Proposition

Let $S \subset X$ and $\mathscr{T}_S = \{ S \cap U \mid U \in \mathscr{T} \}$. Then $\mathscr{T}_S$ is a topology on $S$.

Solution

• $\emptyset$ is in $\mathscr{T}$, so $S \cap \emptyset = \emptyset \in \mathscr{T}_S$.
• $X$ is in $\mathscr{T}$, so $S \cap X = S \in \mathscr{T}_S$.
• Let \(\{ S \cap U_{\alpha} \} \subset \mathscr{T}_S\). Then \(\bigcup (S \cap U_{\alpha}) = S \cap (\bigcup U_{\alpha})\). $\bigcup U_{\alpha} \in \mathscr{T}$, so $S \cap (\bigcup U_{\alpha}) \in \mathscr{T}_S$.
• Let \(\{ S \cap U_1, \cdots, S \cap U_n \} \subset \mathscr{T}_S\). Then $\bigcap (S \cap U_{k}) = S \cap (\bigcap U_{k})$. $\bigcap U_{k} \in \mathscr{T}$, so $S \cap (\bigcap U_{k}) \in \mathscr{T}_S$.

Therefore, $\mathscr{T}_S$ is a topology on $S$.