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The following is a theorem from James Dugundji's Topology.

Theorem: Given any family $\Sigma=\{A_\alpha|\alpha\in\mathcal{A}\}$ of subsets of $X$, there always exists a unique, smallest topology $\tau(\Sigma)\supset\Sigma$. The family $\tau(\Sigma)$ can be described as follows:

It consists of $\emptyset, X,$ all finite intersections of the $A_\alpha,$ and all arbitrary unions of these finite intersections. $\Sigma$ is called a subbasis for $\tau(\Sigma),$ and $\tau(\Sigma)$ is said to be generated by $\Sigma.$

In his proof he has forced $\tau(\Sigma)$ to be the intersection of all topologies containing $\Sigma.$ Now my questions are,

1) why is it the intersection of all such topologies?, 2) what is the crux of this theorem?

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    This is a common construction in mathematics. I can briefly motivate why you might want to work through the details. If there is a "smallest topology" $\mathcal{T}_\Sigma$ containing $\Sigma$, then it is one of those in the family you are intersecting. By taking their intersection we will get $\mathcal{T}_\Sigma$ back again (because "smallest" means all of these topologies contain $\mathcal{T}_\Sigma$, which is itself one of those topologies). – 2017-01-20
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    The "crux" is to show the intersection of a family of topologies on $X$ is indeed a topology on $X$. I'll get you started: every topology for $X$ has $X$ itself as an element (the open set $X$), so the intersection of these topologies is nonempty. Continue and show all parts of the definition are satisfied. – 2017-01-20
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    I think I got it. But is it true that $\Sigma$ is a subbasis for $\tau(\Sigma)$ because $\tau(\Sigma)$ is forced to be the way it is? – 2017-01-20
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    For many authors the above construction *defines* what "subbasis" means. See the Wikipedia article on Subbasis for a discussion of this definition and alternatives used by some authors. – 2017-01-20

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