If $T$ and $T'$ have the same subspace topology on $Cl_T(A)$ then $Cl_{T'}(A)=Cl_T(A)$?

Question

I have a space $X$ with two different topologies $T$ and $T'$. Now consider a set $A\subset X$ and suppose the closure in $T$, $Cl_T(A)$ is such that the subspace topology of $T$ and $T'$ on $Cl_T(A)$ are the same. I was asking myself if necessarily we have $Cl_{T'}(A)=Cl_T(A)$ or if it was possible that they are different. However, I cannot prove it or find a counter-example.

Answer 1

Here is an easy counterexample. Take $X=\mathbb{R}$ and assume that $T$ is the usual topology and $T'$ is the trivial topology. Take a singleton $A=\{x\}$, its closure in $T$ is obviously $A$ itself. The closure in $T'$ is all $\mathbb{R}$. However, the subspace topologies on $A=Cl_T(A)$ are the same, this is just the discrete topology on $A$.