# 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 |