# Exercise 10.2

Let $(X,\mathcal{T})$ be a topological space and let $A$ be a nonempty subset of $X$. Prove that $\mathcal{T}_A = \{ U\cap A: U \in \mathcal{T}\}$ is a topology on $A$.

**Proof:**

$(T1)$ Since $\emptyset,X\in \mathcal{T}$, we have $\emptyset = \emptyset\cap A \in \mathcal{T}_A$ and $A = X\cap A \in \mathcal{T}_A$.

$(T2)$ Let $U,V \in \mathcal{T}_A$. By definition, there exist $U^{'}, V^{'} \in \mathcal{T}$ such that $U = U^{'} \cap A$ and $V = V^{'} \cap A$. Hence, $$U\cap V = (U^{'} \cap A ) \cap (V^{'} \cap A) = (U\cap V) \cap A\in \mathcal{T}_A.$$

$(T3)$ Suppose $\{U_\alpha\}_I$ is a family of sets in $\mathcal{T}_A$. Then we have
$$\bigcup_I U_\alpha = \bigcup_I (U^{'}_\alpha\cap A) = \left(\bigcup_I U^{'}_\alpha \right)\cap A \in \mathcal{T}_A,$$
where $U^{'}_\alpha \in \mathcal{T}$ for all $\alpha \in I$.

###### tags: `Topology` `Solution`