Proving that the subbasis generates a topology. I want to prove that the subbais of a topology really generates a topology. I think I have been able to prove most of it, but I am not able to show that the empty set is in the topology. Can you please tell me how to show that the empty set is in the topology, and also point out any errors in the other 3 parts if you see it?
The definition of a subbasis in my book is:

A subbasis for $\mathcal{S}$ for a topology X is a collection of
  subsets of X whose union equals X. The topology generated by the
  subbasis is defined to be the collection of all unions of finite
  intersections of elements of $\mathcal{S}$.

I need to prove that it generates a topology. That is, I need to prove that what is generated has this properties:


*

*X is in what is generated.


2. $\emptyset$ is in what is generated.


*What is generated is closed under unions.

*What is generated is closed under finite intersections.
My attempt:


*

*Since the union of the subbasis equals X this is clear.


2. I do not see how I get the empty set, how do I know that the finite intersections of some sets will be the empty set?
3 and 4:
I have given the set $\mathcal{S}$, I assume it has indexset $I^{\mathcal{S}}$.
I call the set of all finite intersections of $\mathcal{S}$ for $\mathcal{B}$. Each element in $\mathcal{B}$ is $B_k$, where k is in the index-set $I^{\mathcal{B}}$. Each $B_k=\cap_{i \in I^{B_k}}S_i$, where $S_i \in \mathcal{S}$, $I^{B_k} \subset I^{\mathcal{S}}$, $I^{B_k}$ is a finite set.
I call the topology generated(I don't yet know if it is a topology), $\mathcal{T}$, each element is denoted $T_l$, $l \in I^{\mathcal{T}}$. We have by construction that $T_l=\cup_{k \in I^{T_l}}B_k$, where $I^{T_l}\subset I^{\mathcal{B}}$. $I^{T_l}$ need not be finite.
Now to prove 3:
Assume that I have a union of elements in $\mathcal{T}$, let $V=\cup_{l\in I^V}T_l$, $I^V\subset I^{\mathcal{T}}$. I need to show that V is a union of finite intersections of elements in $\mathcal{S}$.
$V=\cup_{l\in I^V}T_l$
$=\cup_{l \in I^V}[\cup_{k \in I^{T_l}}B_k]$
$=\cup_{l \in I^V}[\cup_{k \in I^{T_l}}\{{\cap_{i \in I^{B_k}} S_i}\}]$
$\cup_{(l,k):l \in I^{V}, k \in I^{T_l}}[\cap_{i \in I^{B_k}}S_i]$.
Hence the element is a union of finite intersections of elements in $\mathcal{S}$.
4.
For 4 I will prove it for two elements, and then it will follow by induction.
I have $T_1$ and $T_2$:
$T_1\cap T_2=[\cup_{k_1 \in I^{T_1}}B_{k_1}]\cap[\cup_{k_2 \in I^{T_2}}B_{k_2}]$
$=\cup_{k_1 \in I^{T_1}}[B_{k_1}\cap \{\cup_{k_2 \in I^{T_2}}B_{k_2}\}]$
$=\cup_{k_1 \in I^{T_1}}[\cup_{k_2 \in I^{T_2}}\{B_{k_1}\cap B_{k_2}\}]$
$=\cup_{(k_1,k_2) \in I^{T_1}\times I^{T_2}}[B_{k_1}\cap B_{k_2}]$. And obviously $B_{k_1}\cap B_{k_2}$ is a finite intersections of elements in $\mathcal{S}$ since each B is.
Is this proof correct, and how do I prove point 2?
 A: A collection $\mathcal{B}$ is a basis of a topology on $X$ if it covers $X$ and: $$\forall P,Q\in\mathcal{B}\forall x\in P\cap Q\exists B_{x}\in\mathcal{B}\left[x\in B_{x}\subseteq P\cap Q\right]$$
Note that this comes to the same as: $$P\cap Q=\cup_{x\in P\cap Q}B_{x}$$
So an intersection can be written as a union.
These condition can shortly be noted as: $$\mathcal{B}^{\stackrel{\bigcap}{f}}\subseteq\mathcal{B}^{\bigcup}$$
where $\mathcal{B}^{\stackrel{\bigcap}{f}}$ denotes the set of finite
intersections of elements of $\mathcal{B}$ and $\mathcal{B}^{\bigcup}$
denotes the set of unions of elements of $\mathcal{B}$.
Here $X$ is defined as empty intersection.
Here $\mathcal{B}^{\bigcup}$ is the topology of wich $\mathcal{B}$
serves as base.
If $\mathcal{B}:=\mathcal{S}^{\stackrel{\bigcap}{f}}$ then $\mathcal{B}^{\stackrel{\bigcap}{f}}=\left(\mathcal{S}^{\stackrel{\bigcap}{f}}\right)^{\stackrel{\bigcap}{f}}=\mathcal{S}^{\stackrel{\bigcap}{f}}=\mathcal{B}\subseteq\mathcal{B}^{\bigcup}$
So this shows that $\mathcal{S}^{\stackrel{\bigcap}{f}}$ is indeed a basis,
and it is the basis of topology $\mathcal{B}^{\bigcup}=\left(\mathcal{S}^{\stackrel{\bigcap}{f}}\right)^{\bigcup}$.
