----- > [!proposition] Proposition. ([[basis for the subspace topology]]) > If $\mathscr{B}$ is a [[basis for a topology|basis]] for the [[topological space|topology]] $\tau$ on [[topological space]] $X$, and $Y \subset X$, then the set $\mathscr{B}_{Y}=\{ B \cap Y : B \in \mathscr{B}\}$ > is a [[basis for a topology|basis]] for the [[subspace topology]] $\tau_{Y}$ on $Y$. > [!proof]- Proof. ([[basis for the subspace topology]]) > We'll use [[condition for obtaining a basis from a topology]]. Let $U \in \tau$ be arbitrary and consider $x \in U \cap Y \in \tau_{Y}$. $\mathscr{B}$ is a [[basis for a topology|basis]] for $\tau$; obtain $B \in \mathscr{B}$ s.t. $x \in B \subset U$. Then since $x\in Y$ we must have $x \in B \cap Y \subset U \cap Y$ and this completes the proof. > [!warning] > The following strategy does NOT work— it just shows we have a basis for *a* topology... not one for *the* topology we want! > We'll just use the [[basis for a topology|basis definition]]. >1. (Covering) Let $x \in Y$. Then $x \in X$ too, and since $\mathscr{B}$ is a [[basis for a topology|basis]] of $X$, this means $\ex B \in \mathscr{B}$ s.t. $x \in B$. Hence $x \in B \cap Y \in \mathscr{B}_{Y}$. >2. (Intersection Nestling) Let $x \in B_{Y_{1}} \cap B_{Y_{2}}$ where $B_{Y_{1}},B_{Y_{2}} \in \mathscr{B}_{Y}$. Then we have $x \in (B_{1} \cap Y) \cap (B_{2} \cap Y)=(B_{1} \cap B_{2}) \cap Y$ for some $B_{1}, B_{2} \in \mathscr{B}$. $\mathscr{B}$ is a [[basis for a topology|basis]] for $X$, meaning that $\ex B_{3}$ s.t. $x \in B_{3} \subset B_{1} \cap B_{2}$. Now $x \in B_{3} \subset B_{1} \cap B_{2} \and x \in Y \implies x \in B_{3} \cap Y \subset (B_{1} \cap B_{2}) \cap Y= B_{Y_{1}} \cap B_{Y_{2}}$ and this completes the proof.