2.3 Continuity and compactness

A limit describes nearby values; continuity also asks those values to agree with the value at the point itself. This small addition is what permits approximation: when a measured input is slightly wrong, a continuous model produces only a small output error. In several variables, the error must be controlled across every available direction, and at a boundary it must be interpreted through the actual domain.

Continuity is a local property, while compactness is a property of an entire set. Their combination gives global conclusions that are not apparent from a formula at one point. A continuous scalar function on a nonempty compact set has a largest and a smallest value, and one input tolerance can control its output error everywhere on the set. The purpose of this section is to explain the logic connecting those conclusions, rather than to treat them as unrelated rules.

Continuity on a specified domain

Definition

Continuity at a point

Let $F:D\subseteq\mathbb R^n\to\mathbb R^m$ and $a\in D$. The function is continuous at $a$ relative to $D$ if, for every $\varepsilon>0$, there is $\delta>0$ such that

$$x\in D,\quad\|x-a\|<\delta \quad\Longrightarrow\quad \|F(x)-F(a)\|<\varepsilon.$$

It is continuous on $D$ if this holds at every point of $D$.

If $a$ is an accumulation point of $D$, the definition is equivalent to $\lim_{x\to a,\ x\in D}F(x)=F(a)$. If $a$ is isolated, continuity holds automatically: choose a ball whose only domain point is $a$. The definition with all domain points covers both cases without requiring a punctured limit at an isolated point.

The sequential criterion gives another useful test: $F$ is continuous at $a$ exactly when $x_k\in D$ and $x_k\to a$ imply $F(x_k)\to F(a)$. Terms equal to $a$ are now allowed. The proof follows the limit criterion; a failure of continuity produces a sequence of inputs within $1/k$ of $a$ whose output errors remain above a fixed positive number.

Theorem

Algebra and composition preserve continuity

Sums, scalar multiples and products of continuous scalar functions are continuous. A quotient is continuous where its denominator is nonzero. If $F:D\to E$ is continuous at $a$ and $G:E\to\mathbb R^p$ is continuous at $F(a)$, then $G\circ F$ is continuous at $a$.

Proof

Passing the tolerance through a composition

Given an output tolerance $\varepsilon$ for $G\circ F$, continuity of $G$ provides an intermediate tolerance $\eta>0$: inputs in $E$ within $\eta$ of $F(a)$ give outputs within $\varepsilon$ of $G(F(a))$. Continuity of $F$ then provides $\delta>0$ so that domain inputs within $\delta$ of $a$ give intermediate values within $\eta$ of $F(a)$. These two implications compose.

The algebraic claims follow from the limit laws at accumulation points and directly from the definition at isolated points. For a quotient, a nonzero denominator value ensures that nearby denominator values stay away from zero, which permits the needed reciprocal bound.

Coordinate functions are continuous because a coordinate difference is no larger than the Euclidean input distance. It follows that polynomials are continuous everywhere. Rational functions are continuous on their natural domains, and familiar one-variable continuous functions can be composed with multivariable expressions. These facts justify direct substitution only at points where the operations involved are defined.

Continuous extensions and boundaries

Worked example

Filling one removable point

For $(x,y)\ne(0,0)$, define

$$f(x,y)=\frac{xy^2}{x^2+y^2}.$$

The denominator is positive there, so $f$ is continuous away from the origin. Moreover,

$$|f(x,y)|\le |x|\le\sqrt{x^2+y^2}.$$

Thus the limit at the origin is zero. Defining $f(0,0)=0$ produces a continuous function on the entire plane.

The assigned value is forced: if the extension were continuous, its value at the origin would have to equal the already determined limit. Assigning any other real number would give a defined but discontinuous function. The extension changes the domain by adding the origin; it does not retroactively make the original quotient defined there.

Worked example

A missing point that cannot be filled continuously

Consider

$$g(x,y)=\frac{x^2-2y^2}{x^2+y^2},\qquad(x,y)\ne(0,0).$$

Along the horizontal axis, $g=1$; along the vertical axis, $g=-2$. Hence there is no limit at the origin, and no single value assigned at that point can make $g$ continuous there.

The quotient is nevertheless continuous at every point of its present domain. “Continuous on the domain” and “extendable continuously to a larger domain” are separate claims. The former follows from a nonzero denominator at each domain point; the latter requires a new limit at each added point.

Worked example

Continuity at the edge of a domain

Let

$$D=\{(x,y):x^2+y^2\le9\},\qquad h(x,y)=\sqrt{9-x^2-y^2}.$$

The polynomial inside the root is continuous, and its values on $D$ lie in $[0,9]$. The square-root function is continuous on that interval, including at zero. Composition therefore proves that $h$ is continuous throughout $D$, including the circle bounding it.

At $a=(3,0)$, write $r=\|(x,y)\|$. For inputs in $D$,

$$|h(x,y)-h(a)|=\sqrt{9-r^2} \le\sqrt{6(3-r)} \le\sqrt{6\|(x,y)-a\|}.$$

The last inequality is the reverse triangle inequality. Thus $\delta=\varepsilon^2/6$ works at this boundary point. No condition is imposed on inputs outside $D$, where the real formula is not defined.

Why closed and bounded sets matter

Definition

Bounded and compact sets in Euclidean space

A set is bounded if it is contained in a ball of finite radius. A set is compact if every cover of it by open sets has a finite subcover. The Heine–Borel theorem states that, in $\mathbb R^n$, compactness is equivalent to being closed and bounded.

Both conditions do distinct work. Boundedness prevents a sequence from escaping to infinity, while closedness keeps the limits of convergent sequences inside the set. A bounded open disc lacks the second property; an entire straight line has the second property but lacks the first. Neither is compact in the ambient Euclidean space.

Theorem

Subsequences in compact sets

Every bounded sequence in $\mathbb R^n$ has a convergent subsequence. Consequently, every sequence in a compact set $K\subseteq\mathbb R^n$ has a subsequence converging to a point of $K$. Conversely, this subsequence property characterises compact sets in $\mathbb R^n$.

Proof

Separating the roles of boundedness and closedness

For the first statement, the first coordinates form a bounded real sequence, so the one-dimensional Bolzano–Weierstrass theorem supplies a subsequence on which that coordinate converges. From this subsequence, extract another on which the second coordinate converges. Continue through the finitely many coordinates. Every coordinate of the final subsequence converges, so the vector subsequence converges.

If all original terms belong to a closed set $K$, its convergent subsequence has its limit in $K$. Combining this with boundedness proves the compact-set conclusion.

Conversely, if a set with the subsequence property were unbounded, choose its $k$th term with norm greater than $k$; no subsequence could converge. If it were not closed, choose a sequence inside it converging to a missing closure point. Every subsequence has the same missing limit, contradicting the requirement that some subsequence converge to a point of the set. Thus the set is closed and bounded, hence compact by Heine–Borel.

This argument also clarifies why a sequence in an open ball still has a convergent subsequence: it is bounded. What may fail is that the subsequence limit belongs to the open ball. For terms moving towards the circumference, the limit lies on a missing boundary point.

Continuous images and attained extrema

Theorem

Compact images and extreme values

If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous, then $F(K)$ is compact. In particular, if $K$ is nonempty and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both its maximum and minimum on $K$.

Proof

Following a subsequence from inputs to outputs

Take any sequence $y_k\in F(K)$ and choose $x_k\in K$ with $F(x_k)=y_k$. Compactness supplies a subsequence $x_{k_j}\to a\in K$. Continuity gives $y_{k_j}\to F(a)\in F(K)$. The subsequence characterisation therefore makes $F(K)$ compact.

For the scalar conclusion, the image is a nonempty closed bounded subset of $\mathbb R$. It has a finite supremum $M$. Choose image values greater than $M-1/k$; these converge to $M$, and closedness places $M$ in the image. Hence some input attains the maximum. The infimum is handled in the same way.

The theorem proves existence, not the locations of the optimisers. It does not assert that every continuous function has an extremum on any chosen domain. Each hypothesis has to be checked for the actual set and function.

Worked example

Finding extrema once existence is established

On the ellipse-filled region

$$K=\{(x,y):x^2+4y^2\le16\},$$

find the extreme values of $f(x,y)=x+2y$. The set is closed and bounded, and $f$ is continuous, so both extremes are attained. Put $u=x$ and $v=2y$. Then $u^2+v^2\le16$ and

$$|u+v|\le\sqrt2\sqrt{u^2+v^2}\le4\sqrt2.$$

The upper bound is attained when $u=v=2\sqrt2$, giving $(x,y)=(2\sqrt2,\sqrt2)$. The lower bound occurs at their negatives. Thus the maximum is $4\sqrt2$ and the minimum is $-4\sqrt2$.

Cauchy–Schwarz supplies the values and equality conditions. Compactness supplies the underlying existence statement; checking the equality points confirms that the bounds are attained in the stated domain.

Removing hypotheses changes the conclusion. On the open unit disc, the continuous function $x^2+y^2$ has supremum one but never reaches it. On the closed unbounded plane, the continuous function $x$ is unbounded. On a compact disc, a function defined as $1/(x^2+y^2)$ away from the origin and assigned any finite value at the origin is unbounded because continuity there fails. These examples isolate three different reasons the extreme-value argument can break.

Uniform continuity

Definition

One radius for every centre

A function $F:D\to\mathbb R^m$ is uniformly continuous if, for every $\varepsilon>0$, there exists $\delta>0$ such that, for every $x,y\in D$,

$$\|x-y\|<\delta\quad\Longrightarrow\quad\|F(x)-F(y)\|<\varepsilon.$$

Ordinary continuity chooses a radius after fixing a centre. Uniform continuity chooses a radius before either input point is specified. The tolerance is still arbitrary, but the resulting radius must work throughout the domain. A bound $\|F(x)-F(y)\|\le C\|x-y\|$ with one constant $C$ is an immediate sufficient condition: use $\delta=\varepsilon/C$ when $C>0$, while a constant function permits any positive radius.

Theorem

Continuity on a compact set is uniform

Every continuous map from a compact subset of $\mathbb R^n$ into $\mathbb R^m$ is uniformly continuous.

Proof

Two inputs become close to the same limit point

If uniform continuity failed, there would be $\varepsilon_0>0$ and pairs $x_k,y_k\in K$ satisfying

$$\|x_k-y_k\|<1/k,\qquad \|F(x_k)-F(y_k)\|\ge\varepsilon_0.$$

Choose a subsequence $x_{k_j}\to a\in K$ using compactness. The triangle inequality forces $y_{k_j}\to a$ as well. Continuity sends both output subsequences to $F(a)$, so their difference tends to zero, contradicting its fixed positive lower bound.

Compactness is sufficient, rather than necessary, for this conclusion. The identity map on all of $\mathbb R^n$ is uniformly continuous because it preserves distances. On the other hand, $t\mapsto1/t$ is continuous on $(0,1)$ but not uniformly continuous: take $x_k=1/k$ and $y_k=1/(k+1)$ for $k\ge2$. Their input difference tends to zero, but their output difference equals one.

The distinction is useful whenever one error budget must work over a whole collection of possible inputs. On a compact set, a continuous model gives such a budget even if an explicit best radius is hard to calculate. At a single point, continuity gives a local budget; the uniform statement explains when these local controls can be made global.

Recognising continuity through inverse images

There is a useful set-based version of continuity. A map on a domain is continuous exactly when the inverse image of every open set in the output space is open relative to that domain. “Open relative to the domain” means the intersection of the domain with an open subset of the ambient input space. This wording is needed when the domain itself contains boundary points.

To see the forward direction, take an input whose output belongs to an open output set. An output ball around that value stays in the open set. Continuity supplies an input ball whose domain points are mapped into that output ball. Thus every point of the inverse image has a relative interior neighbourhood. Conversely, apply the inverse-image condition to any prescribed output ball around the value at a fixed input. A relative input neighbourhood inside its inverse image is precisely the radius required by continuity.

This description explains why equations and inequalities built from continuous functions produce familiar domains. The set where a continuous real function is positive is open; the set where it equals a specified constant is closed. These statements concern inverse images. A continuous map does not generally send every open set to an open set: a constant map sends the entire input space to a singleton. Keeping the direction of the image operation clear prevents a useful criterion from being applied backwards.

Exercises and solutions

Checkpoint

Exercise 1: Find the unique continuous extension.

For $(x,y)\ne(0,0)$, let $q(x,y)=x^2y^2/(x^2+y^2)$. What value at the origin makes the extension continuous?

Solution · Exercise 1 solution

Since $2|xy|\le x^2+y^2$,

$$0\le q(x,y)\le\frac{x^2+y^2}{4}\longrightarrow0.$$

The unique required value is zero. The quotient is already continuous at every other point.

Checkpoint

Exercise 2: Separate existence from calculation.

Does $f(x,y)=3x-y$ attain extrema on $x^2+y^2\le4$? Find their values and points of attainment.

Solution · Exercise 2 solution

The closed disc is compact and nonempty, so existence follows. Cauchy–Schwarz gives $|3x-y|\le2\sqrt{10}$. Equality occurs at $(x,y)=\pm(6,-2)/\sqrt{10}$. The positive sign gives the maximum and the negative sign the minimum.

Checkpoint

Exercise 3: Establish a uniform radius without compactness.

Show that $F(x)=\|x\|$ is uniformly continuous on all of $\mathbb R^n$.

Solution · Exercise 3 solution

The reverse triangle inequality gives $|\|x\|-\|y\||\le\|x-y\|$. Thus $\delta=\varepsilon$ works for every pair of inputs, even though the domain is not compact.

Continuity connects nearby values to a point value. Compactness supplies convergent subsequences whose limits stay in the domain. Together they explain continuous images, attained extrema and uniform continuity. When applying any of these statements, identify the actual domain first; the role of each hypothesis then becomes visible in the proof.