CommeUnJeu · L2 MP
Compactness, connectedness, finite dimension
The chapter Limits and continuity in a normed space equipped us with continuity between normed spaces. Two theorems first met in high school were stated only on a segment \([a,b]\) of the real line: a continuous function on a segment is bounded and attains its bounds, and the intermediate value theorem. Why a segment? This chapter isolates the two structural hypotheses --- one for each theorem --- that make them true in any normed space.
The first hypothesis is compactness: a part on which every sequence has a sub-sequence converging inside it. It is the abstract form of the Bolzano--Weierstrass property, and it carries the extreme value theorem and Heine's theorem. The second is arc-connectedness: a part « of one piece », any two of whose points are joined by a continuous path. It carries the intermediate value theorem in full generality. Section~1 builds compactness and its first properties, Section~2 the two theorems it carries, Section~3 arc-connectedness and the generalised intermediate value theorem.
Section~4 closes the chapter with finite-dimensional normed spaces, where the whole picture becomes the simplest possible: all norms are equivalent, the compact parts are exactly the closed bounded sets, and every linear or multilinear map is automatically continuous --- as is every scalar-valued polynomial map. The continuity criteria of Limits and continuity in a normed space are then satisfied for free.
Standing notation. Throughout, \(\mathbb{K}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\), and \((E,\norme{\cdot})\), \((F,\norme{\cdot})\) denote normed \(\mathbb{K}\)-vector spaces. A part of \(E\) means a subset of \(E\). Open and closed balls, bounded parts, open sets, closed sets and the closure \(\overline{A}\) are those of Normed vector spaces and Topology of a normed space; convergent sequences, sub-sequences (written \((u_{\varphi(n)})\) with \(\varphi\) strictly increasing) and adherence values are those of Normed vector spaces, where the Bolzano--Weierstrass theorem was proved for bounded real sequences; limits and continuity of maps are those of Limits and continuity in a normed space.
The first hypothesis is compactness: a part on which every sequence has a sub-sequence converging inside it. It is the abstract form of the Bolzano--Weierstrass property, and it carries the extreme value theorem and Heine's theorem. The second is arc-connectedness: a part « of one piece », any two of whose points are joined by a continuous path. It carries the intermediate value theorem in full generality. Section~1 builds compactness and its first properties, Section~2 the two theorems it carries, Section~3 arc-connectedness and the generalised intermediate value theorem.
Section~4 closes the chapter with finite-dimensional normed spaces, where the whole picture becomes the simplest possible: all norms are equivalent, the compact parts are exactly the closed bounded sets, and every linear or multilinear map is automatically continuous --- as is every scalar-valued polynomial map. The continuity criteria of Limits and continuity in a normed space are then satisfied for free.
Standing notation. Throughout, \(\mathbb{K}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\), and \((E,\norme{\cdot})\), \((F,\norme{\cdot})\) denote normed \(\mathbb{K}\)-vector spaces. A part of \(E\) means a subset of \(E\). Open and closed balls, bounded parts, open sets, closed sets and the closure \(\overline{A}\) are those of Normed vector spaces and Topology of a normed space; convergent sequences, sub-sequences (written \((u_{\varphi(n)})\) with \(\varphi\) strictly increasing) and adherence values are those of Normed vector spaces, where the Bolzano--Weierstrass theorem was proved for bounded real sequences; limits and continuity of maps are those of Limits and continuity in a normed space.
I
Compact parts
I.1
Definition of a compact part
In Normed vector spaces, the Bolzano--Weierstrass theorem said that a bounded real sequence has a convergent sub-sequence, and a segment \([a,b]\) inherited the property that every sequence of \([a,b]\) has a sub-sequence converging inside \([a,b]\). A compact part is defined to be exactly a part with that property --- it is the right abstraction to carry the great theorems of the next section.
Definition — Compact part
A part \(A\) of a normed space \((E,\norme{\cdot})\) is said to be compact if every sequence of elements of \(A\) admits a sub-sequence converging to a limit belonging to \(A\).
The Borel--Lebesgue property
A second definition of compactness exists, through open covers --- the Borel--Lebesgue property. The Borel--Lebesgue property is out of the program. In this course, « compact » means only the sequential property of the Definition above: every sequence of \(A\) has a sub-sequence converging in \(A\).
The picture is the same as for Bolzano--Weierstrass: the sequence may wander inside \(A\), but a sub-sequence is forced to cluster, and the cluster point cannot escape \(A\). 
A sub-sequence of a sequence of the compact \(A\) converges to a point \(\ell\) of \(A\).
Example — A finite part and a segment
Every finite part \(A\) is compact: a sequence valued in a finite set takes at least one value infinitely often (drawer principle), and the corresponding constant sub-sequence converges to that value, which lies in \(A\). A segment \([a,b]\) of \(\mathbb{R}\) is compact: a sequence of \([a,b]\) is bounded, so by Bolzano--Weierstrass it has a convergent sub-sequence, and since \([a,b]\) is closed the limit stays in \([a,b]\). Example — A closed bounded set that is not compact
Equip \(\mathbb{K}[X]\) with the coefficient sup norm \(\norme{P}_\infty = \max_k |a_k|\) (the largest absolute value among the coefficients of \(P\)). The unit sphere \(S = \{P : \norme{P}_\infty = 1\}\) is closed and bounded. Yet it is not compact: the monomials \(X^n\) all satisfy \(\norme{X^n}_\infty = 1\), so \((X^n)_n\) is a sequence of \(S\), and \(\norme{X^n - X^m}_\infty = 1\) for all \(n \ne m\). No sub-sequence can converge --- two distinct terms always stay at distance \(1\). Closed and bounded is therefore not enough for compactness; Section~4 shows it becomes enough in finite dimension. Method — Show that a part is or is not compact
- To show \(A\) compact from the definition: take an arbitrary sequence of \(A\), extract a sub-sequence, and prove that this sub-sequence converges to a limit lying in \(A\). Section~4 gives a far shorter route once the space is finite-dimensional.
- To show \(A\) not compact: exhibit one sequence of \(A\) with no convergent sub-sequence. The cleanest witness is a sequence \((u_n)\) with \(\norme{u_n - u_p} \ge \alpha > 0\) for all \(n \ne p\) --- no sub-sequence can converge, since any two of its terms stay at distance at least \(\alpha\).
Skills to practice
- Identifying a compact part
I.2
First properties
Compactness immediately forces two shape constraints --- a compact part is closed and it is bounded --- and it is inherited by closed pieces and by finite products. These properties are the everyday toolkit for recognising compacts.
Theorem — A compact part is closed and bounded
Every compact part of a normed space is closed and bounded.
Let \(A\) be a compact part of \(E\).
- \(A\) is closed. Let \((u_n)\) be a sequence of \(A\) converging to a limit \(\ell \in E\). By compactness, \((u_n)\) has a sub-sequence \((u_{\varphi(n)})\) converging to some \(\ell' \in A\). But \((u_{\varphi(n)})\), being a sub-sequence of the convergent sequence \((u_n)\), also converges to \(\ell\). By uniqueness of the limit, \(\ell = \ell' \in A\). So \(A\) contains the limit of every convergent sequence of its points: \(A\) is closed.
- \(A\) is bounded. Suppose not. Then for each \(n \in \mathbb{N}\) there is \(u_n \in A\) with \(\norme{u_n} > n\). By compactness, \((u_n)\) has a convergent sub-sequence \((u_{\varphi(n)})\); a convergent sequence is bounded, so \(\norme{u_{\varphi(n)}} \le M\) for some \(M\) and all \(n\). But \(\norme{u_{\varphi(n)}} > \varphi(n) \ge n\) (a strictly increasing \(\varphi \colon \mathbb{N} \to \mathbb{N}\) satisfies \(\varphi(n) \ge n\)), which exceeds \(M\) for \(n\) large --- a contradiction. Hence \(A\) is bounded.
Proposition — A closed subset of a compact is compact
Let \(A\) be a compact part and \(X \subset A\). Then \(X\) is compact if and only if \(X\) is closed in \(E\) --- equivalently, closed in \(A\), since \(A\) is itself closed.
If \(X\) is compact, it is closed by the Theorem above. Conversely, assume \(X\) is closed in \(E\). Let \((u_n)\) be a sequence of \(X\). Since \(X \subset A\) and \(A\) is compact, \((u_n)\) has a sub-sequence \((u_{\varphi(n)})\) converging to a limit \(\ell \in A\). Each term \(u_{\varphi(n)}\) lies in \(X\), and \(X\) is closed, so \(\ell \in X\). Thus every sequence of \(X\) has a sub-sequence converging in \(X\): \(X\) is compact.
Proposition — Finite product of compacts
Let \(A\) be a compact part of \(E\) and \(B\) a compact part of \(F\). Then \(A \times B\) is a compact part of \(E \times F\) for the product norm \(\norme{(x,y)} = \max(\norme{x},\norme{y})\). More generally, a finite product \(A_1 \times \dots \times A_p\) of compact parts is compact for the product norm \(\norme{(x_1,\dots,x_p)} = \max_i \norme{x_i}\).
For the product norm, \((z_n) = (x_n,y_n)\) converges to \((x,y)\) if and only if \(x_n \to x\) and \(y_n \to y\) --- convergence is coordinatewise.
- Two factors. Let \((z_n) = (x_n,y_n)\) be a sequence of \(A \times B\). The sequence \((x_n)\) is valued in the compact \(A\), so it has a sub-sequence \((x_{\varphi(n)})\) converging to some \(x \in A\). The sequence \((y_{\varphi(n)})\) is valued in the compact \(B\), so it has a sub-sequence \((y_{\varphi(\psi(n))})\) converging to some \(y \in B\). Along the strictly increasing map \(\varphi \circ \psi\), the sequence \(x_{\varphi(\psi(n))}\) still converges to \(x\) (sub-sequence of a convergent sequence) and \(y_{\varphi(\psi(n))}\) converges to \(y\). By coordinatewise convergence, \(z_{\varphi(\psi(n))} \to (x,y) \in A \times B\). So \(A \times B\) is compact.
- Induction on \(p\). The case \(p = 1\) is trivial and \(p = 2\) is the line above. The \((p+1)\)-fold product norm is consistent with the nesting: \(\max_{1 \le i \le p+1} \norme{x_i} = \max\bigl(\max_{1 \le i \le p} \norme{x_i},\, \norme{x_{p+1}}\bigr)\), so \(A_1 \times \dots \times A_{p+1} = (A_1 \times \dots \times A_p) \times A_{p+1}\) is, with its product norm, the product of the two parts \(A_1 \times \dots \times A_p\) and \(A_{p+1}\). If the first is compact (induction hypothesis), this is a product of two compacts, hence compact.
Example — A closed bounded box
The rectangle \([a,b] \times [c,d]\) is a compact part of \(\mathbb{R}^2\) for the product norm, as the product of the two compact segments \([a,b]\) and \([c,d]\). More generally a closed bounded box \(\prod_{i=1}^{n} [a_i,b_i]\) is a compact part of \(\mathbb{R}^n\) for the product norm, the product of \(n\) compact segments. Section~4 will show that in finite dimension the norm does not matter. Skills to practice
- Applying the closed-and-bounded properties
I.3
Sequences valued in a compact
Inside a compact, having an adherence value is automatic --- compactness is precisely that. What remains in order to decide convergence of a sequence is the uniqueness of its adherence value.
Proposition — Convergence by a single adherence value
A sequence valued in a compact part \(A\) converges if and only if it has exactly one adherence value, and the limit is then that value.
Let \((u_n)\) be valued in the compact \(A\); by compactness it has at least one adherence value in \(A\).
- (\(\Rightarrow\)) If \((u_n)\) converges to \(\ell\), then every sub-sequence converges to \(\ell\), so the only adherence value is \(\ell\): exactly one.
- (\(\Leftarrow\)) We argue by contraposition: assume \((u_n)\) does not converge and produce a second adherence value. Let \(\alpha \in A\) be an adherence value. Since \((u_n)\) does not converge to \(\alpha\), there is \(\varepsilon > 0\) such that \(\norme{u_n - \alpha} \ge \varepsilon\) for infinitely many \(n\); let \((u_{\psi(n)})\) be the sub-sequence of those terms. It is valued in the compact \(A\), so it has a sub-sequence \((u_{\psi(\theta(n))})\) converging to some \(\beta \in A\). Each of its terms satisfies \(\norme{u_{\psi(\theta(n))} - \alpha} \ge \varepsilon\); passing to the limit gives \(\norme{\beta - \alpha} \ge \varepsilon\), so \(\beta \ne \alpha\): a second adherence value. Hence a sequence of \(A\) with a single adherence value must converge, and by (\(\Rightarrow\)) its limit is that value.
Example — One adherence value against two
Both sequences below are valued in the compact \([-2,2]\) of \(\mathbb{R}\). The sequence \(u_n = \dfrac{(-1)^n}{n+1}\) has the single adherence value \(0\), and it converges to \(0\). The sequence \(v_n = (-1)^n\!\left(1 + \dfrac{1}{n+1}\right)\) has the two adherence values \(-1\) and \(1\) (along even and odd indices), and it diverges. The criterion is read off directly from the adherence values. Method — Prove that a sequence of a compact converges
For a sequence \((u_n)\) known to be valued in a compact \(A\): - exhibit one adherence value \(\alpha \in A\) --- it exists automatically by compactness, or is guessed from the formula of \(u_n\);
- show that every adherence value equals \(\alpha\), typically by taking an arbitrary convergent sub-sequence and identifying its limit;
- conclude by the Proposition: a single adherence value forces convergence to \(\alpha\).
Skills to practice
- Proving a sequence of a compact converges
II
Continuous maps on a compact
II.1
Image of a compact and the extreme value theorem
Continuity transports compactness: the continuous image of a compact is a compact. When the arrival space is \(\mathbb{R}\), this single fact produces the extreme value theorem --- the general form of the high-school result « a continuous function on a segment is bounded and attains its bounds ».
Theorem — Image of a compact under a continuous map
Let \(A\) be a compact part of \(E\) and \(f \colon A \to F\) a continuous map. Then \(f(A)\) is a compact part of \(F\).
Let \((v_n)\) be a sequence of \(f(A)\). Each \(v_n\) is of the form \(v_n = f(u_n)\) with \(u_n \in A\). The sequence \((u_n)\) is valued in the compact \(A\), so it has a sub-sequence \((u_{\varphi(n)})\) converging to a limit \(\ell \in A\). By the sequential characterisation of continuity (recalled from Limits and continuity in a normed space), \(f(u_{\varphi(n)}) \to f(\ell)\). So the sub-sequence \((v_{\varphi(n)})\) converges to \(f(\ell) \in f(A)\). Every sequence of \(f(A)\) thus has a sub-sequence converging in \(f(A)\): \(f(A)\) is compact.
Theorem — Extreme value theorem
Let \(A\) be a non-empty compact part and \(f \colon A \to \mathbb{R}\) a continuous map. Then \(f\) is bounded and attains its bounds: there exist \(u,v \in A\) with \(f(u) = \inf_{A} f\) and \(f(v) = \sup_{A} f\).
By the previous Theorem, \(f(A)\) is a compact part of \(\mathbb{R}\), hence closed and bounded. Bounded and non-empty, \(f(A)\) has a finite infimum \(m = \inf f(A)\) and a finite supremum \(M = \sup f(A)\). By the characterisation of the infimum and the supremum as limits of points of the set, \(m\) and \(M\) are each the limit of a sequence of points of \(f(A)\). Since \(f(A)\) is closed, those limits stay in \(f(A)\): \(m \in f(A)\) and \(M \in f(A)\). So there exist \(u,v \in A\) with \(f(u) = m\) and \(f(v) = M\): the bounds are attained.
Example — The segment recovered, and a norm on a compact
A continuous function \(f \colon [a,b] \to \mathbb{R}\) attains a maximum and a minimum: the high-school theorem is the extreme value theorem applied to the compact segment \([a,b]\). Likewise, a norm \(\norme{\cdot}\), being a continuous map, attains its maximum on any non-empty compact \(A\) --- a fact used repeatedly in Section~4. Method — Use compactness for an existence or a bound
To prove that a continuous real quantity is bounded, or that an extremum is reached: - write the quantity as \(f(x)\) for a continuous real-valued map \(f\);
- exhibit a non-empty compact part \(A\) as the domain --- in finite dimension, a closed bounded part (Section~4);
- invoke the extreme value theorem: \(f\) is bounded on \(A\) and attains \(\inf_A f\) and \(\sup_A f\).
Skills to practice
- Applying the extreme value theorem
II.2
Heine's theorem
On a compact, continuity is automatically uniform: the modulus \(\alpha\) that answers a challenge \(\varepsilon\) can be chosen once and for all, independently of the point. This is Heine's theorem, the hypothesis silently behind every Riemann-sum and integral approximation.
Theorem — Heine's theorem
Let \(A\) be a compact part of \(E\). Every continuous map \(f \colon A \to F\) is uniformly continuous on \(A\).
Argue by contradiction: suppose \(f\) is not uniformly continuous on \(A\).
- Negating uniform continuity gives an \(\varepsilon > 0\) such that, for every \(\delta > 0\), two points of \(A\) break the implication. Taking \(\delta = \frac{1}{n+1}\) yields points \(x_n, y_n \in A\) with \(\norme{x_n - y_n} \le \frac{1}{n+1}\) and \(\norme{f(x_n) - f(y_n)} \ge \varepsilon\).
- The sequence \((x_n)\) is valued in the compact \(A\), so it has a sub-sequence \((x_{\varphi(n)})\) converging to some \(\ell \in A\). Then $$ \norme{y_{\varphi(n)} - \ell} \le \norme{y_{\varphi(n)} - x_{\varphi(n)}} + \norme{x_{\varphi(n)} - \ell} \le \tfrac{1}{\varphi(n)+1} + \norme{x_{\varphi(n)} - \ell} \xrightarrow[n \to +\infty]{} 0, $$ so \((y_{\varphi(n)})\) also converges to \(\ell\).
- By continuity of \(f\) at \(\ell\), \(f(x_{\varphi(n)}) \to f(\ell)\) and \(f(y_{\varphi(n)}) \to f(\ell)\), hence \(\norme{f(x_{\varphi(n)}) - f(y_{\varphi(n)})} \to 0\). This contradicts \(\norme{f(x_{\varphi(n)}) - f(y_{\varphi(n)})} \ge \varepsilon\).
Example — Why the compactness hypothesis is genuine
The map \(x \mapsto x^2\) is uniformly continuous on the compact \([-1,1]\) by Heine. On all of \(\mathbb{R}\) --- which is not compact --- it is still continuous, but no longer uniformly continuous: take \(x_n = n\) and \(y_n = n + \frac{1}{n}\), so \(|x_n - y_n| = \frac{1}{n} \to 0\), while \(|x_n^2 - y_n^2| = 2 + \frac{1}{n^2} \ge 2\). No single \(\delta\) can work for \(\varepsilon = 2\). Compactness of the domain is exactly what removes this obstruction. Method — Recognise when Heine is needed
Uniform continuity --- not mere continuity --- is the hypothesis required to control \(\norme{f(x) - f(y)}\) by a single \(\delta\) over the whole domain: it is what makes Riemann sums converge to an integral, and what bounds an approximation uniformly. When such a uniform control is needed and the domain is a compact, do not prove uniform continuity by hand: state that \(f\) is continuous on a compact, and invoke Heine. Skills to practice
- Applying Heine's theorem
III
Arc-connectedness
III.1
Paths and arc-connected components
The second structural notion of the chapter answers a different question: when is a part « of one piece »? The answer chosen here is concrete --- any two of its points can be joined by a continuous path that never leaves the part.
Definition — Path and arc-connected part
Let \(A\) be a part of \(E\) and \(a,b \in A\). A path joining \(a\) to \(b\) in \(A\) is a continuous map \(\gamma \colon [0,1] \to E\) with \(\gamma(0) = a\), \(\gamma(1) = b\) and \(\gamma([0,1]) \subset A\). The part \(A\) is arc-connected if any two of its points are joined by a path in \(A\). By convention the empty part, having no two points, is arc-connected. Example — Convex and star-shaped parts
A convex part \(A\) is arc-connected: for \(a,b \in A\) the rectilinear path \(\gamma(t) = (1-t)a + tb\) is continuous, stays in \(A\) by convexity, and joins \(a\) to \(b\). More generally a star-shaped part --- one containing a point \(c\) such that the segment \([c,x]\) lies in \(A\) for every \(x \in A\) --- is arc-connected: any two points \(x,y\) are joined through \(c\), going along \([x,c]\) then \([c,y]\). Proposition — « Joined by a path » is an equivalence relation
On a part \(A\), the relation « \(x\) and \(y\) are joined by a path in \(A\) » is reflexive, symmetric and transitive. - Reflexive. The constant map \(\gamma(t) = x\) is a path joining \(x\) to itself in \(A\).
- Symmetric. If \(\gamma\) joins \(x\) to \(y\) in \(A\), then \(t \mapsto \gamma(1-t)\) is continuous, has image \(\gamma([0,1]) \subset A\), and joins \(y\) to \(x\).
- Transitive. Let \(\gamma_1\) join \(x\) to \(y\) and \(\gamma_2\) join \(y\) to \(z\), both in \(A\). Define \(\gamma\) on \([0,1]\) by \(\gamma(t) = \gamma_1(2t)\) for \(t \le \frac{1}{2}\) and \(\gamma(t) = \gamma_2(2t-1)\) for \(t \ge \frac{1}{2}\). The two formulas agree at \(t = \frac{1}{2}\) (both give \(y\)), so \(\gamma\) is continuous; its image is \(\gamma_1([0,1]) \cup \gamma_2([0,1]) \subset A\); and it joins \(x\) to \(z\).
Definition — Arc-connected components
The equivalence classes of the relation « joined by a path in \(A\) » are called the arc-connected components of \(A\).
The components are the « pieces » of \(A\): each is arc-connected, two different ones cannot be joined by any path. A non-empty part \(A\) is therefore arc-connected if and only if it has a single component --- a single piece. The empty part has no component, which is why this equivalence is stated for a non-empty part. 
Method — Show that a part is arc-connected
To prove a non-empty part \(A\) arc-connected: - recognise it as convex or star-shaped --- the rectilinear or two-segment path is then immediate;
- or fix one base point \(c \in A\) and join every \(x \in A\) to \(c\) by an explicit path, using transitivity to connect any two points through \(c\).
Skills to practice
- Establishing arc-connectedness
III.2
Arc-connected parts of the real line
On the real line, arc-connectedness collapses onto a familiar notion: the arc-connected parts of \(\mathbb{R}\) are exactly its intervals. This is the bridge that turns the next section's image theorem into the intermediate value theorem.
Proposition — The arc-connected parts of \(\mathbb{R}\) are the intervals
A part of \(\mathbb{R}\) is arc-connected if and only if it is an interval --- the empty set and singletons being counted as intervals, consistently with the vacuous-arc-connectedness convention.
An interval of \(\mathbb{R}\) is convex, hence arc-connected by the Example of \S3.1. Conversely, let \(A \subset \mathbb{R}\) be arc-connected; we show \(A\) is convex, which for a part of \(\mathbb{R}\) means being an interval. If \(A\) has at most one point it is the empty set or a singleton --- both intervals --- and there is nothing to prove; assume \(A\) has at least two distinct points. Take \(a,b \in A\) with \(a < b\) and let \(c \in [a,b]\). Since \(A\) is arc-connected, there is a path \(\gamma \colon [0,1] \to \mathbb{R}\), continuous, with \(\gamma(0) = a\), \(\gamma(1) = b\) and \(\gamma([0,1]) \subset A\). The number \(c\) lies between \(\gamma(0)\) and \(\gamma(1)\), so by the (real) intermediate value theorem there is \(t \in [0,1]\) with \(\gamma(t) = c\). Then \(c = \gamma(t) \in A\). So \(A\) contains every point between two of its points: \(A\) is convex, i.e. an interval.
Example — A punctured line is not arc-connected
The part \(\mathbb{R} \setminus \{a\}\) is not an interval, so by the Proposition it is not arc-connected. Its two arc-connected components are the half-lines \(\,]-\infty,a[\,\) and \(\,]a,+\infty[\,\): each is an interval hence arc-connected, and no path can cross from one to the other without passing through \(a\), which has been removed. Skills to practice
- Identifying the arc-connected parts of the line
III.3
Continuous image and the generalized intermediate value theorem
Just as continuity transports compactness, it transports arc-connectedness. When the arrival space is \(\mathbb{R}\), this produces the intermediate value theorem in full generality --- valid on any arc-connected domain, not only on a segment.
Theorem — Image of an arc-connected part
Let \(A\) be an arc-connected part of \(E\) and \(f \colon A \to F\) a continuous map. Then \(f(A)\) is an arc-connected part of \(F\).
Let \(f(a)\) and \(f(b)\) be two points of \(f(A)\), with \(a,b \in A\). Since \(A\) is arc-connected, there is a path \(\gamma\) joining \(a\) to \(b\) in \(A\). The composite \(f \circ \gamma \colon [0,1] \to F\) is continuous (composition of continuous maps), satisfies \((f\circ\gamma)(0) = f(a)\) and \((f\circ\gamma)(1) = f(b)\), and has image \(f(\gamma([0,1])) \subset f(A)\). So \(f \circ \gamma\) is a path joining \(f(a)\) to \(f(b)\) in \(f(A)\): the part \(f(A)\) is arc-connected.
Theorem — Generalized intermediate value theorem
Let \(A\) be an arc-connected part of \(E\) and \(f \colon A \to \mathbb{R}\) a continuous map. Then \(f(A)\) is an interval of \(\mathbb{R}\). In particular, \(f\) takes every value between any two of its values.
By the previous Theorem, \(f(A)\) is an arc-connected part of \(\mathbb{R}\). By the Proposition of \S3.2, an arc-connected part of \(\mathbb{R}\) is an interval. An interval contains every real number lying between two of its elements, which is exactly the statement that \(f\) takes every value between any two of its values.
Example — The unit circle is arc-connected
The unit circle of \(\mathbb{R}^2\) is arc-connected. Indeed it is the image of the segment \([0,1]\) --- convex, hence arc-connected --- under the continuous map \(t \mapsto (\cos 2\pi t, \sin 2\pi t)\). The image theorem applies at once, with no need to study \(\mathbb{R}^2 \setminus \{0\}\). Method — Use the generalized IVT to locate a zero
To prove that a continuous real map \(f\) on an arc-connected domain \(A\) vanishes: - exhibit two points \(x,y \in A\) at which \(f\) has opposite signs --- i.e. \(f(x)\,f(y) \le 0\);
- invoke the generalized intermediate value theorem: \(0\) lies between \(f(x)\) and \(f(y)\), so it is a value of \(f\) --- there is a point of \(A\), along a path joining \(x\) to \(y\), where \(f\) vanishes.
Skills to practice
- Applying the generalized intermediate value theorem
IV
Finite-dimensional normed spaces
IV.1
Equivalence of norms
Recall from Normed vector spaces that two norms \(N_1\) and \(N_2\) on a space are equivalent when each is bounded by a constant times the other: \(\alpha N_1 \le N_2 \le \beta N_1\) for some constants \(\alpha,\beta > 0\). Equivalent norms define the same convergent sequences, the same open and closed sets, the same continuous maps. The headline of this section is that in finite dimension the choice of norm never matters --- all norms are equivalent.
Theorem — In finite dimension all norms are equivalent
On a finite-dimensional \(\mathbb{K}\)-vector space, any two norms are equivalent.
The official program marks this proof « démonstration non exigible ». It is given here because it is short and rests on compactness --- the tool this chapter has just built.
If \(\dim E = 0\) the only norm is the zero map, and the statement is trivial; assume \(\dim E = n \ge 1\) and fix a basis \((e_1,\dots,e_n)\). For \(x = \sum_{k=1}^{n} x_k e_k\), set \(N_\infty(x) = \max_{1 \le k \le n} |x_k|\) --- a norm on \(E\). It suffices to prove every norm \(N\) equivalent to \(N_\infty\); transitivity then makes any two norms equivalent.
If \(\dim E = 0\) the only norm is the zero map, and the statement is trivial; assume \(\dim E = n \ge 1\) and fix a basis \((e_1,\dots,e_n)\). For \(x = \sum_{k=1}^{n} x_k e_k\), set \(N_\infty(x) = \max_{1 \le k \le n} |x_k|\) --- a norm on \(E\). It suffices to prove every norm \(N\) equivalent to \(N_\infty\); transitivity then makes any two norms equivalent.
- (1) Upper bound and continuity. For every \(x\), $$ N(x) = N\Bigl(\sum_k x_k e_k\Bigr) \le \sum_k |x_k|\,N(e_k) \le \Bigl(\sum_k N(e_k)\Bigr) N_\infty(x) = M\,N_\infty(x), $$ with \(M = \sum_k N(e_k) > 0\). Applied to \(x - y\), this gives \(|N(x) - N(y)| \le N(x-y) \le M\,N_\infty(x-y)\): the map \(N\) is \(N_\infty\)-continuous.
- (2) The \(N_\infty\)-unit sphere is compact. Let \(S = \{x \in E : N_\infty(x) = 1\}\), and prove \(S\) compact directly --- not through \S4.2, which would be circular. Let \((x^{(p)})_p\) be a sequence of \(S\). Each coordinate sequence \((x^{(p)}_k)_p\) is bounded, since \(|x^{(p)}_k| \le N_\infty(x^{(p)}) = 1\). Performing \(n\) successive extractions, one coordinate at a time --- each by Bolzano--Weierstrass in \(\mathbb{K}\), applied for \(\mathbb{K} = \mathbb{C}\) to the real and imaginary parts, the real version being the only one assumed --- yields a sub-sequence along which every coordinate converges: \(x^{(\sigma(p))}_k \to c_k\). With \(c = \sum_k c_k e_k\), this means \(N_\infty(x^{(\sigma(p))} - c) = \max_k |x^{(\sigma(p))}_k - c_k| \to 0\), so \(x^{(\sigma(p))} \to c\) for \(N_\infty\). By continuity of \(N_\infty\), \(N_\infty(c) = \lim N_\infty(x^{(\sigma(p))}) = 1\), so \(c \in S\). Thus \(S\) is compact.
- (3) Lower and upper bounds. The set \(S\) is non-empty (it contains \(e_1\)) and, by step (2), compact, while \(N\) is continuous on it by step (1) --- all for one and the same norm \(N_\infty\). So the extreme value theorem of \S2.1, applied in the normed space \((E,N_\infty)\), makes \(N\) attain a minimum \(\alpha\) and a maximum \(\beta\) on \(S\). The minimum is attained at some \(x_0 \in S\), and \(x_0 \ne 0\) since \(N_\infty(x_0) = 1\), so \(\alpha = N(x_0) > 0\). For any \(x \ne 0\), the vector \(x / N_\infty(x)\) lies in \(S\), hence \(\alpha \le N\bigl(x/N_\infty(x)\bigr) \le \beta\), that is \(\alpha\,N_\infty(x) \le N(x) \le \beta\,N_\infty(x)\) --- also true at \(x = 0\). So \(N\) and \(N_\infty\) are equivalent.
- (4) Transitivity. Any two norms \(N_1, N_2\) are each equivalent to \(N_\infty\) by step (3), hence equivalent to each other.
The three usual norms of \(\mathbb{R}^2\) have unit balls of three different shapes --- yet, being equivalent, they cannot be told apart by any topological property. 
The unit balls of \(\norme{\cdot}_1\), \(\norme{\cdot}_2\), \(\norme{\cdot}_\infty\) in \(\mathbb{R}^2\): a diamond, a disc, a square --- three equivalent norms.
Example — The usual norms of \(\mathbb{K}^n\)
On \(\mathbb{K}^n\) the norms \(\norme{x}_1 = \sum_k |x_k|\), \(\norme{x}_2 = \bigl(\sum_k |x_k|^2\bigr)^{1/2}\) and \(\norme{x}_\infty = \max_k |x_k|\) are equivalent, since \(\mathbb{K}^n\) is finite-dimensional. Consequently « open », « closed », « bounded », « convergent » and « continuous » in \(\mathbb{K}^n\) do not depend on which of the three is chosen --- one always picks whichever is most convenient. Method — Exploit the equivalence of norms
In a finite-dimensional space, a topological question --- convergence of a sequence, openness or closedness of a part, boundedness of a part, continuity of a map --- has the same answer for every norm. So: - choose the norm that makes the computation easiest --- often \(\norme{\cdot}_\infty\) in a basis, which reduces convergence to coordinatewise convergence;
- carry out the argument for that norm;
- state that the conclusion transfers to every norm, by equivalence.
Skills to practice
- Exploiting the equivalence of norms
IV.2
Compact parts in finite dimension
The infinite-dimensional counterexample of \S1.1 --- a closed bounded sphere that is not compact --- disappears in finite dimension. There, « compact » and « closed and bounded » coincide exactly.
Theorem — Characterisation of compacts in finite dimension
In a finite-dimensional normed space, a part is compact if and only if it is closed and bounded.
If \(\dim E = 0\) the statement is trivial; assume \(\dim E = n \ge 1\) and fix a basis, with the norm \(N_\infty\) of \S4.1. A compact part is closed and bounded by the Theorem of \S1.2 --- one implication. Conversely, let \(A\) be a closed bounded part. Closedness and boundedness do not depend on the norm (equivalence of norms, \S4.1), so \(A\) is closed and bounded for \(N_\infty\). Let \((u^{(p)})\) be a sequence of \(A\): it is \(N_\infty\)-bounded, say \(N_\infty(u^{(p)}) \le R\), so each coordinate sequence is bounded. As in \S4.1, \(n\) successive extractions --- each by Bolzano--Weierstrass in \(\mathbb{K}\), applied for \(\mathbb{K} = \mathbb{C}\) to the real and imaginary parts --- give a sub-sequence converging coordinatewise, hence converging for \(N_\infty\) to a limit \(\ell\). Since \(A\) is closed, \(\ell \in A\). So \(A\) is compact for \(N_\infty\), hence for every norm by equivalence.
Proposition — Consequences
In a finite-dimensional normed space: - [(i)] the closed unit ball and the unit sphere are compact;
- [(ii)] every bounded sequence has a convergent sub-sequence (Bolzano--Weierstrass);
- [(iii)] a bounded sequence converges if and only if it has a single adherence value.
- (i) The closed unit ball and the unit sphere are closed and bounded, hence compact by the Theorem.
- (ii) A bounded sequence lies inside some closed ball \(B_f(0,R)\), which is closed and bounded hence compact; by definition of compactness it has a convergent sub-sequence.
- (iii) Such a sequence is valued in the compact \(B_f(0,R)\), so the criterion of \S1.3 applies verbatim.
Example — The closed unit ball of \(\mathbb{R}^n\)
The closed unit ball of \(\mathbb{R}^n\) is compact --- closed, bounded, and \(\mathbb{R}^n\) is finite-dimensional. This is in sharp contrast with the unit sphere of \(\mathbb{K}[X]\) in \S1.1, closed and bounded yet not compact: the difference is precisely the infinite dimension of \(\mathbb{K}[X]\). Method — Characterise compacts in finite dimension
In a finite-dimensional space, never return to the definition by sub-sequences to recognise a compact. Instead: - check that the part is bounded --- contained in some ball;
- check that it is closed --- e.g. as the preimage of a closed set by a continuous map, or stable under sequential limits;
- conclude « closed and bounded in finite dimension, hence compact ».
Skills to practice
- Characterising compacts in finite dimension
IV.3
Continuity of linear\(\virgule\) multilinear and polynomial maps
Limits and continuity in a normed space reduced the continuity of a linear map to a single inequality \(\norme{u(x)} \le C\norme{x}\). In finite dimension that inequality always holds: linearity --- and more --- forces continuity for free, with no estimate to find.
Theorem — A linear map from a finite-dimensional space is continuous
Let \(E\) be a finite-dimensional normed space and \(F\) a normed space. Every linear map \(u \colon E \to F\) is continuous.
If \(\dim E = 0\) then \(u\) is the zero map, trivially continuous; assume \(\dim E = n \ge 1\). Fix a basis \((e_1,\dots,e_n)\) of \(E\) and the norm \(N_\infty\) of \S4.1. For \(x = \sum_k x_k e_k\), linearity gives $$ \norme{u(x)} = \norme[\big]{\sum_k x_k\,u(e_k)} \le \sum_k |x_k|\,\norme{u(e_k)} \le \Bigl(\sum_k \norme{u(e_k)}\Bigr) N_\infty(x) = C\,N_\infty(x), $$ with \(C = \sum_k \norme{u(e_k)}\). By the equivalence of norms (\S4.1) there is \(\beta > 0\) with \(N_\infty(x) \le \beta\,\norme{x}\) for the given norm of \(E\). Hence \(\norme{u(x)} \le C\beta\,\norme{x}\) for all \(x\) --- the continuity criterion for a linear map (Limits and continuity in a normed space). So \(u\) is continuous.
Theorem — Multilinear maps on finite-dimensional spaces are continuous
Let \(E_1,\dots,E_p\) be finite-dimensional normed spaces and \(F\) a normed space. Every \(p\)-linear map \(u \colon E_1 \times \dots \times E_p \to F\) is continuous.
Fix a basis of each \(E_i\) and the norm \(N_\infty\) on each. Expanding every argument \(x_i\) in its basis and using \(p\)-linearity, \(u(x_1,\dots,x_p)\) becomes a finite sum of terms, each a product of one coordinate from every argument times a fixed vector \(u(e_{1,k_1},\dots,e_{p,k_p})\). Bounding each coordinate by \(N_\infty(x_i)\) yields a constant \(C\) with $$ \norme{u(x_1,\dots,x_p)} \le C\,N_\infty(x_1)\cdots N_\infty(x_p). $$ By equivalence of norms each \(N_\infty(x_i)\) is bounded by a constant times \(\norme{x_i}\), so \(\norme{u(x_1,\dots,x_p)} \le C'\,\norme{x_1}\cdots\norme{x_p}\) --- the continuity criterion for a multilinear map (Limits and continuity in a normed space), which makes \(u\) continuous for the product norm \(\max_i \norme{x_i}\) on the domain. That product of finite-dimensional spaces is itself finite-dimensional, so all its norms are equivalent: \(u\) is continuous for every norm.
Proposition — Polynomial maps are continuous
Let \(E\) be a finite-dimensional normed space. A polynomial map \(E \to \mathbb{K}\) --- a \(\mathbb{K}\)-linear combination of monomials in the coordinates relative to a fixed basis --- is continuous. A map into a finite-dimensional normed space whose components in a target basis are polynomial is then continuous componentwise; an arbitrary normed codomain is not claimed.
Each coordinate map \(E \to \mathbb{K}\), \(x \mapsto x_k\), is linear, hence continuous by the linear-map Theorem. A monomial is a finite product of such coordinate maps, continuous as a product of continuous maps. A polynomial map is a finite sum of scalar multiples of monomials, continuous as a linear combination of continuous maps. For the vector-valued case, a map into a finite-dimensional normed space is continuous if and only if its components in a target basis are continuous (recalled from Limits and continuity in a normed space); when those components are polynomial maps, each is continuous by the scalar case, hence so is the map.
Proposition — A finite-dimensional subspace is closed
Every finite-dimensional subspace \(F\) of a normed space \(E\) is closed in \(E\).
Let \((e_1,\dots,e_p)\) be a basis of \(F\) and let \(\psi \colon \mathbb{K}^p \to E\), \(c = (c_1,\dots,c_p) \mapsto \sum_k c_k e_k\). The map \(\psi\) is linear and injective with image \(F\); it is continuous, and its inverse \(\psi^{-1} \colon F \to \mathbb{K}^p\) is linear on the finite-dimensional space \(F\), hence continuous. Let \((x_n)\) be a sequence of \(F\) converging to a limit \(\ell \in E\). A convergent sequence is bounded, so \((x_n)\) is bounded; therefore \((c_n) = (\psi^{-1}(x_n))\) is a bounded sequence of \(\mathbb{K}^p\). By Bolzano--Weierstrass in finite dimension (\S4.2), \((c_n)\) has a sub-sequence \((c_{\varphi(n)})\) converging to some \(c \in \mathbb{K}^p\). Then \(x_{\varphi(n)} = \psi(c_{\varphi(n)}) \to \psi(c)\) by continuity of \(\psi\), while \(x_{\varphi(n)} \to \ell\); uniqueness of the limit gives \(\ell = \psi(c) \in F\). So \(F\) contains every sequential limit of its points: \(F\) is closed. No appeal to completeness is made --- the notion of « espace de Banach » is out of the program.
Example — On the space of matrices
On the finite-dimensional space \(\mathcal{M}_n(\mathbb{K})\): the trace and the determinant are continuous, being scalar polynomial maps in the entries; the map \(M \mapsto \chi_M\), viewed as valued in the finite-dimensional space \(\mathbb{K}_n[X]\), is continuous since each coefficient of \(\chi_M\) is polynomial in the entries. The set \(\mathrm{GL}_n(\mathbb{K})\) is open, as \(\det^{-1}(\mathbb{K}^*)\) with \(\det\) continuous and \(\mathbb{K}^* = \mathbb{K} \setminus \{0\}\) the open set of non-zero scalars. The subspace \(\mathcal{S}_n(\mathbb{R})\) of symmetric matrices is closed, being a finite-dimensional subspace. Method — Prove continuity in finite dimension
Once the domain is finite-dimensional: - a linear or multilinear map is continuous with no estimate to write --- state the theorem and stop;
- a polynomial map is continuous when it is scalar-valued, or vector-valued into a finite-dimensional target with polynomial coordinate functions --- the codomain caveat of the Proposition;
- to read off that a part is open or closed, write it as the preimage of an open or closed set by such a continuous map.
Going further
This chapter isolated the two hypotheses --- compactness and arc-connectedness --- that carry the great theorems of continuity, and showed that in finite dimension the topology becomes uniform across all norms. The chapter Vector-valued functions will use finite-dimensional norm-equivalence to differentiate and integrate coordinatewise; Numerical and vector series will rely on the same fact to study convergence in \(\mathbb{K}^n\) and \(\mathcal{M}_n(\mathbb{K})\). In Optimisation, the extreme value theorem on a compact is the existence engine behind every « a minimum is attained » statement. Compactness and arc-connectedness, built here, are from now on standing tools.
Skills to practice
- Proving continuity in finite dimension
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