Proof attempt for collection of all open intervals being a basis of $\Bbb R$ with the standard topology Show that the collection of all open intervals $\{(a,b)\}$ is a basis of $\Bbb R$ with the standard topology:

My attempt:



I believe we want to show two things:

1) All elements, $x\in\Bbb R$ are contained in some basis element:

$\forall x\in\Bbb R$ $x\in(x-1,x+1)$ $\square$

2) If $x\in B_1\cap B_2$ then there is some $B_3$ such that $x\in B_3$ and $B_3\subset B_1\cap B_2$

If $x\in B_1\cap B_2$ and $B_1=(a_1,b_1)$ and $B_2=(a_2,b_2)$ then $B_1\cap B_2 = (\max(a_1,a_2),(\min(b_1,b_2))$, if we let $B_3=(\max(a_1,a_2),(\min(b_1,b_2))$ then $B_3\subset B_1\cap B_2$ and clearly $x\in B_1\cap B_2=B_3 \iff x\in B_3$



Is it acceptable that I simply let $B_3=B_1\cap B_2$ by construction? Or should my $B_3$ be a proper subset, i.e. not equal?

Also is my proof acceptable in general?
What you have shown is that the collection of open intervals is a basis for some topology, but not that it is a basis for the standard topology on $\mathbb R$.  If $X$ is a topological space, then we say that a collection $\mathcal B$ of open sets of $X$ is a basis (of open sets) for the topology if every open set in $X$ can be written as a union of open sets drawn from $\mathcal B$.  

Equivalently, $\mathcal B$ is a basis if the open subsets of $X$ are precisely those sets $U$ such that for every $x\in U$ there exists $V\in\mathcal B$ such that $x\in V\subset U$.  

Using the second version, it should be clear that the collection of open intervals in $\mathbb R$ is a basis for the standard topology.  



Where do the two conditions you talked about come in?  Well, if $X$ is a set, then it can be shown that a collection $\mathcal B\subset\mathcal P(X)$ of subsets of $X$ is a basis of open sets for some topology on $X$ if and only if it satisfies the two conditions you mentioned:


For every $p\in X$, there is $V\in\mathcal B$ such that $x\in V$.
For every $U,V\in\mathcal B$, and any $q\in U\cap V$, there is $W\in\mathcal B$ such that $q\in W\subset U\cap V$.  Equivalently, $U\cap V$ can be written as the union of sets drawn from $\mathcal B$.  


But the topology might not be the one you are interested in.  For instance, the singleton collection $\{\mathbb R\}$ satisfies conditions (1) and (2) (for $X=\mathbb R$), and it is indeed a basis of open sets for a topology on $\mathbb R$ (the indiscrete topology), but it is not a basis for the standard topology.  Similarly, $\mathbb P(\mathbb R)$ satisfies the conditions for a basis, but it is a basis for the discrete topology on $\mathbb R$, not the standard topology.  Finally, the collection $\mathcal B$ of all half-open intervals $[x, y)$ satisfies (1) and (2), but it is a basis for the half-open interval topology, which is not the same as the standard topology (indeed, $[0,1)$ is not open in the standard topology).  

It's a good idea to work through the proofs of the claims I have made.  Namely, that:


Show that the two versions of a definition of a basis for a topology given above are equivalent.
If $X$ is a topological space, and $\mathcal B$ is a basis of open sets for the topology on $X$, show that $\mathcal B$ satisfies (1) and (2).
If $X$ is a set, and $\mathcal B\subset\mathcal P(X)$ satisfies (1) and (2), show that the collection of all unions of sets contained in $\mathcal B$ is a topology on $X$.  


Let us know if you have any trouble with those.
I think you have taken a long winded approach in part 2. Suppose $x \in B_1 \cap B_2$. That means $B_1 \cap B_2 \neq \phi$. If you say $B_1=(a_1,b_1)$, $B_2=(a_2,b_2)$ then it must hold that $a_2 &lt; b_1$. So $B_1 \cap B_2 = (a_2,b_1)$.