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\def\msk{\medskip}
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\def\u{\underbar}

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\ctln{\bf Math 970 Homework}

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\ctln{{\bf Due: }Sept. 18}

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\item{1.} If $(X,{\Cal T})$ is a topological space, $Y$ is a set, and $f:X\rightarrow Y$ is a function, show that


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\ctln{${\Cal T}^\prime$ = $\bigl\{{\Cal U}^\prime\subseteq Y : f^{-1}({\Cal U}^\prime)\in {\Cal T}\bigr\}$}

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is the finest topology on $Y$ for which $f:(X,{\Cal T})\rightarrow (Y,{\Cal T}^\prime)$ is continuous.

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(Note that this problem is actually asking you to show \u{three} things...)

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\item{2.} If $(X,{\Cal T})$ is a topological space, and $A\subseteq X$,  then $A\in {\Cal T}$ if and only if

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\ctln{for all $x\in A$, there is a $U\in {\Cal T}$ so that $x\in U\subseteq A$}

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\item{3.} Show that ${\Cal B} = \{(a,\infty)\times(b,\infty) : a,b\in{\Bbb R}\}$ is a basis for a topology ${\Cal T}$
on ${\Bbb R}^2={\Bbb R}\times {\Bbb R}$, which is coarser than the usual Euclidean topology on ${\Bbb R}^2$.
Show that ${\Cal B}^\prime = \{[a,\infty)\times[b,\infty) : a,b\in{\Bbb R}\}$ is a basis for a topology ${\Cal T}^\prime$ which is 
strictly finer than ${\Cal T}$, and not comparable to the usual Euclidean topology.

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\item{4.} Show that, in general, if ${\Cal B}$ and ${\Cal B}^\prime$ are both bases for topologies 
on $X$, that
${\Cal B}\cap{\Cal B}^\prime$ and ${\Cal B}\cup{\Cal B}^\prime$ \u{need} \u{not} \u{be}. Show,
however, that ${\Cal B}^{\prime\prime} = \{B\cap B^\prime : B\in {\Cal B}, B^\prime \in{\Cal B}^\prime \}$
\u{is} a basis for a topology, and ${\Cal T}({\Cal B}^{\prime\prime})$ is the coarest topology containing both 
${\Cal B}$ and ${\Cal B}^\prime$.

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\item{5.} Show that the topology generated by a basis $\Cal B$ is the coarsest topology \underbar{containing}
$\Cal B$ (i.e., it is the intersection of all such topologies).



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