What Does It Mean to Be Near a Point?
In topology, a neighborhood is a set that contains some open set around the point in question. The neighborhood itself need not be open. This distinction matters: openness is a property of a set, while being a neighborhood is a relationship between a set and a specified point. The same set can be a neighborhood of one point and not of another.
Throughout, \((X,\mathcal{T})\) denotes a topological space, and \(x\in X\). An open neighborhood of \(x\) is an open set containing \(x\). We use the following broader definition for a neighborhood.
The point must be specified when we ask whether a set is a neighborhood: \(N\) might be a neighborhood of \(x\) but not of \(y\). Also, merely containing \(x\) is not enough. The definition requires an open set containing \(x\) to fit inside \(N\).
Testing the Definition with Interior
For \(A\subseteq X\), its interior, written \(\operatorname{int}A\), is the union of all open subsets of \(X\) contained in \(A\). Equivalently, it is the largest open set contained in \(A\). This gives a concise test for neighborhoods.
Proof. Suppose \(N\) is a neighborhood of \(x\). By definition, there is an open set \(U\) with \(x\in U\subseteq N\). Since \(U\) is an open subset of \(N\), it is contained in \(\operatorname{int}N\). Therefore \(x\in\operatorname{int}N\).
Conversely, suppose \(x\in\operatorname{int}N\). The interior is open and is contained in \(N\), so \(U=\operatorname{int}N\) is an open set satisfying \(x\in U\subseteq N\). Thus \(N\) is a neighborhood of \(x\). \(\square\)
This test highlights why a set containing \(x\) might still fail to be a neighborhood: \(x\) must lie in the interior, not merely in the set. In practical terms, there must be an open set around \(x\) that stays entirely inside \(N\).
Worked Example: Neighborhoods in the Usual Real Line
Give the topology on \(\mathbb{R}\) its standard meaning, so open intervals form a basis. For \(x\in\mathbb{R}\), a set \(N\subseteq\mathbb{R}\) is a neighborhood of \(x\) exactly when it contains an interval \((x-\varepsilon,x+\varepsilon)\) for some \(\varepsilon>0\).
Why this test works. If \(N\) is a neighborhood, it contains an open set \(U\) with \(x\in U\). By the basis property of open intervals, there is an \(\varepsilon>0\) such that \(x\in(x-\varepsilon,x+\varepsilon)\subseteq U\subseteq N\). Conversely, each interval \((x-\varepsilon,x+\varepsilon)\) is open and contains \(x\), so if it is contained in \(N\), then \(N\) is a neighborhood by definition.
Take \(x=0\) and \(N=[-1,2]\). The interval \((-1,1)\) is open, contains \(0\), and is contained in \(N\), so \(N\) is a neighborhood of \(0\). But \(N\) is not open: any open interval containing \(2\) also contains points greater than \(2\), so no such interval is contained in \(N\). Thus this is a neighborhood that is not an open neighborhood.
Now take \(M=[0,1]\) and again ask about \(0\). Although \(0\in M\), \(M\) is not a neighborhood of \(0\). For every \(\varepsilon>0\), the interval \((-\varepsilon,\varepsilon)\) contains negative numbers, none of which belong to \(M\). The interval test therefore fails, and \(0\notin\operatorname{int}M\).
Check: the open set \((-\tfrac12,\tfrac12)\) is contained in \([-1,2]\), so the first set passes the test. No interval \((-\varepsilon,\varepsilon)\) is contained in \([0,1]\), so the second set fails it.
The interval test depends on the standard topology on \(\mathbb{R}\). A different topology on the same underlying set can change which sets count as neighborhoods. The next two examples make this dependence explicit.
Discrete Spaces: Every Set Is Open
A set \(X\) with the discrete topology has \(\mathcal{T}=\mathcal{P}(X)\), meaning every subset of \(X\) is open. Consequently, the distinction between neighborhoods and open neighborhoods disappears in a discrete space.
Proof. If \(N\) is a neighborhood of \(x\), its definition gives an open set \(U\) with \(x\in U\subseteq N\), so \(x\in N\). Conversely, if \(x\in N\), then \(N\) itself is open in the discrete topology. It is therefore an open neighborhood of \(x\), and hence a neighborhood of \(x\). \(\square\)
Worked Example: A Three-Point Discrete Space
Let \(X=\{p,q,r\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). Determine the neighborhoods of \(p\), and decide whether \(\{p\}\), \(\{p,q\}\), and \(\{q,r\}\) are open neighborhoods of \(p\).
The subsets of \(X\) containing \(p\) are exactly \(\{p\}\), \(\{p,q\}\), \(\{p,r\}\), and \(X\). By the proposition, these are precisely the neighborhoods of \(p\). Each is open, so each is also an open neighborhood of \(p\).
The set \(\{q,r\}\) is open because every subset is open, but it is not a neighborhood of \(p\): it does not contain \(p\), so it cannot contain any open set that contains \(p\). It is an open neighborhood of \(q\), however, since it is open and contains \(q\).
Check: \(\{p\}\) itself is an open set containing \(p\), so it witnesses that \(\{p\}\) is a neighborhood of \(p\). The set \(\{p,q\}\) also contains \(p\) and is open, so it witnesses that \(\{p,q\}\) is a neighborhood. The set \(\{q,r\}\) has no point \(p\), so it cannot be a neighborhood of \(p\).
The Sierpinski Space: Neighborhoods Depend on the Point
Consider the Sierpinski space \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This is a topology: it contains \(\varnothing\) and \(X\), and unions and finite intersections of its members remain in the displayed family. Unlike the discrete example, not every subset is open. We can therefore inspect the open neighborhoods of each point separately.
Worked Example: Neighborhoods at Each Sierpinski Point
For \(a\), the open sets containing \(a\) are \(\{a\}\) and \(X\). Any neighborhood of \(a\) must contain one of these open neighborhoods. The sets in \(X\) that contain \(\{a\}\) are \(\{a\}\) and \(X\); the sets that contain \(X\) contribute only \(X\). Thus the neighborhoods of \(a\) are exactly \(\{a\}\) and \(X\), and both are open neighborhoods.
For \(b\), the only open set containing \(b\) is \(X\). A neighborhood of \(b\) must contain this open neighborhood, so the only neighborhood of \(b\) is \(X\) itself. In particular, \(\{b\}\) contains \(b\), but it is not a neighborhood: the only open set containing \(b\) is \(X\), and \(X\not\subseteq\{b\}\).
The set \(\{a\}\) shows why the point matters. It is an open neighborhood of \(a\), but it is not a neighborhood of \(b\), since it does not contain \(b\). In this particular space, every neighborhood of each point happens to be open; that coincidence does not hold in the usual real line.
Check: for \(a\), the open set \(\{a\}\) witnesses that \(\{a\}\) is a neighborhood. For \(b\), no open neighborhood other than \(X\) is available, so any neighborhood of \(b\) must contain \(X\), and therefore must equal \(X\).
A Common Mistake
Pitfall: treating “\(x\in N\)” as sufficient to conclude that \(N\) is a neighborhood of \(x\). For example, in \(\mathbb{R}\) with its standard topology, the calculation \(0\in[0,1]\), therefore \([0,1]\) is a neighborhood of \(0\), is wrong. Membership gives only the first condition. The interval test shows the correction: for every \(\varepsilon>0\), \((-\varepsilon,\varepsilon)\) contains negative numbers, so it is not contained in \([0,1]\). Thus \(0\notin\operatorname{int}[0,1]\), and \([0,1]\) is not a neighborhood of \(0\).
There is a second, related error: assuming that a neighborhood must be open. The set \([-1,2]\) is a neighborhood of \(0\) because it contains the open neighborhood \((-1,1)\), even though \([-1,2]\) is not open. Keep the two questions separate: does the set contain the point, and does it contain an open neighborhood of that point?
Neighborhoods provide a local way to describe the topology around an individual point. The examples show the two ingredients to check every time: which open sets contain the point, and which sets contain at least one of those open neighborhoods.
Check Your Understanding
For each question, state the topology being used and justify your answer from the definition.
- In the usual topology on \(\mathbb{R}\), is \((-2,3]\) a neighborhood of \(0\)? Is it an open neighborhood?
- In the usual topology on \(\mathbb{R}\), is \((0,2]\) a neighborhood of \(0\)? Explain why containing \(0\) is or is not relevant here.
- In a discrete space \(X\), list all neighborhoods of a point \(x\) when \(X=\{x,y,z\}\).
- In the Sierpinski space \(\{\varnothing,\{a\},X\}\), is \(\{a\}\) a neighborhood of \(b\)? Is \(X\) a neighborhood of \(b\)?
- Give an example of a neighborhood in \(\mathbb{R}\) that is not open, and identify an open neighborhood it contains.