Topology induced by Basis B I have recently started studying topology on my own. I have a confusion regarding defination of topology of set X induced by basis $\mathfrak{B}$. ${\tau} = \{ U: x\in U \implies \exists B :x \in B \in \mathfrak B, B \subset U   \}$
Here defination of topology $\tau$ means i need to include all sets such that x belongs to some B in basis. It means i need to include B, B+atleast 1 element in X .... X
Does it mean finally $\tau$ becomes powerset of X? please correct if wrong
 A: It means that the topology is defined by all sets $U$ that satisfy the property that each of its elements has at least a basis element $B$ that pass through that point and is entirely included in that set $U$.
Note that there is a typo in what you write. The first inclusion symbol must be a "belongs to".
A: Another way to decribe the same topology is $$\{O\mid \exists \mathfrak{B}' \subseteq \mathfrak{B}: O= \bigcup \mathfrak{B}'\}$$
which is the set of all unions of subfamilies of the base $\mathfrak{B}$. (Note that all $B \in \mathfrak{B}$ will always be in the generated topology and topologies are closed under unions so all these sets are for sure in that topology.)
If $O$ is such as in my description, clearly every $x \in O$ is in some $B \in \mathfrak{B}'$ by the definition of a union.
And if $U$ is open as in your description, pick $B_x \in \mathfrak{B}$ such that $x \in B_x \subseteq U$ for every $x \in U$. Then $U = \bigcup \{B_x\mid x \in U\}$ and so we can take $\mathfrak{B}' = \{B_x\mid B_x \in U\}$ for my description etc.
A simple example: $X = \Bbb R$ with base $\mathfrak{B}  =\{(x, \rightarrow): x \in \Bbb R\}$; all unions of base elements are also of the form $(x, \rightarrow)$, as is easilt checked plus $\Bbb R$ and $\emptyset$ (for the union of the empty subfamily $\mathfrak{B}'$; in your description this set is trivially open too as  there are no elements so the implication condition is always satisfied). So the topology is not the power set, and in fact a set of the form $\{x\}$ will only be open if it that set $\{x\}$ is already in the base $\mathfrak{B}$.
