Basis determining a unique topology When I read Munkres' Topology, I have a feeling that if we have a basis $\mathscr{B}$ on a set $X$, then the basis determines uniquely a topology on $X$; that is, if we have two topologies $\mathscr{T}_1, \mathscr{T}_2$ with the same basis $\mathscr{B}$, then $\mathscr{T}_1=\mathscr{T}_2$. I'm not sure if I'm right because I can't see this in the definition, which is as follows:

If $X$ is set, a basis for a topology on $X$ is a collection $\mathscr{B}$ of subsets of $X$ (called basis elements) such that for each $x\in X$, there is at least one $B\in \mathscr{B}$ such that $x\in B$ and if $x\in B_1\cap B_2$, where $B_1, B_2\in \mathscr{B}$, then there exists $B_3\in \mathscr{B}$ such that $x\in B_3\subset B_1\cap B_2$.

Moreover, the basis $\mathscr{B}$ generates a topology

$\mathscr{T}_\mathscr{B}=\left\{ U\subset X: \text{for each $x\in U$, there exists $B\in \mathscr{B}$ such that $x\in B\subset U$}\right\}$,

which is the smallest topology containing $\mathscr{B}$. Hence, I guess it's likely that those topologies whose bases are $\mathscr{B}$ should be equal to $\mathscr{T}_\mathscr{B}$.
By the way, I have consulted the article
Uniqueness of Topology and Basis and one of comments (left by Henno) seems to justify my hunch and they mentioned any open set $O$ is a union of the elements of $\mathscr{B}$, so $O$ is already in the topology $\mathscr{T}_\mathscr{B}$, but how could they know $O$ can be written this way just by the definition of a basis? I mean, in Munkres' book, he mentioned in lemme 13.1, from my understanding, that $\mathscr{T}_\mathscr{B}=\left\{\cup_\alpha B_\alpha:B_\alpha \in \mathscr{B}\right\}$, as opposite to saying it holds for any topology with basis $\mathscr{B}$. Perhaps I'm misunderstanding at this point.
Any help is really appreciated!!
 A: We say that topology $\mathcal T$ has basis $\mathcal B$ if $\mathcal T_{\mathcal B}=\mathcal T$.
Thus, it's immediate that if two topologies have the same basis then they coincide.
Saying that for every $x\in U$ there's a $B_x\in\mathcal B$ such that $x\in B_x\subseteq U$ is equivalent to saying that $U$ is the union of elements of $\mathcal B$, specifically $U=\bigcup_{x\in U}B_x$.
What you might be missing is that

A set $\mathcal B$ of subsets of $X$ is a basis for a topology (meaning $\mathcal T_{\mathcal B}=\left\{\bigcup \mathcal D:\mathcal D\subseteq\mathcal B\right\} $ is a topology) if and only if the given conditions hold, i.e. $\forall x\in X\,\exists B\in\mathcal B: x\in B$ and $\forall x\in X\,\forall B_1,B_2\in\mathcal B\ x\in B_1\cap B_2\implies \exists B\in\mathcal B: x\in B\subseteq B_1\cap B_2$.

A: I would start from the definition of topology as the collection of all open sets. Note now that every single open set can be written as the set-theoretic union of every basis element that contains a point $x \in U$, that is, $U = \bigcup_{x\in U} B_x $. Note now that, by the assumptions of a basis of a topology, you can always take two basis elements $B_1, B_2$ with nonempty intersection and find a third basis element in them (call it $B_3$). Nevertheless, the topology generated by the collection without $B_3$ and the one with $B_3$ is exactly the same and this comes from the fact that the set-theoretical union is not changing if we add a set that is already taken into account considering the sets $B_1$ and $ B_2$. This is the meaning for when Munkres writes that a basis for a topology is not like a basis for a vector space. So, from this point of view you can see that since the set-theoretic union of all (fixed) open sets is a unique object, then you can say that a basis determines the topology but not the converse.
