subspace of a subspace

Theorem 1.

Suppose $X\mathrm{\subseteq }Y\mathrm{\subseteq }Z$ are sets and $Z$ is a topological space with topology ${\tau }_Z$. Let ${\tau }_{Y\mathrm{,}Z}$ be the subspace topology in $Y$ given by ${\tau }_Z$, and let ${\tau }_{X\mathrm{,}Y\mathrm{,}Z}$ be the subspace topology in $X$ given by ${\tau }_{Y\mathrm{,}Z}$, and let ${\tau }_{X\mathrm{,}Z}$ be the subspace topology in $X$ given by ${\tau }_Z$. Then ${\tau }_{X\mathrm{,}Z}\mathrm{=}{\tau }_{X\mathrm{,}Y\mathrm{,}Z}$.

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
Canonical name	SubspaceOfASubspace
Date of creation	2013-03-22 15:17:53
Last modified on	2013-03-22 15:17:53
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	5
Author	matte (1858)
Entry type	Theorem
Classification	msc 54B05