Induced topology base Let $X$ be a topological space, $B$ an open base for $X$ and $Y \subseteq X$. Prove that $B_Y = \{U \cap Y: U \in B\}$ is an open base for $Y$ (considering, over $Y$, the subspace topology).
We must show that given an open $A$ relative to the topology of $Y$, we have to show that there is an element of $B_Y$ such that $ A = \cup (U \cap Y) $. Since $ B $ is a base for X, we have that Y can be written as a union of elements of $B$. That is, there is $W \in B$ such that $ Y = \cup W $. Is this demonstration absurd?

Use the fact that the subspace toplogy $\tau_Y$ is the coarsest one for which the inclusion $i:Y\to X$ is continuous.
If $A$ is open in $Y$ then there is a $V$ open in $X$ such that $i^{-1}(V)=A$. Since $V$ is open in $X$ there is a basis element $B\in \tau_X$   such that $B\subseteq V.$ Then, $i^{-1}(B)=B\cap Y\subseteq A$ and $B\cap Y$ is open (in $Y$) by definition of $\tau_Y.$ It follows that the collection $\{B\cap Y\}_{B\in \tau_X}$ is a base for $\tau_Y.$
