Neighborhoods

From Open Balls to Neighborhoods

In “Open Balls,” a ball \(B_r(a)\) collected all points whose distance from \(a\) is less than a positive radius \(r\). A neighborhood uses that same idea to describe a more general set: the set need not itself be a ball, but it must contain some ball around the point in question. This lets us describe what happens locally without specifying the entire shape of the set.

The radius \(r\) in this definition may depend on both the set \(N\) and the point \(x\). A set can be a neighborhood of one of its points but fail to be a neighborhood of another. The definition asks for a whole ball around \(x\) to lie in \(N\), not merely for \(x\) itself to belong to \(N\).

A neighborhood does not have to be an open ball, or even be open. For example, it can include one or both endpoints of a larger interval, or include additional points far from the center. What matters is that at least one positive-radius ball around the specified point is contained in it. In contrast, the ball \(B_r(x)\) itself always is a neighborhood of \(x\), since \(B_r(x)\subseteq B_r(x)\).

Neighborhoods Need Room on Both Sides

On the real line, a ball around \(x\) is an interval extending both to the left and to the right of \(x\). Consequently, a set that contains \(x\) but only has nearby points on one side may fail to be a neighborhood of \(x\). The next example makes the role of a positive radius explicit.

A useful way to test a proposed neighborhood is to look for a radius that works, rather than to assume that a large interval or a maximal radius is needed. If \(B_r(x)\subseteq N\), then every smaller positive-radius ball \(B_s(x)\), where \(0<s\leq r\), is also contained in \(N\), by the nesting property of balls. A single successful radius is enough.

Basic Operations on Neighborhoods

The definition immediately gives a useful rule about enlarging sets. If \(N\) is a neighborhood of \(x\) and \(N\subseteq M\), then \(M\) is also a neighborhood of \(x\). Indeed, if \(B_r(x)\subseteq N\), then \(B_r(x)\subseteq M\) as well. Thus adding points to a neighborhood cannot destroy its neighborhood property at the same point.

Intersections behave differently from enlargements: intersecting sets can remove points, so it is necessary to check that a ball remains. For a finite number of neighborhoods of the same point, this can always be done by choosing the smallest of their radii.

Proof. For each \(j\in\{1,\ldots,m\}\), the definition gives a radius \(r_j>0\) such that \(B_{r_j}(x)\subseteq N_j\). Since there are finitely many positive radii, their minimum \(r=\min\{r_1,\ldots,r_m\}\) is positive. If \(z\in B_r(x)\), then \(d(z,x)<r\leq r_j\) for every \(j\). Hence \(z\in B_{r_j}(x)\subseteq N_j\) for every \(j\), so \(z\in\bigcap_{j=1}^{m}N_j\). We have shown that \(B_r(x)\subseteq\bigcap_{j=1}^{m}N_j\), with \(r>0\). Therefore the intersection is a neighborhood of \(x\). \(\square\)

The finiteness assumption matters to this proof: the minimum of finitely many positive radii is positive, but an infinite collection of positive radii need not have a positive lower bound. For example, the radii \(1,1/2,1/3,\ldots\) are all positive, while no positive number is less than or equal to all of them. The theorem makes no claim about arbitrary intersections.

A Neighborhood Can Serve Nearby Points

A neighborhood of \(x\) contains not just \(x\), but an entire ball around \(x\). As a result, points sufficiently close to \(x\) also have room around them inside the same set. The amount of room can be estimated by subtracting the distance from \(x\) to the nearby point from the radius already known to fit inside the set.

Proof. Since \(y\in B_R(x)\), we have \(d(x,y)<R\), so \(\rho=R-d(x,y)>0\). Let \(z\in B_\rho(y)\). Then \(d(y,z)<\rho\). By the triangle inequality,

Therefore \(z\in B_R(x)\), and the assumed inclusion \(B_R(x)\subseteq N\) gives \(z\in N\). This proves \(B_\rho(y)\subseteq N\) for a positive \(\rho\), so \(N\) is a neighborhood of \(y\). \(\square\)

Why the Definition Matters

Neighborhoods separate local information from the global shape of a set. To know that \(N\) is a neighborhood of \(x\), it is unnecessary to describe every point in \(N\): it is enough to find one ball around \(x\) contained in \(N\). The set may extend far beyond that ball, have included endpoints, or have a complicated shape elsewhere.

This language is also useful when several local conditions must hold at once. If finitely many sets each contain a ball around the same point, the finite-intersection theorem provides a ball that satisfies all those conditions together. The nearby-point theorem gives another local tool: once a ball is known to lie in a set, each point inside that ball also has a smaller ball lying in the set. These facts prepare the way for the next topic, where sets are described by requiring this kind of local room at each of their points.

A common pitfall is to confuse “contains the point” with “is a neighborhood of the point.” A set can include \(x\) but omit points arbitrarily close to \(x\), as in the example \(\{0\}\cup(2,3)\) at \(0\). Another is to overlook the strict inequality in \(B_r(x)\): endpoints at distance exactly \(r\) are not part of the ball. To prove a neighborhood claim, exhibit a positive radius and verify the full inclusion; to disprove one, show that every positive radius has a point in its ball missing from the set.