Proposition

Suppose $X$ is a topological space and $U \subset S \subset X$.

  1. Show that the closure of $U$ in $S$ is equal to $\overline{U} \cap S$.
  2. Show that the interior of $U$ in $S$ contains $\Int{U} \cap S$; give an example to show that they might not be equal.

Solution

1

Let $\mathscr{C}$ be the set of all closed sets in $X$. Let $\mathscr{C}_S$ be the set of all closed sets in $S$. Then $\mathscr{C}_S = \{ C \cap S \mid C \in \mathscr{C} \}$.

\[\begin{align*} \cl_S(U) &= \bigcap_{C \in \mathscr{C}_S, U \subset C} C \\ &= \bigcap_{C \in \mathscr{C}, U \subset C \cap S} (C \cap S) \\ &= [\bigcap_{C \in \mathscr{C}, U \subset C \cap S} C] \cap S \\ &= [\bigcap_{C \in \mathscr{C}, U \subset C} C] \cap S \\ &= \overline{U} \cap S. \end{align*}\]

2

Let $x \in \Int{U} \cap S$. Since $x \in \Int{U}$, $x$ has a neighborhood $N$ of $x$ in $X$ such that $N \subset U$. Then $N \cap S$ is open in $S$ and it contains $x$. Thus $N \cap S$ is contained in the interior of $U$ in $S$. Thus $x$ is in the interior of $U$ in $S$.

Consider $X = \mathbb{R}, S = U = [0, 1]$. Then $\Int{U} \cap S = (0, 1) \cap [0, 1] = (0, 1)$. On the other hand, the interior of $U$ in $S$ is $U$ because $U$ is open in $S$. This is an example where they might not be equal.