CommeUnJeu · L2 MP

Topology of a normed space

⌚ ~67 min ▢ 8 blocks ✓ 15 exercises ➣ Prerequisites : Normed vector spaces

Example

Show that $\lim_{x \to 0} \mathrm{e}^{1/x}$ does not exist.

We compute the two one-sided limits.

As $x \to 0^+$, $1/x \to +\infty$, hence (composition with $\exp$, P4.5) $\mathrm{e}^{1/x} \to +\infty$.
As $x \to 0^-$, $1/x \to -\infty$, hence $\mathrm{e}^{1/x} \to 0$ (Standard functions: $\exp(-\infty) = 0$).

The two one-sided limits ($+\infty$ on the right, $0$ on the left) differ, so by P2.4, $\lim_{x \to 0} \mathrm{e}^{1/x}$ does not exist.

I Open sets and closed sets

I.1 Neighbourhoods and open sets

Method — Lifting an indeterminate form

When the algebra of operations on limits (P4.1, P4.2) yields one of the four indeterminate forms ($\infty - \infty$, $0 \cdot \infty$, $\infty / \infty$, $0/0$), proceed in three steps:

Identify the dominant term. The largest power of $x$ (or the slowest-going-to-zero factor in a $0/0$).
Factor it out of numerator and denominator (or of each summand in $\infty - \infty$).
Take the limit of each remaining factor by P4.1; the indeterminate form is now a determinate operation.

This is a survival technique; Asymptotic comparison (later chapter) gives the systematic toolkit (Taylor expansions, Landau notation).

Ex 1 Ex 2 Ex 3 Ex 4 Ex 5 Ex 6

Skills to practice

Identifying open sets

I.2 Closed sets and the sequential characterisation

Ex 7 Ex 8 Ex 9

Definition — Continuity at a point

Let $f : E \to \mathbb{R}$ and $a \in E$. We say $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$. Equivalently: $$ \forall \varepsilon > 0, \ \exists \delta > 0, \ \forall x \in E, \ |x - a| \le \delta \Rightarrow |f(x) - f(a)| \le \varepsilon. $$

Definition — Left- and right-continuity

Let $f : E \to \mathbb{R}$ and $a \in E$. We say $f$ is left-continuous at $a$ if $\lim_{x \to a^-} f(x) = f(a)$ (assuming $a$ is adherent to $E \cap \,]-\infty, a[$), and right-continuous at $a$ if $\lim_{x \to a^+} f(x) = f(a)$ (assuming $a$ is adherent to $E \cap \,]a, +\infty[$).

Proposition — Continuity via one-sided limits

Let $I \subset \mathbb{R}$ be an interval, $f : I \to \mathbb{R}$, and $a \in I$.

If $a$ is an interior point of $I$: $\textcolor{colorprop}{f}$ is continuous at $a$ if and only if $f$ is both left-continuous and right-continuous at $a$.
If $a$ is the left endpoint of $I$: $\textcolor{colorprop}{f}$ is continuous at $a$ if and only if $f$ is right-continuous at $a$.
If $a$ is the right endpoint of $I$ (symmetric): $\textcolor{colorprop}{f}$ is continuous at $a$ if and only if $f$ is left-continuous at $a$.

Definition — Continuous extension

Let $E' \subset \mathbb{R}$ with $a \notin E'$ adherent to $E'$, and $f : E' \to \mathbb{R}$ admitting a finite limit $\ell$ at $a$. The continuous extension (prolongement par continuité) of $f$ at $a$ is the function $\tilde{f} : E' \cup \{a\} \to \mathbb{R}$ defined by $$ \tilde{f}(x) = f(x) \text{ for } x \in E', \qquad \tilde{f}(a) = \ell. $$ By construction, $\tilde{f}$ is continuous at $a$ and equal to $f$ on $E'$.

Proposition — Sequential characterization of continuity

Let $f : E \to \mathbb{R}$ and $a \in E$. Then $\textcolor{colorprop}{f}$ is continuous at $a$ if and only if for every sequence $(u_n) \in E^{\mathbb{N}}$ with $u_n \to a$, $f(u_n) \to f(a)$. Direct consequence of Heine (T3.1) applied with $\ell = f(a)$.

Skills to practice

Applying the sequential characterisation

I.3 Stability under union and intersection

Proposition — Operations on continuity at a point

Let $f, g : E \to \mathbb{R}$ continuous at $a \in E$. Then:

$\textcolor{colorprop}{\lambda f + \mu g}$ is continuous at $a$ for any $\lambda, \mu \in \mathbb{R}$;
$\textcolor{colorprop}{f g}$ is continuous at $a$;
if $g(a) \ne 0$, then $\textcolor{colorprop}{f / g}$ is continuous at $a$ (well-defined on a neighborhood of $a$ by the same $\varepsilon = |g(a)|/2$ argument as in P4.1);
if $f : E \to F \subset \mathbb{R}$ is continuous at $a$ and $h : F \to \mathbb{R}$ is continuous at $f(a)$, then $\textcolor{colorprop}{h \circ f}$ is continuous at $a$.

Example

Show that $f(x) = (\sin x) / x$, defined on $\mathbb{R}^*$, admits a continuous extension at $0$ with value $1$.

Answer

By Ex4.1, $\lim_{x \to 0} (\sin x)/x = 1$, a finite limit. By D5.3, $\tilde{f} : \mathbb{R} \to \mathbb{R}$ defined by $\tilde{f}(x) = (\sin x)/x$ for $x \ne 0$ and $\tilde{f}(0) = 1$ is the continuous extension.

Example — Jump discontinuity

The Heaviside-style function $H : \mathbb{R} \to \mathbb{R}$ defined by $H(x) = 0$ for $x \le 0$ and $H(x) = 1$ for $x > 0$ is left-continuous at $0$ ($\lim_{x \to 0^-} H = 0 = H(0)$) but not right-continuous ($\lim_{x \to 0^+} H = 1 \ne H(0) = 0$). Hence $H$ is discontinuous at $0$, with a jump of amplitude $1$. The point $0$ is interior to $\mathbb{R}$, so by P5.1 the failure of right-continuity alone breaks continuity.

Method — Establishing continuity by direct $\varepsilon$-$\delta$ argument

To prove $f$ continuous at $a$ from the definition:

Fix $\varepsilon > 0$. Treat it as a generic positive number; the choice of $\delta$ will depend on it.
Bound $|f(x) - f(a)|$ by an expression in $|x - a|$. Use the triangle inequality, factoring of polynomial expressions, mean-value-type bounds, or Lipschitz constants.
Choose $\delta$ so that the bound is $\le \varepsilon$. Often $\delta = \varepsilon / L$ where $L$ is a Lipschitz constant.

For continuous functions of class $C^k$ encountered in the rest of the program, this direct argument is rarely needed: continuity will follow from operations (P5.3) applied to functions known to be continuous (Standard functions).

Skills to practice

Applying the stability rules

II Interior$\virgule$ closure and frontier

II.1 Interior and closure

Ex 10 Ex 11 Ex 12 Ex 13 Ex 14 Ex 15 Ex 16

When $f$ is continuous at every point of an interval, three properties become available: (i) the image of the interval is an interval (théorème des valeurs intermédiaires and image of an interval), (ii) the image of a segment is a segment (see the next section, Continuous functions on a segment), and (iii) bijectivity onto the image transfers monotonicity (same section). The TVI is proved by dichotomy --- a constructive bisection scheme that converges to a zero of $f$ when the signs at the endpoints differ.

Definition — Continuous on an interval

Let $I \subset \mathbb{R}$ be an interval. We say $f : I \to \mathbb{R}$ is continuous on $I$ if $f$ is continuous at every $a \in I$. The set of such functions is denoted $\textcolor{colordef}{C^0(I, \mathbb{R})}$, or $\textcolor{colordef}{C^0(I, \mathbb{C})}$ for $\mathbb{C}$-valued continuous functions.

Skills to practice

Determining interior and closure

II.2 Frontier

Theorem — Théorème des valeurs intermédiaires (TVI)

Let $f \in C^0([a, b], \mathbb{R})$ with $a < b$ and $f(a) f(b) \le 0$. Then there exists $\textcolor{colorprop}{c \in [a, b]}$ such that $f(c) = 0$.

Proof

The proof is by dichotomy.

Endpoint cases. If $f(a) = 0$, take $c = a$ and stop. If $f(b) = 0$, take $c = b$ and stop. Otherwise $f(a) f(b) < 0$ (strict).
Bisection. Build two sequences $(a_n), (b_n)$ with $a_0 = a$, $b_0 = b$. At step $n$, let $m_n = (a_n + b_n)/2$. If $f(m_n) = 0$, take $c = m_n$ and stop. Otherwise $f(a_n) f(m_n)$ and $f(m_n) f(b_n)$ have opposite signs; replace the endpoint of $[a_n, b_n]$ on the side where the strict sign change persists, i.e. set $a_{n+1} = a_n$, $b_{n+1} = m_n$ if $f(a_n) f(m_n) < 0$, and $a_{n+1} = m_n$, $b_{n+1} = b_n$ otherwise.
Convergence. The sequences are adjacent: $(a_n)$ is increasing, $(b_n)$ is decreasing, $b_n - a_n = (b - a)/2^n \to 0$. By the theorem on adjacent sequences (admis ; cours Suites réelles), both converge to a common limit $c \in [a, b]$.
Conclusion. By continuity of $f$ at $c$, $f(a_n) \to f(c)$ and $f(b_n) \to f(c)$. By construction $f(a_n) f(b_n) \le 0$ for every $n$; passing to the limit (P4.3 applied to the product), $f(c)^2 \le 0$, hence $f(c) = 0$.

Proposition — TVI form 2

Let $f \in C^0([a, b], \mathbb{R})$ and $y$ a real number lying between $f(a)$ and $f(b)$. Then there exists $\textcolor{colorprop}{c \in [a, b]}$ such that $f(c) = y$.

Proof

Apply T6.1 to $g(x) = f(x) - y \in C^0([a, b], \mathbb{R})$: $g(a) = f(a) - y$ and $g(b) = f(b) - y$ have opposite signs (or one is zero) since $y$ lies between $f(a)$ and $f(b)$. Hence $g$ vanishes at some $c \in [a, b]$, i.e. $f(c) = y$.

Theorem — Image of an interval is an interval

Let $I \subset \mathbb{R}$ be an interval and $f \in C^0(I, \mathbb{R})$. Then $\textcolor{colorprop}{f(I)}$ is an interval.

Proof

We use the characterization « $J$ is an interval iff for any $u, v \in J$ with $u \le v$, $[u, v] \subset J$ ». Take $u, v \in f(I)$ with $u \le v$, and $y \in [u, v]$. Pick $\alpha, \beta \in I$ with $f(\alpha) = u$ and $f(\beta) = v$. The closed segment $[\alpha, \beta]$ (or $[\beta, \alpha]$ if $\beta < \alpha$) is contained in $I$ since $I$ is an interval, and $y$ lies between $f(\alpha) = u$ and $f(\beta) = v$. By the TVI (form 2) applied to $f$ on this segment, $y \in f(I)$. Hence $[u, v] \subset f(I)$, so $f(I)$ is an interval.

Example

Show that any real polynomial of odd degree has at least one real root.

Answer

Let $P \in \mathbb{R}[X]$ of odd degree $n$, with leading coefficient $a_n \in \mathbb{R}^*$. Up to replacing $P$ by $-P$ (which has the same roots), we may assume $a_n > 0$. Then $$ \lim_{x \to +\infty} P(x) = +\infty, \qquad \lim_{x \to -\infty} P(x) = -\infty, $$ because $n$ is odd. There exists $A < 0$ with $P(A) < 0$ and $B > 0$ with $P(B) > 0$. The polynomial $P$ is continuous on $[A, B]$, with $P(A) P(B) < 0$. By the TVI, there exists $c \in [A, B]$ with $P(c) = 0$.

Method — Proving an equation has a solution via TVI

To show that an equation $f(x) = 0$ has a solution in $[a, b]$:

Continuity. Verify $f$ is continuous on $[a, b]$. As polynomial, exponential, trigonometric, rational (with no zero of the denominator in the interval), or composition of continuous functions, this is automatic.
Sign at the endpoints. Compute $f(a)$ and $f(b)$. If their signs differ ($f(a) f(b) < 0$, or one is zero), the TVI applies.
Conclude. There exists $c \in [a, b]$ such that $f(c) = 0$.

The TVI gives existence, not uniqueness. Adding strict monotonicity (T7.2, in the next section) upgrades existence to existence and uniqueness.

Ex 17

Skills to practice

Computing the frontier

II.3 Density

Ex 18 Ex 19 Ex 20 Ex 21 Ex 22 Ex 23

Skills to practice

Proving density

III Relative topology

III.1 Relative neighbourhoods

On a closed bounded interval $[a, b]$ --- a segment --- continuous functions are particularly well-behaved: they are bounded, attain their bounds (théorème des bornes atteintes), and the image of a segment is a segment. The proof of the bornes-atteintes theorem uses Bolzano-Weierstrass on a maximizing (resp. minimizing) sequence; the image-of-a-segment statement then follows as a clean corollary. The closing pair --- continuous injective on an interval is strictly monotone, and the continuous strictly monotone bijection theorem --- are admitted (program: « démonstration non exigible ») but used in every subsequent analysis chapter.

Theorem — Théorème des bornes atteintes

Let $f \in C^0([a, b], \mathbb{R})$ with $a < b$. Then $f$ is bounded on $[a, b]$ and there exist $\textcolor{colorprop}{x_*, x^* \in [a, b]}$ such that $$ f(x_*) = \min_{[a, b]} f \qquad \text{and} \qquad f(x^*) = \max_{[a, b]} f. $$

Proof

The proof has two steps.

Step 1 --- boundedness. We show $f$ is bounded above (boundedness below is symmetric). By contradiction, suppose $f$ unbounded above. For every $n \in \mathbb{N}$, there exists $u_n \in [a, b]$ with $f(u_n) \ge n$. The sequence $(u_n)$ lies in $[a, b]$, hence is bounded. By Bolzano-Weierstrass (admis; cours Suites réelles), $(u_n)$ admits a convergent subsequence $u_{\varphi(n)} \to x_0 \in [a, b]$. By continuity of $f$ at $x_0$, $f(u_{\varphi(n)}) \to f(x_0) \in \mathbb{R}$. But $f(u_{\varphi(n)}) \ge \varphi(n) \to +\infty$, contradiction.
Step 2 --- maximum attained. Set $M = \sup_{[a, b]} f \in \mathbb{R}$ (well-defined by Step 1). Build a maximizing sequence $(u_n)$ in $[a, b]$ with $f(u_n) \to M$: take $u_n \in [a, b]$ such that $M - 1/(n+1) \le f(u_n) \le M$ (possible by definition of the supremum). By Bolzano-Weierstrass, extract $u_{\varphi(n)} \to x^* \in [a, b]$. By continuity at $x^*$, $f(u_{\varphi(n)}) \to f(x^*)$. The same subsequence still tends to $M$ (by extraction), hence $f(x^*) = M$.
Symmetric step --- minimum attained. Same argument with a minimizing sequence and $m = \inf_{[a, b]} f$, giving $x_* \in [a, b]$ with $f(x_*) = m$.

Proposition — Image of a segment is a segment

Let $f \in C^0([a, b], \mathbb{R})$ with $a < b$. Then $\textcolor{colorprop}{f([a, b]) = [\min_{[a, b]} f, \max_{[a, b]} f]}$.

Proof

By T7.1, $\min$ and $\max$ are attained, hence $f([a, b]) \supset \{\min f, \max f\}$. By T6.2 ($f([a, b])$ is an interval), $f([a, b])$ is a sub-interval of $[\min f, \max f]$ containing both endpoints, hence equal to $[\min f, \max f]$.

Theorem — Continuous strictly monotone bijection

Let $I \subset \mathbb{R}$ be an interval and $f \in C^0(I, \mathbb{R})$ strictly monotone. Then:

$f$ is a bijection from $I$ onto $J := f(I)$;
$J$ is an interval;
the reciprocal $\textcolor{colorprop}{f^{-1} : J \to I}$ is continuous on $J$ and strictly monotone of the same sense as $f$.

Admis (programme : « La démonstration n'est pas exigible. »).

Hors programme --- énoncés admis

The following sentence from the official 2021 program (p.~14) justifies admitting the proofs of P7.2 and T7.2: « Une fonction continue sur un intervalle, à valeurs réelles et injective, est strictement monotone. La démonstration n'est pas exigible. Toute fonction réelle strictement monotone, définie et continue sur un intervalle, admet une fonction réciproque de même monotonie, définie et continue sur un intervalle. La démonstration n'est pas exigible. »

Skills to practice

Handling relative neighbourhoods

III.2 Relative open and closed sets

Example

Application of T7.2 to $\arctan$. The restriction of $\tan$ to $\,]-\pi/2, \pi/2[$ is continuous and strictly increasing. By T7.2, it is a bijection from $\,]-\pi/2, \pi/2[$ onto $\mathbb{R}$, and the inverse $\arctan : \mathbb{R} \to \,]-\pi/2, \pi/2[$ is continuous on $\mathbb{R}$ and strictly increasing. (Forward reference to the Standard functions chapter, where $\arctan$ is studied in detail.)

Example

Show that $f(x) = x^2$ on $[-1, 2]$ attains its bounds, locating $x_*$ and $x^*$ explicitly.

Answer

The function $f$ is continuous on $[-1, 2]$ as a polynomial. By T7.1, $f$ attains its bounds. We compute them by hand: $f$ is decreasing on $[-1, 0]$ and increasing on $[0, 2]$, so the minimum is at $0$ with $f(0) = 0$, and the maximum is at the endpoint with the largest $|x|$, namely $x = 2$ with $f(2) = 4$. Hence $x_* = 0$, $x^* = 2$, $\min f = 0$, $\max f = 4$.

Example

Application of T7.2 to $\sqrt{\cdot}$. The restriction of $x \mapsto x^2$ to $\mathbb{R}_+$ is continuous and strictly increasing. By T7.2, it is a bijection from $\mathbb{R}_+$ onto $\mathbb{R}_+$, and the inverse $\sqrt{\cdot} : \mathbb{R}_+ \to \mathbb{R}_+$ is continuous on $\mathbb{R}_+$ and strictly increasing. (Forward reference to the Standard functions chapter.)

Forward pointer --- uniform continuity and Lipschitz functions

The 2021 program lists a finer notion of continuity on a segment --- uniform continuity --- together with Heine's uniform-continuity theorem (every continuous function on a segment is uniformly continuous) and the example of Lipschitz functions. We do not treat these here: the program introduces them « uniquement en vue de la construction de l'intégrale », and we therefore reserve them for the chapter Intégration sur un segment, where they power the uniform approximation of continuous functions by step functions.

Ex 24 Ex 25 Ex 26

Skills to practice

Identifying relative open and closed sets