Describe the topology generated by a set Let $A$ be a subset of $X$. I want to define a topology of $A$ called $\sigma(A)$ such that every element of $A$ is in $\sigma(A)$. I thought the following:
$B = \{C \in \mathcal{P}(X) / C=\cap A_i$ with $A_i \in A\}$
$\sigma(A) = \{U \in \mathcal{P}(X) / C=\cup B_i$ with $B_i \in B\}$
It wasn't hard to prove that $\varnothing, A \in \sigma(A)$ and $\cup U_i \in \sigma(A)$. But I'm not finding a way to prove that $U \cap V \in \sigma(A)$ if $U,V \in \sigma(A)$.
 A: If $\sigma(A)$ is the smallest topology on $X$ such that $A\subset\sigma(A)$, then $A$ is called a subbase for the topology on $X$. 
First, a topology should be closed under finite intersections. Therefore, if $A$ is not closed under finite intersections, we should first do so. This is probably what you tried to do with your set $B$, but as the range of indices is not clear from your definition, it is not clear whether you mean finite intersections or arbitrary intersections.
I would define $B$ as follows:

$B:=\left\{C\in\mathcal P(X)\mid \exists C'\subset A\ (\ C=\bigcap C'\text{ and $C'$ is finite}\ )\ \right\}$

Here I also use that the intersection of nothing is equal to everything, that is, $\bigcap \varnothing=X$. This is important to be able to show that $X$ is in the topology. If you disagree with this use of empty intersection, you could remedy this by adding $X$ to $B$.
The set $B$ that I now have, is a base for the topology $\sigma(A)$. We can now define $\sigma(A)$ by closing $B$ under arbitrary unions:

$\sigma(A):=\left\{C\in\mathcal P(X)\mid \exists C'\subset B\ (\ C=\bigcup C'\ )\ \right\}$

Note that here it is not specified how many sets of $B$ are in the union. Similar to with the definition of $B$, I use that the union of nothing is equal to nothing, that is, $\bigcup \varnothing=\varnothing$. Alternatively, if this bothers you, you could simply add $\varnothing$ afterwards.

The proof that $\varnothing\in\sigma(A)$ is done by using an empty union. Since $X\in B$ by using an empty intersection, we have that $X\in\sigma(A)$ as well (Note that you said you proved "$A\in\sigma(A)$"; this is not correct, we need $X\in\sigma(A)$ as well. Actually $A\in\sigma(A)$ usually makes no sense, since $A$ is a set of subsets of $X$, and a set of subsets of $X$ does not have to be a subset of $X$ itself).
To show that $\sigma(A)$ is closed under arbitrary unions is trivial from its definition.
Finally, if $U,V\in\sigma(A)$, then let $C_U,C_V\subset B$ be such that $U=\bigcup C_U$ and $V=\bigcup C_V$. Then $$U\cap V=\bigcup C_U\cap \bigcup C_V=\bigcup_{c_U\in C_U\,\land\, c_V\in C_V} c_U\cap c_V.$$
The last equality follows from the infinite distributive law $X\cap \bigcup Y_i=\bigcup (X\cap Y_i)$. For each $c_U\in C_U$ and $c_V\in C_V$ we have $c_U,c_V\in B$, and thus the finite intersection $c_U\cap c_V\in B$ by how we defined $B$. Therefore $U\cap V$ is a union of elements of $B$, giving $U\cap V\in \sigma(A)$.
