Definition of basis in topology In the definition of a basis on a set $X$ in topology, one of the properties is that, "for any two basis elements $B_1, B_2$ and any point $x \in B_1 \cap B_2$, there is a third basis element $B_3$ containing $x$ and contained in $B_1 \cap B_2$". I am wondering what happens if we change this property to the stronger property of requiring closure under pairwise intersections (and hence, inductively, by finite intersections)? In other words, what if we changed the requirement to "for all basis elements $B_1, B_2$, their intersection $B_1 \cap B_2$ is a basis element"?
Are there any simple/important examples in which the original (i.e. weaker) property holds but the latter (i.e. stronger) property doesn't?
 A: Well, if $\mathcal{B}$ is closed under finite intersections then it certainly obeys the property that you mention. But not all bases do: e.g. in $\Bbb R^2$ the Euclidean-balls form a base for the topology but they are not closed under intersections.
The point of these properties is that they are necessary and sufficient to define a base.  Being closed under binary intersections is sufficient only.
A: Consider $\mathbb{R}^2$ with the usual euclidean topology.
Consider $B_{\epsilon}^q=\{x \in \mathbb{R}^2: \|x-q\|<\epsilon  \}$ where $q \in \mathbb{R}^2$.
It is clear that $\{B_{\epsilon}^q  : q \in \mathbb{R}^2 \mbox{ and } \epsilon>0\}$ is a basis but you do not have that property.
A: There are easy examples of bases that have the first property but not the second; open balls in $\Bbb R^n$ for any $n>1$ will do. On the other hand, if you take the closure of any base under finite intersections, you get a base for the same topology that has the stronger property, so in that sense it doesn’t really matter which property you use to define bases. The main advantage to using the weaker property, apart from the fact that it is necessary as well as sufficient, is that it often allows simpler descriptions of a base.
A: The set of open discs in the plane is a basis for the usual topology, but the intersection of two discs is a disc only in trivial cases.
A: This is not an important example. Suppose that the underlying set for the topology is $\mathbb{R}^{2}$. Let the original basis be the collection of open squares with arbitrary orientation. Depending on the two open squares their intersection will be empty or some open polygon, which might have as few as three sides or as many as eight sides.
