# characterization of subspace topology

###### Theorem.

Let $X$ be a topological space^{} and $Y\mathrm{\subset}X$ any subset. The subspace topology on $Y$ is the weakest topology making the inclusion map^{} continuous^{}.

###### Proof.

Let $\mathcal{S}$ denote the subspace topology on $Y$ and $j:Y\hookrightarrow X$ denote the inclusion map.

Suppose $\{{\mathcal{T}}_{\alpha}\mid \alpha \in J\}$ is a family of topologies on $Y$ such that each inclusion map ${j}_{\alpha}:(Y,{\mathcal{T}}_{\alpha})\hookrightarrow X$ is continuous. Let $\mathcal{T}$ be the intersection^{} ${\bigcap}_{\alpha \in J}{\mathcal{T}}_{\alpha}$. Observe that $\mathcal{T}$ is also a topology on $Y$. Let $U$ be open in $X$. By continuity of ${j}_{\alpha}$, the set ${j}_{\alpha}^{-1}(U)={j}^{-1}(U)$ is open in each ${\mathcal{T}}_{\alpha}$; consequently, ${j}^{-1}(U)$ is also in $\mathcal{T}$. This shows that there is a weakest topology on $Y$ making inclusion continuous.

We claim that any topology strictly weaker than $\mathcal{S}$ fails to make the inclusion map continuous. To see this, suppose ${\mathcal{S}}_{0}\u228a\mathcal{S}$ is a topology on $Y$. Let $V$ be a set open in $\mathcal{S}$ but not in ${\mathcal{S}}_{0}$. By the definition of subspace topology, $V=U\cap Y$ for some open set $U$ in $X$. But ${j}^{-1}(U)=V$, which was specifically chosen not to be in ${\mathcal{S}}_{\mathcal{0}}$. Hence ${\mathcal{S}}_{\mathcal{0}}$ does not make the inclusion map continuous. This completes^{} the proof.
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Title | characterization of subspace topology |
---|---|

Canonical name | CharacterizationOfSubspaceTopology |

Date of creation | 2013-03-22 15:40:32 |

Last modified on | 2013-03-22 15:40:32 |

Owner | mps (409) |

Last modified by | mps (409) |

Numerical id | 6 |

Author | mps (409) |

Entry type | Theorem |

Classification | msc 54B05 |