Weak (coherent) topology determined by a finite family of subsets Let X be a set and $\{A_j \, : \, j\in J\}$ a family of subsets of $X$ such that $X=\bigcup_{j\in J} A_j$. Assume that

*

*Each $A_j$ is a topological space.

*For every $j,k\in J$, the topology on $A_j\cap A_k$ induced from $A_j$ is the same as the one induced from $A_k$.

*For every $j,k\in J$, the intersection $A_j\cap A_k$ is closed in both $A_j$ and $A_k$.

Then, the weak topology on $X$ determined by $\{A_j \, : \, j\in J\}$ is the topology on which the closed sets are the sets $F\subseteq X$ such that $F\cap A_j$ is closed in $A_j$ for every $j\in J$. This topology is also known as the coherent topology determined by $\{A_j \, : \, j\in J\}$.
In Rotman's An Introduction to Algebraic Topology the autor says (page 196): "If the index set $J$ is finite, then there is only one topology on $X$ compatible with conditions 1, 2, and 3, and so it must be the weak topology."
I was trying to figure it out: Does it means that if $\tau$ and $\tau'$ are two topologies on $X$ which induce on each $A_j$ its original topology and the collection $\{A_j \, : \, j\in J\}$ satisfies conditions 1, 2 and 3, then $\tau=\tau'$, provided $J$ is finite?
If the answer is no, what does it mean by "compatible  with conditions , 2 and 3"? If the answer is yes, I am unsuccesfully trying to prove that $\tau = \tau'$ (or, which amounts to the same, that if $\tau$ is a topology compatible with conditions 1, 2 and 3, then $\tau$ is the weak topology determined by the finite collection $\{A_j \, : \, j\in J\}$).
Finally, what would be a counterexample of a space $X$ with an infinite collection $\{A_j \, : \, j\in J\}$ and two (or more) distinct topologies compatible with conditions 1, 2 and 3?
 A: The basic question is what "a topology on $X$ compatible with conditions 1, 2, and 3" should be.
I think that Rotman's approach is suboptimal since 1 is not a condition on the family $\{A_j \, : \, j\in J\}$, but means that we are given additional structural components. I would do it as follows:
Let $X$ be a set and $\{A_j \, : \, j\in J\}$ a family of subsets  $A_j \subset X$ each of which is endowed with a topology $\tau_j$ such that $X=\bigcup_{j\in J} A_j$. Assume that


*For every $j,k\in J$, the topology on $A_j\cap A_k$ induced from $A_j$ is the same as the one induced from $A_k$.

*For every $j,k\in J$, the intersection $A_j\cap A_k$ is closed in both $A_j$ and $A_k$.

We call $\{(A_j,\tau_j) \, : \, j\in J\}$ an admissible family on $X$ (this is just an adhoc notation)
Then, the weak topology on $X$ determined by an admissible family $\{(A_j,\tau_j) \, : \, j\in J\}$ is the topology whose closed sets are the sets $F\subseteq X$ such that $F\cap A_j$ is closed in $A_j$ for every $j\in J$.
The basic features of the weak topology are that each $A_j$ becomes closed subset of $X$ and that each $A_j$ , as a subspace of $X$, retains its original topology.
Now observe that if $(X,\tau)$ is a topological space and $\{A_j \, : \, j\in J\}$ a family of subsets  $A_j \subset X$ as above, then each $A_j$ can be endowed with the subspace topology $\tau_j$ induced by $\tau$. The family $\{(A_j,\tau_j) \, : \, j\in J\}$ trivially satisfies 2, but we cannot expect that 3 is true.
The obscure phrase "a topology on $X$ compatible with conditions 1, 2, and 3" should be replaced by the following definition:
A topology $\tau$ on $X$ is compatible with the admissible family $\{(A_j,\tau_j) \, : \, j\in J\}$ if
(a) Each $\tau_j$ is the subspace topology induced by $\tau$ on $A_j$.
(b) Each $A_j$ is $\tau$-closed in $X$.
Note that condition 3 is an immediate consequence of (b).
Clearly the weak topology determined by $\{(A_j,\tau_j) \, : \, j\in J\}$ is compatible with $\{(A_j,\tau_j) \, : \, j\in J\}$. In fact, it is the finest  topology compatible with $\{(A_j,\tau_j) \, : \, j\in J\}$. To see this, let $\tau$ be any compatible topology on $X$. Then for a $\tau$-closed $F \subset X$ all intersections $F \cap A_j$ are $\tau_j$-closed in $A_j$. That is, $F$ is closed in weak topology. In other words, the weak topology is finer than $\tau$.
Now let us prove that for a finite $J$ the weak topology is the only compatible topology. Let $\tau$ be any compatible topology and let $F \subset X$ be closed in the weak topology. Then all intersections $F \cap A_j$ are $\tau_j$-closed in $A_j$. Since the $A_j$ are closed in $X$, the $F \cap A_j$ are $\tau$-closed in $X$. Hence $F = \bigcup_{j \in J} F \cap A_j$ is $\tau$-closed in $X$. We conclude that $\tau$ agrees with the weak topology.
Let us conclude with an example showing that this is in general false for infinite $J$.
Let $X$ be any set and let $J = X$, $A_x = \{x\}$. Note that singletons always have a unique topology. Then the weak topology on $X$ is the discrete topology. However, any $T_1$-topology on $X$ is compatible. Therefore, if $X$ is infinite (like $X = \mathbb R$), there are many compatible topologies.
