Finite-dimensional compactness and attained extrema

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

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

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Worked example: the sphere needed for the real spectral theorem. The sphere $S^{n-1}=\{x\in\mathbb R^n:\|x\|=1\}$ is nonempty because it contains $e_1$, and it is bounded. The reverse triangle inequality $|\|x\|-\|y\||\le\|x-y\|$ proves norm continuity, so $S^{n-1}$ is closed as the inverse image of the closed set $\{1\}$. Heine–Borel makes it compact. For a real matrix $A$, the polynomial $x\mapsto x^TAx$ is continuous, so its restriction to the sphere attains both extrema. Symmetry of $A$ is not needed for this existence claim; it is needed in the subsequent spectral argument and the gradient formula $\nabla(x^TAx)=2Ax$.

Exercises

P01. Explain why the compactness assertion fails for the open unit ball even though that ball is bounded.

P02. Which step of the cube proof fails if $K$ is not closed? Give an example.

P03. For $A=\operatorname{diag}(2,5)$ on the unit circle, find the extreme values of $x^TAx$ and every point where equality holds. Distinguish the existence theorem from the calculation.

Full solutions

P01. The open unit ball is not closed: $(1-1/j)e_1$ lies in it for $j\ge2$ and converges to $e_1$ outside it. A compact set would be closed by the proved necessity, so the ball is not compact. Boundedness alone is insufficient.

P02. The limit of $x_j\in K$ may be outside $K$, at a point $p$, so no member of the cover need contain $p$. Explicitly take $K=(0,1)$, the cover $O_m=(1/m,2)$ for integers $m\ge2$, and the nested closed intervals $Q_j=[0,2^{-j}]$ for integers $j\ge0$. These open intervals cover $K$, but any finite collection has a positive smallest left endpoint and therefore fails to cover $K\cap Q_j$. The intervals $Q_j$ have common point zero, which is outside $K$; consequently no member of the cover contains $0$. This is exactly where the closed-set proof can fail.

P03. Writing $x=(u,v)^T$ with $u^2+v^2=1$ gives $x^TAx=2u^2+5v^2=2+3v^2$. The minimum is two, attained exactly at $(\pm1,0)^T$, and the maximum is five, attained exactly at $(0,\pm1)^T$. Compactness and continuity guarantee that extrema exist; the displayed identity determines their values and all equality cases.