Topological Basis: Equivlance of Definitions I want to prove the equivalence of the following two definitions:


*

*A set $\mathcal B\subseteq\mathcal T$ is called a topology over $\Omega$ if and only if for every $O\in\mathcal T$, there exists a $\mathcal B'\subseteq\mathcal B$ such that $O = \bigcup_{B'\in\mathcal B'}B'$.

*A set $\mathcal B\subseteq 2^\Omega$ is called a basis for a topology $\mathcal T$ if and only if for every finite $\mathcal B'\subseteq\mathcal B$, there exists a $\mathcal B''\subseteq\mathcal B$ such that $\bigcap_{B'\in\mathcal B'}B' = \bigcup_{B''\in\mathcal B''}B''$. (using the convention $\bigcup_{i\in\emptyset}i=\emptyset$ and $\bigcap_{i\in\emptyset}i=\Omega$)
The "$\Rightarrow$" direction is pretty easy: For any finite $\mathcal B'\subseteq\mathcal B$ it holds that $\bigcap_{B'\in\mathcal B'}B'\in\mathcal T$ (since $\mathcal B\subseteq\mathcal T$ and $\mathcal T$ is a topology). But since $\mathcal B$ is a basis, it follows 2.
I struggle with the "$\Leftarrow$" direction. I googled and found the following proof idea: Put $\mathcal T := \{\bigcup_{V\in\mathcal V}V:\mathcal V\subseteq\mathcal V\}$ and show that this is indeed a topology. $\emptyset,\Omega\in\mathcal T$ is easy using the conventions. It remains to show that for $\mathcal A\subseteq\mathcal T$ we have that $\bigcup_{A\in\mathcal A}A\in\mathcal T$ and for finite $\mathcal A\in\mathcal T$ we have that $\bigcap_{A\in\mathcal A}A\in\mathcal T$.
For union I had the following idea: Let $\mathcal A\subseteq\mathcal T$. By construction of $\mathcal T$, ther exists a $\mathcal V_A\subseteq\mathcal B$ such that $A = \bigcup_{V_A\in\mathcal V_A}V_A$ for every $A\in\mathcal A$. Taking unions yields 
$$\begin{align*}\bigcup_{A\in\mathcal A}A &= \bigcup_{A\in\mathcal A}\bigcup_{V_A\in\mathcal V_A}V_A\\
&= \bigcup_{W\in\bigcup_{A\in\mathcal A}\mathcal V_A}W\end{align*}$$
and since $\bigcup_{A\in\mathcal A}\mathcal V_A\subseteq\mathcal B$ it follow that $\bigcup_{A\in\mathcal A}A\in\mathcal T$. However, I am not really sure wether it was okay how I treated the indices.
For the intersection I had no idea. I could show it with induction but that made the proof so lengthy. Does someone have an idea wether there is way to show it in a similar way as the union or at least without induction?
 A: Let $\mathcal{V}_1, \ldots, \mathcal{V}_n$ be subsets of of $\mathcal{B}$.
\begin{align}S =\left(\bigcup_{V_1 \in \mathcal{V}_1} V_1\right) \cap \left(\bigcup_{V_2 \in \mathcal{V}_2} V_2\right) \cap \cdots \cap \left(\bigcup_{V_n \in \mathcal{V}_1} V_n\right) &= \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} V_1 \cap \cdots \cap V_n
\end{align}
Now, because of $(2)$, each $V_1 \cap \cdots \cap V_n$ is in fact equal to a union of $\mathcal{B}_{V_1, \ldots, V_n} \subseteq \mathcal{B}$:
$$V_1 \cap \cdots \cap V_n = \bigcup_{B \in \mathcal{B}_{V_1, \ldots, V_n}} B$$
Now we have:
$$S = \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} V_1 \cap \cdots \cap V_n = \bigcup_{V_1 \in \mathcal{V_1}}\cdots \bigcup_{V_n \in \mathcal{V_n}} \bigcup_{B \in \mathcal{B}_{V_1, \ldots, V_n}} B \in \mathcal{T} = \bigcup_{V_1 \in \mathcal{V}_1, \ldots, V_n \in \mathcal{V_n}, B \in \mathcal{B}_{V_1, \ldots, V_n}} B$$
Therefore, $\mathcal{T}$ is a topology and $\mathcal{B}$ is a basis for $\mathcal{T}$.
A: For intersection note that $$(\bigcup_{V\in\mathcal V_1}V)\cap(\bigcup_{W\in\mathcal V_1}V)=\bigcup_{V\in\mathcal V_1,W\in\mathcal V_2}(V\cap W)\tag1$$
Here $V\cap W=\bigcap_{B'\in\mathcal B'}B' = \bigcup_{B''\in\mathcal B''}B''$ for $\mathcal B'=\{V,W\}\subseteq\mathcal B$.
So every intersection $V\cap W$ in $(1)$ can be replaced by a union of elements of $\mathcal B$. 
That implies that the LHS of $(1)$ is an element of $\mathcal T$.
If the intersection concerns more than $2$ sets then you can just repeat.

Side note:
I respected your notation, but can be done much nicer. 
Observe that e.g. $\bigcup_{V\in\mathcal V_1}V$ can be written as $\bigcup\mathcal V_1$.
This because $$\bigcup A:=\{x\mid\exists a\in A[x\in a]\}=\bigcup_{a\in A}a$$
