Need for basis for a topology Why did we need to define a basis for a topology?
I read that it is difficult to specify topology on a bigger set, so we define topology on a smaller collection.
Now I want to ask that what were the requirements to accomplish this plan?
When this question was in front of us, what actually we wanted to find?
 A: The biggest advantage to talking about bases is that they control the entirety of a space's topological information while being easy to work with. The prototypical example would be something like open balls or open rectangles in $\mathbb{R}^n$. Open subsets of $\mathbb{R}^n$ can be very strange, but open balls are simple to imagine and easy to work with. If I want to, say, prove that some function $f: X \rightarrow \mathbb{R}^n$ is continuous I would, in principal, need to prove that $f^{-1}(U)$ is open for every single open subset of $\mathbb{R}^n$. However, in actuality, what I will typically prove is that $f^{-1}(B_r(x))$ is open for all $x$ and for all sufficiently small $r$. This is sufficient though, because preimages are well behaved with unions, and because I can represent any arbitrary open subset of $\mathbb{R}$ as a union of open balls.
A: *

*There are various methods for constructing  (defining) topologies. Many "special" spaces have been very useful for some general theorems, as well as for producing examples and counter-examples for many conjectures. The Tychonoff product topology has been especially valuable. One widely used method to define a topology is to use a base.
A collection $B$ of subsets of a set $X$ is a base for a topology on $X$ iff (i) $\cup B=X$ (i.e. every $p\in X$ belongs to at least one $b\in B$), and (ii) if $b_1,b_2\in B$ and $p\in b_1\cap b_2$ then there exists $b_3 \in B$ such that $p\in b_3\subset b_1\cap b_2.$
Note that if $b_1\cap b_2\in B$ whenever $b_1,b_2\in B$ then condition (ii) is satisfied by $b_3=b_1\cap b_2$. An example of a base not satisfying (ii), for the standard topology on $\Bbb R,$ is $$\{(x-r,x+s): x\in \Bbb R \land r,s>0 \land (\,[r\in \Bbb Q\land s \not \in \Bbb Q]\lor [r\not \in \Bbb Q\land s\in \Bbb Q]\,)\}.$$


*Any topology is a base for itself but a smaller or simpler base may help to reveal much  about the space. Example: The standard topology  on $\Bbb R^n$ (for any $n\in \Bbb N$) is an uncountable set but it has a countable base. If a space $X$ has a countable base then (iii) $X$ has a countable dense subset , and (iv) if $F$ is a family of pair-wise disjoint open subsets of $X$ then $F$ is countable, and (v) $X$ is Lindelof, i.e. if $G$ is a family of open subsets of $X$ and $\cup G=X$ then there exists a countable $H\subset G$ such that $\cup H=X,$ and (vi) every subspace of $X$ also has a countable base.

