how do we find the topology generated by a given subbasis? I am new to Topology, and I was reading up the definition of subbasis, and a point that says that "a Topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of the subbasis S."
I do not understand this. so for example if we have X={a,b,c,d} and a subbasis S={(a,b},{a,c},{a,b,d},{a}}, how do we find T(S)?
 A: Just follow the definition. Since $S$ is finite, the family of all finite intersections of elements of $S$ is just the family of all intersections of elements of $S$. Intersecting a single element of $S$, we just get $S$ back, so we get the sets $\{a,b\},\{a,c\},\{a,b,d\}$, and $\{a\}$. There are six pairs of sets in $S$ that we can intersect:
$$\begin{align*}
&\{a,b\}\cap\{a,c\}=\{a\}\;,\\
&\{a,b\}\cap\{a,b,d\}=\{a,b\}\;,\\
&\{a,b\}\cap\{a\}=\{a\}\;,\\
&\{a,c\}\cap\{a,b,d\}=\{a\}\;,\\
&\{a,c\}\cap\{a\}=\{a\}\;,\text{ and}\\
&\{a,b,d\}\cap\{a\}=\{a\}\;.
\end{align*}$$
As you can see, this did not give us anything new: $\{a\}$ and $\{a,b\}$ were already in the subbase $S$.
There are four triples of sets that we can intersect:
$$\begin{align*}
&\{a,b\}\cap\{a,c\}\cap\{a,b,d\}=\{a\}\;,\\
&\{a,b\}\cap\{a,c\}\cap\{a\}=\{a\}\;,\\
&\{a,b\}\cap\{a,b,d\}\cap\{a\}=\{a\}\;,\text{ and}\\
&\{a,c\}\cap\{a,b,d\}\cap\{a\}=\{a\}\;.
\end{align*}$$
Once again we got nothing new. Finally, we can intersect all four members of $S$, but that also just gives us $\{a\}$ back. Thus, $S$ is already closed under taking finite intersections and is therefore a base for a topology on $X$. That topology is the family of all unions of members of the base, i.e., in this case the family of all unions of members of $S$.


*

*The union of no members of $S$ is $\varnothing$.  

*The unions of one-member families from $S$ are the members of $S$: $\{a,b\},\{a,c\},\{a,b,d\}$, and $\{a\}$.  

*The unions of two-member families from $S$ are: $$\begin{align*}&\{a,b\}\cup\{a,c\}=\{a,b,c\}\;,\\&\{a,b\}\cup\{a,b,d\}=\{a,b,d\}\;,\\&\{a,b\}\cup\{a\}=\{a,b\}\;,\\&\{a,c\}\cup\{a,b,d\}=X\,\\&\{a,c\}\cup\{a\}=\{a,c\}\;,\text{ and}\\&\{a,b,d\}\cup\{a\}=\{a,b,d\}\;.\end{align*}$$ 


I’ll leave it to you to check that the unions of the four $3$-member families from $S$ are $X,\{a,b,c\}$, and $\{a,b,d\}$, and that the union of all four members of $S$ is $X$. The topology generated by the subbase $S$ is therefore
$$\big\{\varnothing,\{a\},\{a,b\},\{a,c\},\{a,b,c\},\{a,b,d\},X\big\}\;.$$
