This question is from Exercise 5.1 of Wayne Patty's topology.
Let (X,T) be a $T_1$ space and let (D,U) be a decomposition space of X. Prove that (D,U) is $T_1$ iff each member of D is a closed subset of X.
The following definition might be useful:Let(X,T) be a topological space and let D be a partition of X. Define a function $p: X \to D$ as follows: For each $x\in X$ , p(x) is the member of D that contains x. If U is the quotient topology on D induced by p, then (D,U) is a quotient space of X. The function p is called natural map of X onto D. The set D is also called a decomposition of X and the quotient space (D,U) is also called a decomposition space.
Attempt for this question: Let (D,U) is $T_1$ , then for each distinct x,y in X there exists disjoint open sets such that $x\in U$ and $y\in V$ and x doesn't lies in V and y doesn't lies in U .
Why must each member of D be a closed subset of X? I have no idea why it must hold true and what result I should use. Also ,I couldn't make any progress while proving the converse.
The problem is that I am very bad in questions related to Quotient Topology although I have read theory from Wayne Patty but still I need help in solving such problems.
So, can you please help in this question?
Most likely the book contains this characterization of $T_1$-spaces: one-point sets are closed. Thus "$F$ is closed in $D$ is closed iff $p^{-1}[F]$ is closed in $X$" should make the problem doable.