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[parent] subspace of a subspace (Theorem)
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

$\displaystyle U_Z\cap X=U_Z\cap Y\cap X=U_Y\cap X=U_X,$    

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}$. $ \qedsymbol$



"subspace of a subspace" is owned by matte. [ full author list (2) ]
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Cross-references: equality, inclusion, open set, subspace topology, topological space
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This is version 2 of subspace of a subspace, born on 2005-05-22, modified 2005-05-30.
Object id is 7096, canonical name is SubspaceOfASubspace.
Accessed 1289 times total.

Classification:
AMS MSC: 54B05 (General topology :: Basic constructions :: Subspaces)
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