# subspace of a subspace

###### 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

 $U_{Z}\cap X=U_{Z}\cap Y\cap X=U_{Y}\cap X=U_{X},$

so $U_{X}\in\tau_{X,Z}$ and thus $\tau_{X,Y,Z}\subseteq\tau_{X,Z}$.

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

Title subspace of a subspace SubspaceOfASubspace 2013-03-22 15:17:53 2013-03-22 15:17:53 matte (1858) matte (1858) 5 matte (1858) Theorem msc 54B05