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Lemma:  If ${\cal S}$ is a subbasis for a topology on $X$, then ${\cal B } = 
\{ \cap_{i=1}^n S_i ~|~ S_i \in {\cal S} \}$
is a basis for a
topology.



(1)  Show for each $x \in X$, there is at least one basis element $B$ containing
$x$.



(2) Show that if $x \in B_1 \cap B_2$ where $B_1, B_2 \in {\cal B}$, then there
exists $B_3 \in {\cal B}$ such that $$x \in B_3 \subset B_1 \cap B_2.$$

Proof of (2): \hfil \break
 Suppose that $x \in B_1 \cap B_2$ where $B_1, B_2 \in {\cal
B}$

Find $B_3 \in {\cal B}$ such that $x \in B_3 \subset B_1 \cap B_2.$

Let $B_3 = B_1 \cap B_2$.  

Note $x \in B_3 = B_1 \cap B_2\subset B_1
\cap B_2$.

$B_1, B_2 \in {\cal B}$ implies that $B_i$ is a finite intersection of
subbasis elements for $i = 1, 2$.  Thus since $B_1 \cap B_2$ is the
intersection of
two sets each of which is a finite intersection of subbasis elements,
$B_1 \cap B_2$ is a finite intersection of subbasis elements.  Thus
$B_3 = B_1 \cap B_2 \in {\cal B}$.

OR

$B_1, B_2 \in {\cal B}$ implies that $B_1 = \cap_{i=1}^n S_i$ and
\break
$B_2
= \cap_{i=1}^m U_i$ where $S_i, U_i \in {\cal S}$ for all $i$.  Thus,
$B_3 = B_1 \cap B_2 = (\cap_{i=1}^n S_i) \cap (\cap_{i=1}^m U_i) =
\cap_{i=1}^{m+n} V_i$ 
\vskip 3pt
\centerline{where $V_i = \cases{S_i, & $i = 1, ..., n$ \cr
U_i & $i = n+1, ..., m+n$}$ for all $i$.}  
\vskip -3pt
Thus $V_i \in {\cal S}$ and hence
$B_3 = B_1
\cap B_2 \in {\cal B}$.

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Defn:  The {\bf topology generated by the
subbasis ${\cal S}$} is the topology generated by the basis ${\cal B } =     
\{ \cap_{i=1}^n S_i ~|~ S_i \in {\cal S} \}$.


Corollary:  The {topology generated by the
subbasis ${\cal S}$ } is the collection ${\cal T}$ of all unions of
finite intersections of elements of ${\cal S}$.



\end