Topological spaces · Definition
Neighborhood of a point
Before you start you need to know
Neighborhoods give a precise meaning to «near a point» without distances: they are the sets that surround the point with some room to spare.
Definition
A neighborhood of a point \(x\) in a topological space \((X,\mathcal{T})\) is a set \(N\subseteq X\) that contains some open set \(U\in\mathcal{T}\) to which \(x\) belongs.
A neighborhood need not be open: it only has to contain an open set around the point. So \([0,2]\) is a neighborhood of \(1\) in \(\mathbb{R}\) even though it is not open, while it is not one of \(0\), since no open set containing \(0\) fits inside \([0,2]\). On the other hand, every open set is a neighborhood of each of its points; this is why a set is open exactly when it is a neighborhood of all its points.
The neighborhoods of a point \(x\) form a stable family: any set containing a neighborhood of \(x\) is also a neighborhood of \(x\), and the intersection of two neighborhoods of \(x\) is again one. When one neighborhood lies inside all the others it is called the minimal neighborhood; in the trivial topology it is \(X\), and in the discrete topology it is \(\{x\}\).
A common mistake is to think that «neighborhood» means «open ball» or «small set». In a metric space the balls \(B(x,r)\) form a neighborhood base, but a neighborhood can be huge (for instance, \(X\) itself). This concept underlies the definition of the interior of a set, accumulation points and continuity at a point.
Think about it
In \(\mathbb{R}\) with the usual topology, is \((0,1)\cup\{5\}\) a neighborhood of \(5\)?
No: an open set containing \(5\) would have to include an interval \((5-\varepsilon,5+\varepsilon)\), which does not fit inside the set.
Why is \(X\) a neighborhood of all its points in any topology?
Because \(X\) is open and \(x\in X\subseteq X\) for every \(x\).
Exercises 10
- In \(X=\{1,2,3\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{1,2\},\{1,2,3\}\}\), which of these sets is a neighborhood of the point \(2\)?
- In \(X=\{1,2,3\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{1,2\},\{1,2,3\}\}\), which is the smallest neighborhood of the point \(1\)?
- In \(X=\{1,2,3\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{1,2\},\{1,2,3\}\}\), how many neighborhoods does the point \(3\) have?
- In \(X=\{a,b,c,d\}\) with \(\mathcal{T}=\{\varnothing,\{a\},\{b\},\{a,b\},\{a,b,c\},\{a,b,c,d\}\}\), how many subsets of \(X\) are neighborhoods of the point \(c\)?
- In \(X=\{1,2,3,4\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{3\},\{1,3\},\{1,2,3\},\{1,3,4\},\{1,2,3,4\}\}\), which is the smallest neighborhood of the point \(2\)?
- In \(X=\{a,b,c\}\) with \(\mathcal{T}=\{\varnothing,\{a,b,c\}\}\), how many neighborhoods does the point \(a\) have?
- Let \(U\) be an open set of \(X\) and \(x\in U\). Which statement is correct?
- Which statement about the neighborhoods of a point \(x\) is true?
- In \(X=\{1,2,3,4\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{3\},\{1,3\},\{1,2,3\},\{1,3,4\},\{1,2,3,4\}\}\), of which points of \(X\) is \(\{1,2\}\) a neighborhood?
- In \(X=\{1,2,3,4\}\) with \(\mathcal{T}=\{\varnothing,\{1\},\{3\},\{1,3\},\{1,2,3\},\{1,3,4\},\{1,2,3,4\}\}\), which of these sets is a neighborhood of at least one point?
What comes next
References
- James R. Munkres, Topology, 2nd edition, Pearson, 2000. §17
- Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990. cap. 3