Proving that a set is a topology

Question

Given $X\neq \emptyset$, $Y\subset X$ and $\tau$ a topology on $X$, prove that $\rho = \{G\cap Y:G\in \tau\}$ is a topology on Y. I'm struggling to prove that if $\emptyset \neq \mathbb{A}\subset \rho$ then $\cup \mathbb{A} \in \rho$. Any help will be greatly appreciated. Thanks in advance!

Answer 1

Hint: Write elements of $\rho$ as $G \cap Y$ for some $G \in \tau$, and use the distributive property of union over intersection to talk about unions of elements of $\rho$.

Any collection of sets in $\rho$ can be written as $\{G_\alpha \cap Y\}_\alpha$ where $G_\alpha \in \tau$ for all $\alpha$, so $$\bigcup_\alpha (G_\alpha \cap Y) = \left(\bigcup_\alpha G_\alpha\right) \cap Y.$$

Answer 2

Let $O_i$, $i \in I$ be any family of sets from $\rho$. This means that each is of the form $U_i \cap G$ for $U_i \in \tau$.

Then $$\bigcup_{i \in I} O_i = \bigcup_{i \in I} (U_i \cap G) = \left(\bigcup_{I \in I} U_i\right) \cap G \in \rho$$ as $\tau$ is closed under unions so $\bigcup_{i \in I} U_i \in \tau$.

The last equal sign (the distributivity property) can be seen by simple element chasing (double inclusion). If $x \in \bigcup_{i \in I} (U_i \cap G)$, then for soem $j \in I$, $x \in U_j \cap G$, so $x \in U_j$ and hence $x \in \bigcup_{i \in I} U_i$, and $x \in G$ so $x \in \left(\bigcup_{I \in I} U_i\right) \cap G$. And the reverse is similarly clear.