Definition 12.6.
Let \((X, \tau)\) be a topological space, and let \(a \in X\text{.}\) A subset \(N\) of \(X\) is a neighborhood of \(a\) if \(N\) contains an open set that contains \(a\text{.}\)
Section Neighborhoods in Topological Spaces
Recall that we defined a neighborhood of a point \(a\) in a metric space to be a subset of the space that contains an open ball centered at \(X\text{.}\) Every open ball is an open set, so we can extend the idea of neighborhood to topological spaces.
Let’s look at some examples.
Activity 12.7.
Let \(X = \{a,b,c,d\}\) and let \(\tau = \{\emptyset, \{a\}, \{b\}, \{a,b\}, X \}\text{.}\)
(a)
Find all of the neighborhoods of the point \(a\text{.}\)
(b)
Find all of the neighborhoods of the point \(c\text{.}\)
In metric spaces, an open set was a neighborhood of each of its points. This is also true in topological spaces.
Theorem 12.7.
Let \((X, \tau)\) be a topological space. A subset \(O\) of \(X\) is open if and only if \(O\) is a neighborhood of each of its points.
Proof.
Let \((X, \tau)\) be a topological space, and let \(O\) be a subset of \(X\text{.}\) First we demonstrate that if \(O\) is open, then \(O\) is a neighborhood of each of its points. Assume that \(O\) is an open set, and let \(a \in O\text{.}\) Then \(O\) contains the open set \(O\) that contains \(a\text{,}\) so \(O\) is a neighborhood of \(a\text{.}\)
The reverse containment is the subject of the next activity.
Activity 12.8.
Let \((X, \tau)\) be a topological space. Let \(O\) be a subset of \(X\text{.}\) Assume \(O\) is a neighborhood of each of its points.
(a)
What do we need to do to show that \(O\) is an open set?
(b)
Let \(a \in O\text{.}\) Why must there exist an open set \(O_a\) such that \(a \in O_a \subseteq O\text{?}\)
(c)
Complete the proof that \(O\) is an open set.
