Theorem 1   Suppose $X\subseteq Y \subseteq Z$ are sets and $Z$ is a topological space with topology $\tau_Z$ . Let $\tau_{Y,Z}$ be the subspace topology in $Y$ given by $\tau_Z$ , and let $\tau_{X,Y,Z}$ be the subspace topology in $X$ given by $\tau_{Y,Z}$ , and let $\tau_{X,Z}$ be the subspace topology in $X$ given by $\tau_Z$ . Then $\tau_{X,Z}=\tau_{X,Y,Z}$ .

Proof. Let $U_X\in\tau_{X,Z}$ , then there is by the definition of the subspace topology an open set $U_Z\in\tau_Z$ such that $U_X=U_Z\cap X$ . Now $U_Z\cap Y\in\tau_{Y,Z}$ and therefore $U_Z\cap Y\cap X\in\tau_{X,Y,Z}$ . But since $X\subseteq Y$ , we have $U_Z\cap Y\cap X=U_Z\cap X=U_X$ , so $U_X\in\tau_{X,Y,Z}$ and thus $\tau_{X,Z}\subseteq\tau_{X,Y,Z}$ .

To show the reverse inclusion, take an open set $U_X\in\tau_{X,Y,Z}$ . Then there is an open set $U_Y\in\tau_{Y,Z}$ such that $U_X=U_Y\cap X$ . Furthermore, there is an open set $U_Z\in\tau_Z$ such that $U_Y=U_Z\cap Y$ . Since $X\subseteq Y$ , we have \begin{equation*} U_Z\cap X=U_Z\cap Y\cap X=U_Y\cap X=U_X, \end{equation*}so $U_X\in\tau_{X,Z}$ and thus $\tau_{X,Y,Z}\subseteq\tau_{X,Z}$ .

Together, both inclusions yield the equality $\tau_{X,Z}=\tau_{X,Y,Z}$ .