}_2 = \{\emptyset, \{a\}, \{a, b, c\}\}$. A
subbasis for the largest topology containing ${\cal T}_1$ and ${\cal
T}_2$ = ${\cal T}_1 \cup {\cal T}_2 = \{\emptyset, \{a\}, \{a, b\},
\{b, c\}\{a, b, c\}\}$.  Thus, the largest topology containing ${\cal
T}_1$ and ${\cal T}_2 = \{\emptyset, \{a\}, \{b\}, \{a, b\}, \{b, c\},
\{a, b, c\}\}$.  


8a.)  Let ${\cal B} = \{(a,b) ~|~ a < b, a$ and $b$ rational$\}$.
Since
$(a, b)$ is open for every $a < b, a$ and $b$ rational, ${\cal B}$ is a
collection of open sets of $R$ with the standard topology.  Suppose
that $U$ is an open set in $R$ and $x \in U$.  Since ${\cal B}' =
\{(a,b) ~|~ a < b, a$ and $b$ real numbersl$\}$ is a basis for the
standard topology and $U$ is open, there exists $a, b \in R$, $a < b$
such that $x \in (a, b) \subset U$.  Since the rationals are dense in
$R$, there exists $c, d$ such that $a < c < x < d < b$.  Thus $x \in
(c, d) \subset U$.  Since $(c, d) \in {\cal B}$, ${\cal B}$ is a basis
for the standard topology.

8b.)  Let ${\cal T}$ be the topology generated by ${\cal C}$.  $[\pi,
4)$ is open in the lower limit topology since it is a basis element,
but $[\pi,  
4)$ is not open in ${\cal T}$.  $\pi \in [\pi,
4)$.  If $\pi \in [a, b)$ where $a, b$ are rational, then $a \leq \pi <
b$.  Since $a$ is rational and $\pi$ is irrational, $a \not= \pi$.
Thus $a < \pi < b$.   Hence ${a + \pi \over 2} \in [a, b)$, but ${a +
\pi \over 2} \not\in [\pi,
4)$.  Thus $[a, b) \not\subset [\pi,
4)$.  Hence there does not exists a basis element, $[a, b)$ in ${\cal
C}$ such
that $\pi \in [a, b) \subset[\pi, 4)$.  Thus $[\pi, 4)$ is not open in
${\cal T}$.



\end