Induced topology in nested subsets Consider the sets $D \subset C \subset B \subset A$. 
If does not make much sense to say "$D$ is open" unless we specify in which subset $D$ is open relative to which other subset.
For example, if $D$ is open in $C$ relative to $A$, we mean that $D = C \cap O_{A}$ where $O_{A}$ is an open set in $A$. 
Are the 4 statements here equivalent?
1) $D$ is open in $C$ relative to $B$
2) $D$ is open in $C$ relative to $A$
3) $D$ is open in $B$ relative to $A$
4) $D$ is open in $A$
 A: No. They are not equivalent. 
If $A\subset B\subset C$, then $A$ open in $C$ $\implies$ $A$ open in $B$ relative to $C$. But converse is not true (it will be true if $B$ is also open in $A$). For example, consider$A=B=[0,1], C=(-1,2)$. 
So $4\implies 1,2,3$ but 1,2,3 can't imply 4. Similarly, $3\implies1,2$, but neither 1 nor 2 implies 3.
But 1,2 are equivalent, i.e. $1\iff 2$.
$2\implies 1$
$D$ is open in $C$ relative to $A$, means $D = C \cap O_{A}$ where $O_{A}$ is an open set in $A$. Then $D = C \cap O_{B}$,where $O_{B}=B\cap O_{A}$ is an open set in $B$ with subspace topology.
$1\implies 2$
$D$ is open in $C$ relative to $B$, means $D = C \cap O_{B}$ where $O_{B}$ is an open set in $B$ with subspace topology. Hence $O_B = B \cap O_{A}$ where $O_{A}$ is an open set in $A$, Then $D = C \cap O_{B}=C\cap B \cap O_{A}=C \cap O_{A}$,where $O_{A}$ is an open set in $A$
So in general, $A\subset B\subset C\subset D$, you can fix the nearest layer and change the further layers, but when you want to change the nearest layer, you need to be careful.
