2.1 Domains and point sets

A formula for a function of two variables does not usually allow every point of the plane as an input. A square root may restrict the input to a disc; a logarithm may select one side of a curve; a denominator may remove a line from an otherwise solid region. Before asking how the function changes near a point, we need to know which nearby inputs are available. This is why the geometry of a domain belongs at the beginning of multivariable analysis.

The central tool is distance. An interval measures closeness on the real line; a ball performs the same job in Euclidean space. We will use balls to distinguish points inside a set, points outside it, and points on its boundary. These definitions also explain why a region can be neither open nor closed, why a circle has no interior in the plane, and why an omitted point can still influence a limit.

Functions and their domains

Definition

Domain, codomain and range

A function $F:D\to\mathbb R^m$, with $D\subseteq\mathbb R^n$, assigns exactly one output $F(x)$ to each $x\in D$. The set $D$ is its domain, $\mathbb R^m$ is the stated codomain, and $F(D)=\{F(x):x\in D\}$ is its range. Writing $F=(F_1,\ldots,F_m)$ describes its scalar component functions.

The input dimension and output dimension answer different questions. A temperature function on space has the form $T:\mathbb R^3\to\mathbb R$, whereas a parametrised spatial curve has the form $\gamma:I\subseteq\mathbb R\to\mathbb R^3$. The latter has one input variable and three output coordinates.

When a domain is explicitly stated, it is part of the function. When only an expression is given, its natural real domain consists of all inputs for which every operation is defined. Intersect the restrictions: square-root arguments must be nonnegative, logarithm arguments positive, and denominators nonzero. Simplifying an expression can reveal a possible extension, but does not automatically change the original domain.

Worked example

Intersecting three restrictions

Find the natural domain of

$$f(x,y)=\frac{\sqrt{16-x^2-y^2}}{\log(y-x)}.$$

The square root requires $x^2+y^2\le16$. The logarithm requires $y-x>0$, and its occurrence in the denominator also requires $y-x\ne1$. Thus

$$D=\{(x,y):x^2+y^2\le16,\ y>x,\ y-x\ne1\}.$$

Geometrically, start with the closed disc of radius $4$, retain the side above $y=x$, and remove the segment on $y=x+1$. A point on the circular arc may belong to $D$, because a zero numerator is allowed. A point on either removed line does not. Checking one restriction at a time prevents the common error of forgetting where the logarithm vanishes.

Balls and local descriptions

Definition

Open balls and neighbourhoods

For $a\in\mathbb R^n$ and $r>0$, define

$$B(a,r)=\{x:\|x-a\|<r\},\qquad \overline B(a,r)=\{x:\|x-a\|\le r\}.$$

A neighbourhood of $a$ is any set containing some $B(a,r)$. A punctured ball is $B(a,r)\setminus\{a\}$.

In the plane, an open ball is a filled disc without its circumference; in three dimensions, it is the solid interior of a sphere. The word “open” specifies the strict inequality, not a hollow shape. A neighbourhood need not itself be a ball or an open set: it only has to contain an entire sufficiently small ball around its centre.

Definition

Interior and exterior

A point $a$ is interior to $A$ if some $B(a,r)$ lies in $A$. The interior $A^\circ$ is the set of all such points. The exterior is $\operatorname{ext}A=(\mathbb R^n\setminus A)^\circ$. A set is open if every one of its points is interior, equivalently $A=A^\circ$.

A point is interior only when movement in every sufficiently small direction stays in the set. For instance, a point on a straight line in $\mathbb R^2$ has small movements perpendicular to the line that leave it. Thus a line has empty interior in the plane. The same line, regarded as the entire ambient space in its own one-dimensional geometry, behaves differently; the ambient space must be stated.

Theorem

Open balls are open

Every open ball in $\mathbb R^n$ is an open set. Arbitrary unions of open sets are open, and finite intersections of open sets are open.

Proof

Choosing a radius that stays inside

If $x\in B(a,r)$, set $\rho=r-\|x-a\|>0$. Whenever $\|y-x\|<\rho$, the triangle inequality gives

$$\|y-a\|\le\|y-x\|+\|x-a\|<r.$$

Hence $B(x,\rho)\subseteq B(a,r)$.

If a point belongs to a union of open sets, it belongs to one of them; an interior ball from that set also lies in the union. For a finite intersection, choose an interior radius $r_i>0$ for each participating set and take their minimum. This minimum is positive and works for every set simultaneously.

Finiteness matters in the intersection statement. The intersection of $(-1/k,1/k)$ over all positive integers $k$ is $\{0\}$, which is not open in $\mathbb R$. The corresponding radii shrink to zero, so there is no positive minimum radius left.

Closure, accumulation and boundary

Definition

Closure, accumulation points and closed sets

The closure $\overline A$ consists of points $a$ for which every $B(a,r)$ meets $A$. An accumulation point of $A$ is a point for which every punctured ball meets $A$. A point $a\in A$ is isolated if some ball meets $A$ only at $a$. A set is closed if its complement is open.

The distinction between closure and accumulation is the centre of the ball. An isolated point belongs to the closure because the ball already contains that point; it is not an accumulation point. An accumulation point need not belong to the original set. For $\{1/k:k\ge1\}\subset\mathbb R$, zero is an accumulation point outside the set, while each displayed reciprocal is isolated.

Theorem

Three descriptions of closure and boundary

For every $A\subseteq\mathbb R^n$,

$$\overline A=A\cup\{\text{accumulation points of }A\},\qquad \partial A=\overline A\setminus A^\circ.$$

A set is closed exactly when it contains all its accumulation points, equivalently when $A=\overline A$. The interior, boundary and exterior form a disjoint partition of $\mathbb R^n$.

Here a boundary point means that every ball centred there meets both $A$ and its complement. This does not decide whether the boundary point itself belongs to $A$.

Proof

Why the boundary formula holds

If $a\in\partial A$, every ball meets $A$, so $a\in\overline A$. Every ball also meets the complement, so none lies entirely in $A$; therefore $a\notin A^\circ$. Conversely, if $a\in\overline A\setminus A^\circ$, each ball meets $A$ by closure, and each ball meets the complement because $a$ is not interior. Thus $a$ is a boundary point.

If a point is not on the boundary, some ball fails to meet one of the two sets. It is therefore interior to $A$ or interior to the complement. These alternatives are disjoint, giving the partition.

For the closed-set criterion, an accumulation point outside a closed set would have an entire ball in the open complement, a contradiction. Conversely, if all accumulation points lie in $A$, a point outside $A$ is not an accumulation point and has a ball disjoint from $A$. Thus the complement is open.

Worked example

An annulus with one included boundary

Let $c=(-2,1)$ and

$$A=\{p\in\mathbb R^2:1<\|p-c\|\le3\}.$$

Its interior has $1<\|p-c\|<3$; its closure has $1\le\|p-c\|\le3$. Its boundary is the union of the two circles with radii $1$ and $3$, and

$$\operatorname{ext}A=\{\|p-c\|<1\}\cup\{\|p-c\|>3\}.$$

Points on the outer circle belong to $A$ but are not interior, so $A$ is not open. Points on the inner circle are limits of points in $A$ but are absent, so $A$ is not closed. It is bounded because $\|p\|\le3+\sqrt5$. The exterior includes the central hole as well as the unbounded outside region.

Worked example

A discrete set with a missing limit point

Consider $E=\{(2/k,0):k=1,2,\ldots\}$. Every point of $E$ is isolated: the distances from a fixed term to its neighbouring terms are positive, and a sufficiently small ball excludes them. The only accumulation point is $(0,0)$. To see why there are no others, any point away from the horizontal axis has positive distance from $E$, and any point of the axis away from zero has a small neighbourhood meeting only finitely many terms.

Consequently $E^\circ=\varnothing$, $\overline E=E\cup\{(0,0)\}$, and $\partial E=\overline E$. It is bounded and neither open nor closed. Adding $(0,0)$ makes it closed without creating any interior.

Common mistake

Not open does not mean closed

Open and closed are separate conditions. The half-open annulus satisfies neither; the empty set and the whole space satisfy both. Check interior balls and missing accumulation points separately instead of treating the two labels as opposite choices.

Paths and global shape

Local properties do not determine how a set hangs together. Two disjoint open discs and one open disc are all open, but only the latter can be travelled through without leaving the set.

Definition

Path-connectedness

A set $A$ is path-connected if, for any $p,q\in A$, there is a continuous map $\gamma:[0,1]\to A$ with $\gamma(0)=p$ and $\gamma(1)=q$. A set is convex if the straight segment $(1-t)p+tq$, $0\le t\le1$, belongs to $A$ for every $p,q\in A$.

Every convex set is path-connected, since the straight segment supplies the required path. The annulus above is path-connected but not convex. Join each point radially to the circle of radius $2$ about $c$, follow an arc of that circle, and reverse the second radial segment. These pieces stay within radii strictly greater than $1$ and at most $3$. In contrast, a segment joining two opposite points can pass through the removed hole.

The two regions $x\ge1$ and $x\le-1$ form a set that is not path-connected. A path from one side to the other would have a continuous first coordinate changing sign. The one-variable intermediate value theorem forces that coordinate to be zero somewhere, outside the set. This argument shows how a familiar theorem about real functions controls the geometry of paths.

Theorem

Jordan curve theorem in the plane

A simple closed curve in $\mathbb R^2$ has a complement with exactly two path-connected components: one bounded and one unbounded. The curve is the boundary of each component.

“Simple closed” means that a continuous traversal returns to its starting point and has no other self-intersections. The theorem concerns the complement of the curve, rather than the interior of the curve as a point set. For an ellipse, every ball centred on the curve contains points off the curve, so its point-set interior is empty. Nevertheless, its complement contains a bounded region enclosed by the curve.

The simplicity condition matters. A figure-eight traversal crosses itself and can enclose two separate bounded regions. One cannot apply the theorem to that traversal and conclude that there is only one bounded complementary component. Nor does the statement extend to every curve in three-dimensional space: the ambient plane is part of its hypothesis. The theorem supplies a global conclusion from the shape of a closed curve; the ball definitions supply the local classification of its points.

From geometric definitions to proof routines

The definitions above are useful because they turn drawings into arguments with explicit witnesses. To prove that a point is interior, produce one positive radius. To prove it is not interior, explain why every radius admits a point outside the set. To prove a point is in the closure, find an admissible point in every ball, or exhibit a sequence of admissible points converging to it. A picture suggests where these points should be; the quantified statement establishes the conclusion.

For example, consider the half-plane $H=\{(x,y):2x-y<5\}$. At a point $p=(p_1,p_2)\in H$, the number $s=5-(2p_1-p_2)$ is positive. Cauchy–Schwarz gives

$$|2(x-p_1)-(y-p_2)|\le\sqrt5\,\|(x,y)-p\|.$$

Thus the radius $s/\sqrt5$ works: every point strictly closer to $p$ satisfies $2x-y<5$. The radius depends on the point, which is exactly what the interior definition allows. There is no requirement that one fixed radius work throughout the whole half-plane.

Closure also has a useful minimality property: $\overline A$ is the smallest closed set containing $A$. One way to establish that it is closed is to examine a point $p\notin\overline A$. Some ball $B(p,r)$ misses $A$. Each point sufficiently close to $p$ also has a small ball missing $A$, so the complement of $\overline A$ is open. If a closed set $F$ contains $A$, any point outside $F$ has a ball missing $F$ and therefore missing $A$. Such a point cannot belong to $\overline A$, proving $\overline A\subseteq F$.

The relationship with sequences will soon make these arguments shorter. A set $F$ in Euclidean space is closed precisely when every convergent sequence of points of $F$ has its limit in $F$. To use this test for an equation such as $x^2+y^2=7$, take a convergent sequence on the equation and pass to the limit in both coordinates. To disprove closedness, find just one sequence whose terms stay inside but whose limit is omitted.

Finally, boundedness asks for a global distance bound, rather than a local ball around each point. An infinite strip can be open and still extend arbitrarily far. A circle can be bounded and closed while having no interior. Keeping the local and global questions separate prevents a misleading picture from doing the work of several distinct definitions.

Practice

Checkpoint

Exercise 1: Find the natural domain of the function below.

$$g(x,y)=\frac{\log(25-x^2-y^2)}{\sqrt{y+2}}.$$

State separately which inequality is strict and why.

Solution · Exercise 1 solution

The logarithm requires $x^2+y^2<25$. The square root in the denominator requires $y+2>0$, because zero is not allowed there. Hence the domain is $\{(x,y):x^2+y^2<25,\ y>-2\}$. The numerator may vanish when its logarithm is zero; this causes no extra exclusion.

Checkpoint

Exercise 2: Classify a punctured disc.

For $S=\{(x,y):0<x^2+y^2<9\}$, find its interior, closure and boundary. Is it open, closed, bounded or path-connected?

Solution · Exercise 2 solution

$S^\circ=S$, $\overline S=\{x^2+y^2\le9\}$, and $\partial S=\{(0,0)\}\cup\{x^2+y^2=9\}$. It is open and bounded, but not closed. It is path-connected: join points radially to the circle of radius $1$, then along that circle. All path radii stay positive and below $3$.

Checkpoint

Exercise 3: Compare a curve with its enclosed region.

Let $C=\{(x,y):x^2/9+y^2/4=1\}$. Find $C^\circ$, $\overline C$, $\partial C$ and $\operatorname{ext}C$.

Solution · Exercise 3 solution

The ellipse itself contains no two-dimensional ball, so $C^\circ=\varnothing$. It is closed, so $\overline C=C$ and $\partial C=C$. Its exterior is $\mathbb R^2\setminus C$, containing both the bounded inside and the unbounded outside. The interior of the curve as a set is different from the region enclosed by the curve.

The useful order is to determine the domain first, then use balls to classify local points, and finally study global properties such as boundedness and paths. In the next section these same balls will express convergence and the requirement that every sufficiently close admissible input produce a close output.