\magnification 2300 \parindent 0pt \parskip 9pt \hsize 7truein \vsize 9.7 truein \def\u{\vskip -7pt} \def\v{\vskip -6pt} 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 $B_3 = B_1 \cap B_2 \in {\cal B}$. \vskip 10pt \hrule 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