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?

Observe that if $B = \{B_i: i \in I \}$ is an open base of $X$ if and only if for any $x$ in open set $G$ there exist $B_i$ in $B$ such that $x \in B_i \subseteq G$.
Now let $A$ be open in $Y$ then  $A = Y \cap G$ where $G$ is open in $X$ and let $x$ be in $A$. Clearly $x \in G$ which implies there is a $B_i$ in $B$ such that $x \in B_i$ and $B_i \subseteq G$. This also implies that $x \in Y \cap B_i$ and clearly $Y \cap B_i \in B_Y$. So we conclude that $B_Y$ is an open base for subspace $Y$.
