Section The Interior of a Set in a Topological Space
We have seen that topologies define the open sets in a topological space. As in metric spaces, open sets can be characterized in terms of their interior points. We defined interior points in metric spaces in terms of neighborhoods β the same holds true in topological spaces.
Remember that a set is a neighborhood of a point if the set contains an open set that contains the point. By definition, every open set is a neighborhood of each of its points, so every point of an open set is an interior point of Conversely, if every point of a set is an interior point, then is a neighborhood of each of its points and is open. This argument is summarized in the next theorem.
Theorem 12.9.
Let be a topological space. A subset of is open if and only if every point of is an interior point of
The collection of interior points in a set form a subset of that set, called the interior of the set.
Definition 12.10.
Activity 12.9.
(a)
Consider where is the standard topology (by standard in this situation, we mean the metric topology determined by the Euclidean metric). Let in What is What is the largest open subset of contained in
(b)
Consider where is the discrete topology (the one where all subsets are open). Let in What is What is the largest open subset of contained in
(c)
Consider where is the finite complement topology (the one where the open sets are the empty set along with all subsets of such that is finite). Let in What is What is the largest open subset of contained in
(d)
One might expect that the interior of a set is an open set, as it was in metric spaces. This is true, but we can say even more. In Activity 12.9 we saw that in our examples that was the largest open subset of contained in That this is always true is the subject of the next theorem.
Theorem 12.11.
Let be a topological space, and let be a subset of The interior of is the largest open subset of contained in
Proof.
Let be a topological space, and let be a subset of We need to prove that is an open set in and that is the largest open subset of contained in First we demonstrate that is an open set. Let Then is an interior point of so is a neighborhood of This implies that there exists an open set containing so that But is a neighborhood of each of its points, so every point in is an interior point of It follows that Thus, is a neighborhood of each of its points and, consequently, is an open set.
Activity 12.10.
(a)
(b)
Suppose that is an open subset of that is contained in and let What does the fact that is open tell us? Then complete the proof that
One consequence of Theorem 12.11 is the following.