Proposition

Suppose $X$ is a topological space and $Y$ is an open subset of $X$. Show that the collection of all open subsets of $X$ that are contained in $Y$ is a topology on $Y$.

Solution

Let $\mathcal{T}$ be the topology on $X$, and $\mathcal{T}’ = \{ U \in \mathcal{T} \mid U \subset Y \}$.

• $\emptyset \in \mathcal{T}’$.
• $Y \in \mathcal{T}’$.
• Let $\{ U_{\alpha} \} \subset \mathcal{T}’$. For all $\alpha$, $U_{\alpha} \in \mathcal{T}$ and $U_{\alpha} \subset Y$. Thus $\bigcup_{\alpha} U_{\alpha} \in \mathcal{T}’$.
• Let $U_{\alpha_1}, \cdots, U_{\alpha_n} \in \mathcal{T}’$. For all $i = 1, \cdots, n$, $U_{\alpha_i} \in \mathcal{T}$ and $U_{\alpha_i} \subset Y$. Thus $\bigcap_{i=1}^n U_{\alpha_i} \in \mathcal{T}’$.

Therefore, $\mathcal{T}’$ is a topology on $Y$.