Why the interior of $U$ in $S$ contains $\mathrm{int\,}U\cap S$

Question

(Exercise 3.7. John Lee’s Introduction to Topological Manifolds) Suppose $X$ is a topological space and $U\subseteq S\subseteq X$. The interior of $U$ in $S$ contains $\mathrm{int } \, U\cap S$.

My attempt:

I can see why they are not equal: an open set in the subspace topology may not necessarily be open with respect to the topology from which the subspace inherits it, and thus the two interiors may not be equal. But, why the word “contained?”

I can think of an example, but not prove it in general: take the set $S = [0,1]\cup (2,3)$ and a subset $[0,1]$ which is open in $S$, but not in $\mathbb{R}$. Thus, it follows that the interior of $[0,1]$ in $S$ contains $\mathrm{int}\,[0,1]\cap S$. So then, how can I compete my proof?

Answer 1

$\operatorname{int}(U) \cap S$ is open in $S$ (it's open in $S$ by the definition of the subspace topology and the fact that $\operatorname{int}(U)$ is open), and it's clearly a subset of $U$. As the interior of $U$ w.r.t. $S$ (denoted by $\operatorname{int}_S(U)$) is the largest open (in $S$) set that is contained in $U$, we immediately get $$\operatorname{int}(U) \cap S \subseteq \operatorname{int}_S(U)$$

The reverse need not hold, as witnessed by, e.g. $U = S = \mathbb{Q}$ in the reals, which has empty interior in the reals but is its own interior in itself of course.

Answer 2

Suppose $p\in\operatorname{int}(U)\cap S$. Because $p\in\operatorname{int}(U)$, there exists some open set $O$ of $X$ such that $p\in O\subset U$. Now remember how the open sets of $S$ in its subspace topology are defined... do you see how to finish the problem?