Metric Spaces & Topology

A metric space is a pair $(X, d)$ consisting of a set $X$ and a function $d : X \times X \to [0, \infty)$ called a metric (or distance function) satisfying:

1. Positivity: $d(x, y) = 0$ if and only if $x = y$, and $d(x, y) \geq 0$ for all $x, y$.
2. Symmetry: $d(x, y) = d(y, x)$ for all $x, y \in X$.
3. Triangle inequality: $d(x, z) \leq d(x, y) + d(y, z)$ for all $x, y, z \in X$.

For $1 \leq p < \infty$, the $\ell^p$ metric on $\mathbb{R}^n$ is

$d_p(x, y) = \left(\sum_{i=1}^n |x_i - y_i|^p\right)^{1/p}.$

For $p = \infty$ the formula degenerates and we use

$d_\infty(x, y) = \max_{1 \leq i \leq n} |x_i - y_i|.$

All three axioms are satisfied for every $p \in [1, \infty]$ (the triangle inequality is Minkowski’s inequality). Three values of $p$ are worth remembering: $p = 2$ gives the usual Euclidean distance, $p = 1$ gives the “Manhattan” or taxicab distance, and $p = \infty$ gives the Chebyshev (max-coordinate) distance. In $\mathbb{R}^2$, the “unit ball” $\{x : d_p(x, 0) \leq 1\}$ is a diamond for $p = 1$, a circle for $p = 2$, and a square for $p = \infty$ — the same set $\mathbb{R}^2$ acquires three visibly different geometries depending on which metric we choose.

The space $C([a, b])$ of continuous real-valued functions on $[a, b]$ becomes a metric space under

$d_\infty(f, g) = \|f - g\|_\infty = \sup_{x \in [a,b]} |f(x) - g(x)|.$

Positivity and symmetry are immediate. The triangle inequality follows from the pointwise inequality $|f(x) - h(x)| \leq |f(x) - g(x)| + |g(x) - h(x)|$ by taking suprema on both sides. This is the same sup-norm that appeared in Uniform Convergence: a sequence $f_n$ converges to $f$ in this metric if and only if $f_n \to f$ uniformly. Section 5 will prove that $(C([a, b]), d_\infty)$ is complete — every Cauchy sequence converges — which is why the sup-norm is the right tool for function-space fixed-point arguments.

On any set $X$, define

$d(x, y) = \begin{cases} 0 & \text{if } x = y, \\ 1 & \text{if } x \neq y. \end{cases}$

All three axioms are trivial to verify. Under this metric every subset is open, every convergent sequence is eventually constant, and the “unit ball” $B(x, 1)$ contains only the single point $x$. The discrete metric shows that metric-space theory applies to exotic settings — there is no requirement that $X$ look anything like $\mathbb{R}^n$ — and it is a useful source of counterexamples whenever you suspect a theorem secretly needs continuity.

On the set $\{0, 1\}^n$ of binary strings of length $n$, the Hamming distance is

$d_H(x, y) = \#\{\, i : x_i \neq y_i\,\},$

the number of coordinates in which $x$ and $y$ differ. Symmetry is clear, positivity is clear, and the triangle inequality holds because “coordinates where $x$ and $z$ differ” is a subset of “coordinates where $x$ and $y$ differ” $\cup$ “coordinates where $y$ and $z$ differ.” Hamming distance is the metric behind error-correcting codes and is directly relevant to discrete ML: training a classifier to produce outputs within Hamming distance $k$ of a target is a metric-space constraint.

The same underlying set can carry many different metrics, and the choice shapes everything else: which sequences converge, which sets are open, which functions are continuous. In $\mathbb{R}^2$ the $\ell^1$, $\ell^2$, and $\ell^\infty$ metrics all generate the same notion of “open set” (they are topologically equivalent — we will define this precisely in Section 8), so they all agree on which sequences converge. But the Hamming metric on $\{0, 1\}^n$ is qualitatively different from the Euclidean metric on the same set viewed as a subset of $\mathbb{R}^n$: under Hamming distance, consecutive binary strings (differing in a single bit) are always at distance 1, but under Euclidean distance their positions depend on which bit changed. The metric determines the geometry, and the geometry determines the analysis.

Let $(X, d)$ be a metric space, let $x \in X$, and let $r > 0$. The open ball of radius $r$ centered at $x$ is

$B(x, r) = \{\, y \in X : d(x, y) < r\,\}.$

The closed ball of radius $r$ centered at $x$ is

$\overline{B}(x, r) = \{\, y \in X : d(x, y) \leq r\,\}.$

A set $U \subseteq X$ is open if for every point $x \in U$ there exists some $r > 0$ such that the open ball $B(x, r)$ is entirely contained in $U$:

$\forall x \in U \ \exists r > 0 : B(x, r) \subseteq U.$

A set $F \subseteq X$ is closed if its complement $X \setminus F$ is open.

In any metric space $(X, d)$:

1. $\emptyset$ and $X$ are both open and closed.
2. Arbitrary unions of open sets are open.
3. Finite intersections of open sets are open.
4. Arbitrary intersections of closed sets are closed.
5. Finite unions of closed sets are closed.

For a subset $A \subseteq X$:

• The interior $A^\circ$ is the union of all open sets contained in $A$; equivalently, it is the largest open set inside $A$.
• The closure $\overline{A}$ is the intersection of all closed sets containing $A$; equivalently, it is the smallest closed set that contains $A$.
• The boundary is $\partial A = \overline{A} \setminus A^\circ$.
• A point $x \in X$ is a limit point of $A$ if every open ball $B(x, r)$ contains a point of $A$ other than $x$.

In $C([0, 1])$ with the sup-norm, the open ball $B(f, \varepsilon)$ consists of all continuous functions $g$ whose graph lies within the horizontal band $\{(x, y) : f(x) - \varepsilon < y < f(x) + \varepsilon\}$. A set $\mathcal{U} \subseteq C([0, 1])$ is open precisely when for every $f \in \mathcal{U}$ we can find a small enough $\varepsilon$ so that every continuous function inside the $\varepsilon$-tube around $f$ is also in $\mathcal{U}$. This is exactly the “$\varepsilon$-tube” picture from Topic 4’s visualization of uniform convergence — the only difference is that now we are thinking of a tube as an open ball in a metric space.

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. A function $f : X \to Y$ is continuous at a point $x_0 \in X$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that

$d_X(x, x_0) < \delta \quad \Longrightarrow \quad d_Y(f(x), f(x_0)) < \varepsilon.$

The function $f$ is continuous (on all of $X$) if it is continuous at every point of $X$.

Let $f : (X, d_X) \to (Y, d_Y)$ be a function between metric spaces. The following are equivalent:

1. $f$ is continuous at every point of $X$ (the $\varepsilon$–$\delta$ definition).
2. For every convergent sequence $x_n \to x$ in $X$, we have $f(x_n) \to f(x)$ in $Y$.
3. For every open set $V \subseteq Y$, the preimage $f^{-1}(V) \subseteq X$ is open in $X$.

We prove the three implications $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$.

$(1) \Rightarrow (2)$. Assume $f$ is continuous at every point, let $x_n \to x$ in $X$, and fix $\varepsilon > 0$. Continuity at $x$ gives some $\delta > 0$ such that $d_X(y, x) < \delta$ implies $d_Y(f(y), f(x)) < \varepsilon$. Since $x_n \to x$, there exists $N$ such that for all $n \geq N$ we have $d_X(x_n, x) < \delta$, and therefore $d_Y(f(x_n), f(x)) < \varepsilon$. This is exactly the statement that $f(x_n) \to f(x)$.

$(2) \Rightarrow (3)$. Assume sequential continuity, let $V \subseteq Y$ be open, and let $x \in f^{-1}(V)$. We need to show some ball $B(x, r)$ is contained in $f^{-1}(V)$. Suppose not. Then for every $n \geq 1$ there exists a point $x_n$ with $d_X(x_n, x) < 1/n$ but $f(x_n) \notin V$. By construction $x_n \to x$, so by sequential continuity $f(x_n) \to f(x)$. But $f(x) \in V$ and $V$ is open, so some ball $B(f(x), \varepsilon)$ is contained in $V$, which means $f(x_n) \in V$ for all sufficiently large $n$ — contradicting $f(x_n) \notin V$.

$(3) \Rightarrow (1)$. Assume preimages of open sets are open, let $x_0 \in X$, and fix $\varepsilon > 0$. The open ball $B(f(x_0), \varepsilon)$ is open in $Y$, so by hypothesis its preimage $f^{-1}(B(f(x_0), \varepsilon))$ is open in $X$ and contains $x_0$. By the definition of “open” we can find some $\delta > 0$ with $B(x_0, \delta) \subseteq f^{-1}(B(f(x_0), \varepsilon))$, which rewrites as: $d_X(x, x_0) < \delta \Rightarrow d_Y(f(x), f(x_0)) < \varepsilon$. That is the $\varepsilon$–$\delta$ condition at $x_0$. Since $x_0$ was arbitrary, $f$ is continuous.

Characterization (3) is the conceptual leap: it describes continuity entirely in terms of the structure of open sets, with no reference to distances or sequences. This is what lets the definition generalize to topological spaces in Section 8, where there may be no metric at all. In metric spaces all three characterizations are equivalent — we just proved it — so the preimage formulation looks like one more way to say the same thing. The power of (3) only becomes apparent when we leave metrics behind: in an arbitrary topological space, the $\varepsilon$–$\delta$ definition is not even expressible, and the sequential definition can fail in strange ways. The preimage characterization is the universally correct one.

A function $f : (X, d_X) \to (Y, d_Y)$ is Lipschitz continuous with constant $L \geq 0$ if

$d_Y(f(x), f(y)) \leq L \cdot d_X(x, y) \quad \text{for all } x, y \in X.$

The smallest such $L$ is called the Lipschitz constant of $f$. If $L < 1$ the map is called a contraction, and this case will drive Section 7.

A sequence $(x_n)$ in a metric space $(X, d)$ is Cauchy if for every $\varepsilon > 0$ there exists $N$ such that

$d(x_m, x_n) < \varepsilon \quad \text{for all } m, n \geq N.$

A metric space $(X, d)$ is complete if every Cauchy sequence in $X$ converges to a limit that is also in $X$.

$\mathbb{R}$ is complete by the Cauchy-completeness theorem from Topic 3, Section 1. $\mathbb{R}^n$ is complete under any of the $\ell^p$ metrics: if $(x^{(k)})$ is a Cauchy sequence in $\mathbb{R}^n$, then each coordinate sequence $(x^{(k)}_i)$ is Cauchy in $\mathbb{R}$ (because $|x^{(k)}_i - x^{(\ell)}_i| \leq d_p(x^{(k)}, x^{(\ell)})$), so each coordinate converges to some limit $x^*_i$, and coordinate-wise convergence in $\mathbb{R}^n$ implies $\ell^p$ convergence. This is the same argument Topic 3 ran for the Euclidean metric; all it needed was the coordinate inequality, which holds for every $p \geq 1$.

The space $C([a, b])$ with the sup-norm metric $d_\infty(f, g) = \|f - g\|_\infty$ is complete.

Let $(f_n)$ be a Cauchy sequence in $C([a, b])$. We need to exhibit a continuous function $f$ such that $\|f_n - f\|_\infty \to 0$.

Step 1: Pointwise limit exists. For each $x \in [a, b]$, the sequence of real numbers $(f_n(x))$ is Cauchy in $\mathbb{R}$, because

$|f_n(x) - f_m(x)| \leq \sup_{t \in [a,b]} |f_n(t) - f_m(t)| = \|f_n - f_m\|_\infty.$

Since $\mathbb{R}$ is complete, $(f_n(x))$ converges to some real number, which we define to be $f(x)$. This gives a function $f : [a, b] \to \mathbb{R}$ as the pointwise limit.

Step 2: Convergence is uniform. Given $\varepsilon > 0$, the Cauchy condition gives $N$ such that $\|f_n - f_m\|_\infty < \varepsilon$ for all $m, n \geq N$. Fix $x \in [a, b]$ and any $n \geq N$. Letting $m \to \infty$ in the inequality $|f_n(x) - f_m(x)| < \varepsilon$, the right side stays $\leq \varepsilon$ and the left side converges to $|f_n(x) - f(x)|$. Therefore $|f_n(x) - f(x)| \leq \varepsilon$ for all $x$, which means $\|f_n - f\|_\infty \leq \varepsilon$. Since $\varepsilon$ was arbitrary, $f_n \to f$ uniformly.

Step 3: The limit is continuous. This is exactly the Uniform Limit Theorem from Uniform Convergence: a uniform limit of continuous functions is continuous. So $f \in C([a, b])$, and the previous step says $\|f_n - f\|_\infty \to 0$. The sequence converges in the sup-norm metric, as needed.

The same set $C([0, 1])$ is not complete under the integral-based metric

$d_2(f, g) = \sqrt{\int_0^1 (f(x) - g(x))^2 \, dx}.$

Consider the sequence of continuous functions $f_n$ on $[0, 1]$ defined to be $0$ on $[0, 1/2 - 1/n]$, linear from $0$ to $1$ on $[1/2 - 1/n, 1/2 + 1/n]$, and $1$ on $[1/2 + 1/n, 1]$ — a “ramp” of width $2/n$ around $x = 1/2$. For large $m, n$, the difference $f_m - f_n$ is supported in a shrinking interval of length $2/\min(m, n)$, so $d_2(f_m, f_n) \to 0$ as $m, n \to \infty$: the sequence is Cauchy in $d_2$. But its pointwise limit is the step function that jumps from $0$ to $1$ at $x = 1/2$, which is discontinuous — and no continuous function can be the $d_2$-limit of this Cauchy sequence. So $(C([0, 1]), d_2)$ is not complete. The completeness of a space depends on the metric, not just the set.

Every metric space $(X, d)$ has a completion: a complete metric space $(\hat{X}, \hat{d})$ together with an isometric embedding $\iota : X \hookrightarrow \hat{X}$ such that $\iota(X)$ is dense in $\hat{X}$. The construction mirrors how $\mathbb{R}$ is built from $\mathbb{Q}$: form equivalence classes of Cauchy sequences in $X$, where two Cauchy sequences are equivalent if their interleaving is also Cauchy. The completion of $(C([0, 1]), d_2)$ is a larger space containing limits like our ramp-sequence step function — a space that analysts call $L^2[0, 1]$ and that shows up constantly in modern machine learning. We state the completion theorem as a fact rather than proving it in detail; Munkres, Chapter 3, has the full construction.

A metric space $(X, d)$ is compact if every open cover of $X$ has a finite subcover: whenever $X = \bigcup_{\alpha \in I} U_\alpha$ with each $U_\alpha$ open, there is a finite set $\{\alpha_1, \ldots, \alpha_N\} \subseteq I$ such that $X = U_{\alpha_1} \cup \cdots \cup U_{\alpha_N}$. A subset $K \subseteq X$ is compact if it is compact as a metric space under the restricted metric.

A metric space $(X, d)$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

A metric space $(X, d)$ is totally bounded if for every $\varepsilon > 0$ there exist finitely many points $x_1, \ldots, x_N \in X$ such that the corresponding $\varepsilon$-balls cover $X$:

$X = \bigcup_{i=1}^N B(x_i, \varepsilon).$

The collection $\{x_1, \ldots, x_N\}$ is called a finite $\varepsilon$-net for $X$.

For a metric space $(X, d)$, the following three conditions are equivalent:

1. $X$ is compact (every open cover has a finite subcover).
2. $X$ is sequentially compact (every sequence has a convergent subsequence).
3. $X$ is complete and totally bounded.

We sketch the three directions. Full proofs are in Munkres, Chapter 3, or Rudin, Chapter 2.

$(1) \Rightarrow (2)$. Suppose $X$ is compact and let $(x_n)$ be a sequence with no convergent subsequence. Then no point $x \in X$ is a cluster point of $(x_n)$, so each $x$ has an open ball $B(x, r_x)$ containing only finitely many terms of the sequence. These balls form an open cover of $X$. Compactness gives a finite subcover, which therefore contains only finitely many terms in total — but the full sequence has infinitely many terms, a contradiction.

$(2) \Rightarrow (3)$. Sequential compactness implies completeness: a Cauchy sequence has a convergent subsequence by hypothesis, and a Cauchy sequence with a convergent subsequence converges to the same limit (standard Cauchy-subsequence lemma). For total boundedness, suppose $X$ has no finite $\varepsilon$-net for some $\varepsilon > 0$. Pick any $x_1 \in X$. Since $B(x_1, \varepsilon)$ is not all of $X$, pick $x_2 \notin B(x_1, \varepsilon)$. Since $B(x_1, \varepsilon) \cup B(x_2, \varepsilon)$ is not all of $X$, pick $x_3$ outside both balls. Continuing, we construct a sequence with pairwise distances $\geq \varepsilon$, which has no convergent subsequence — contradicting sequential compactness.

$(3) \Rightarrow (1)$ (the hardest direction). Suppose $X$ is complete and totally bounded but has an open cover $\{U_\alpha\}$ with no finite subcover. Total boundedness gives a finite $\varepsilon_1$-net; at least one of the balls $B(x_i, \varepsilon_1)$ intersected with $X$ has no finite subcover (otherwise combining them would give a finite subcover of the whole space). Pick one such ball, shrink $\varepsilon$ to $\varepsilon_1 / 2$, and apply total boundedness inside that ball to get a smaller uncovered ball $B(x_2, \varepsilon_1/2)$. Iterating, we obtain a nested sequence of sets with diameters shrinking to zero; picking one point from each gives a Cauchy sequence. By completeness it converges to some limit $x^*$, which belongs to some $U_\alpha$ in the cover. Because $U_\alpha$ is open, $U_\alpha$ contains a ball around $x^*$, so $U_\alpha$ covers all the tail sets in our nested sequence once they are small enough — giving a finite subcover for that tail after all, contradicting our construction.

In $\mathbb{R}^n$, the Heine–Borel theorem from Completeness & Compactness in Rⁿ says compact ⟺ closed and bounded. This is a special case of the general three-way equivalence we just proved: in $\mathbb{R}^n$, bounded and totally bounded are the same thing (because $\mathbb{R}^n$ is finite-dimensional, so a single bounded ball contains only finitely many small balls’ worth of volume). And a closed subset of a complete space is complete. So in $\mathbb{R}^n$, closed + bounded is the same as complete + totally bounded, which by Theorem 3 is the same as compact. The equivalence “bounded ⟺ totally bounded” fails in infinite dimensions — and we are about to see exactly how badly.

The closed unit ball $\mathcal{B} = \{f \in C([0, 1]) : \|f\|_\infty \leq 1\}$ in the sup-norm is closed and bounded, but not compact. This is the counterexample promised in Topic 3.

Proof. Consider the sequence $f_n(x) = \sin(n\pi x)$ for $n = 1, 2, 3, \ldots$. Each $f_n$ is continuous and $\|f_n\|_\infty = 1$, so every $f_n \in \mathcal{B}$. We claim that no subsequence of $(f_n)$ is Cauchy in the sup-norm — and therefore no subsequence converges, so $\mathcal{B}$ is not sequentially compact and (by Theorem 3) not compact.

To see why, fix distinct indices $m \neq n$. The functions $\sin(m\pi x)$ and $\sin(n\pi x)$ are orthogonal in $L^2[0, 1]$:

$\int_0^1 \sin(m\pi x) \sin(n\pi x) \, dx = 0.$

Expanding $\int_0^1 (\sin(m\pi x) - \sin(n\pi x))^2 \, dx$ using orthogonality and $\int_0^1 \sin^2(k\pi x)\,dx = 1/2$:

$\int_0^1 (\sin(m\pi x) - \sin(n\pi x))^2 \, dx = \tfrac{1}{2} + \tfrac{1}{2} - 0 = 1.$

Now use the bound $\|f\|_2^2 \leq \|f\|_\infty^2$ on $[0, 1]$: if the sup-norm of the difference were less than 1, the integral above would be less than 1. But the integral equals 1, so

$\|f_m - f_n\|_\infty \geq \|f_m - f_n\|_2 = 1 \quad \text{for all } m \neq n.$

Every pair of terms in the sequence is at sup-norm distance at least 1 — so no subsequence is Cauchy, and no subsequence converges.

The lesson. In an infinite-dimensional space, closed and bounded is not enough for compactness. The missing ingredient is equicontinuity: the family $\sin(n\pi x)$ is bounded, but the functions “wiggle faster and faster” — for large $n$ there is no single $\delta$ that controls $|f_n(x) - f_n(y)|$ uniformly across the family. We will name this obstruction and package it into the Arzelà–Ascoli theorem in Section 9.

Let $(X, d)$ be a complete metric space and let $T : X \to X$ be a contraction with Lipschitz constant $0 \leq k < 1$, meaning

$d(T(x), T(y)) \leq k \cdot d(x, y) \quad \text{for all } x, y \in X.$

Then:

1. $T$ has a unique fixed point $x^* \in X$ — a point with $T(x^*) = x^*$.
2. For any starting point $x_0 \in X$, the iterates $x_{n+1} = T(x_n)$ converge to $x^*$.
3. The convergence rate is geometric: $d(x_n, x^*) \leq \dfrac{k^n}{1 - k} \cdot d(x_0, x_1)$.

Fix a starting point $x_0 \in X$ and define the iterates $x_{n+1} = T(x_n)$. The contraction property applied to consecutive iterates gives

$d(x_{n+1}, x_n) = d(T(x_n), T(x_{n-1})) \leq k \cdot d(x_n, x_{n-1}).$

By induction, this bound extends to

$d(x_{n+1}, x_n) \leq k^n \cdot d(x_1, x_0).$

Now we show $(x_n)$ is Cauchy. For any $m > n$, the triangle inequality stacks up $m - n$ consecutive-iterate bounds:

$d(x_m, x_n) \leq \sum_{j = n}^{m-1} d(x_{j+1}, x_j) \leq d(x_1, x_0) \sum_{j = n}^{m-1} k^j.$

The geometric-series tail bound gives $\sum_{j = n}^{m-1} k^j \leq \dfrac{k^n}{1 - k}$, so

$d(x_m, x_n) \leq \frac{k^n}{1 - k} \cdot d(x_1, x_0).$

Since $k < 1$, the right-hand side tends to $0$ as $n \to \infty$ uniformly in $m$ — the sequence $(x_n)$ is Cauchy. Because $X$ is complete, $(x_n)$ converges to some limit $x^* \in X$.

Next we show $x^*$ is a fixed point. Contractions are Lipschitz (with constant $k$), hence continuous, so we can pass $T$ through the limit:

$T(x^*) = T\big(\lim_{n \to \infty} x_n\big) = \lim_{n \to \infty} T(x_n) = \lim_{n \to \infty} x_{n+1} = x^*.$

Uniqueness: suppose $y^*$ is another fixed point of $T$. Then

$d(x^*, y^*) = d(T(x^*), T(y^*)) \leq k \cdot d(x^*, y^*).$

Since $k < 1$, this forces $d(x^*, y^*) = 0$, so $y^* = x^*$.

Finally, the convergence rate. Letting $m \to \infty$ in the Cauchy-tail bound $d(x_m, x_n) \leq \frac{k^n}{1-k} \cdot d(x_1, x_0)$, the left side converges to $d(x^*, x_n)$, proving

$d(x_n, x^*) \leq \frac{k^n}{1 - k} \cdot d(x_0, x_1).$

The Picard–Lindelöf existence-and-uniqueness theorem for first-order ODEs from First-Order ODEs rewrites the initial-value problem $y'(t) = f(t, y(t))$, $y(0) = y_0$ as the fixed-point equation

$y(t) = y_0 + \int_0^t f(s, y(s)) \, ds.$

Defining the Picard operator $\Phi[y](t) = y_0 + \int_0^t f(s, y(s)) \, ds$, a fixed point of $\Phi$ is exactly a solution of the ODE. Topic 21 showed that when $f$ is Lipschitz in its second argument with constant $L$ and we work on a short interval $[0, h]$ with $Lh < 1$, the Picard operator is a contraction on $C([0, h])$ with the sup-norm — constant $Lh$ — and $C([0, h])$ is complete by Theorem 2. The Banach theorem then produces a unique fixed point, which is the ODE solution. The abstract theorem we just proved is what makes Topic 21’s concrete argument go through.

The implicit Euler method from Numerical Methods for ODEs requires solving the nonlinear equation $y_{n+1} = y_n + h \cdot f(t_{n+1}, y_{n+1})$ at each time step. Newton’s method reduces this to iterating $g(y) = y - [I - h f_y(t_{n+1}, y)]^{-1}(y - y_n - h f(t_{n+1}, y))$. For sufficiently small $h$, the derivative of $g$ at the root is bounded in norm by some $k < 1$, so by the mean value theorem $g$ is a contraction on a small neighborhood of the root and Banach’s theorem guarantees convergence. Topic 24 invoked this conclusion in its convergence analysis without proof; we have now proved the general result.

The proof of the Inverse Function Theorem in Topic 12 constructs a local inverse by setting up the fixed-point equation $x = \phi(x)$ where $\phi$ is a map on a small ball in $\mathbb{R}^n$ whose Jacobian is bounded below 1 in operator norm. Banach’s theorem then gives a unique fixed point, which is the preimage of the target point. The whole construction is a single application of Theorem 4 to a carefully chosen contraction — the argument was specialized to $\mathbb{R}^n$ in Topic 12, but the contraction-mapping framework developed here shows why it works in full generality.

Topic 20 proved that Bernstein operators $B_n f$ converge uniformly to $f$ on $[0, 1]$ — a statement that lives entirely in the metric space $(C([0, 1]), d_\infty)$. But $B_n$ is not a contraction: it does not uniformly shrink sup-norm distances between arbitrary functions. Bernstein convergence is therefore not an application of Theorem 4; it is a different kind of approximation phenomenon that happens to share the same ambient metric space. The general lesson: sitting inside a complete metric space does not automatically make every self-map a contraction, and density-of-polynomials results require separate machinery (probabilistic moment arguments in the Bernstein case). What Topic 20 does share with this topic is the underlying metric-space structure — the sup-norm distance between $B_n f$ and $f$ is how “convergence” is defined there, and that distance is a metric in exactly the sense defined here.

A topological space is a pair $(X, \tau)$ where $X$ is a set and $\tau$ is a collection of subsets of $X$ — called open sets — satisfying:

1. $\emptyset \in \tau$ and $X \in \tau$.
2. Arbitrary unions of elements of $\tau$ are in $\tau$.
3. Finite intersections of elements of $\tau$ are in $\tau$.

The collection $\tau$ is called a topology on $X$. A metric space $(X, d)$ induces a topology $\tau_d = \{\, U \subseteq X : U \text{ is open in the metric-space sense}\,\}$ by Proposition 1, so every metric space is a topological space. But there are topological spaces with no compatible metric, and those are strictly more general.

Let $X$ be an uncountable set — for concreteness, take $X = \mathbb{R}$. Define $\tau$ to consist of the empty set together with all subsets $U \subseteq X$ whose complement $X \setminus U$ is finite:

$\tau = \{\emptyset\} \cup \{\, U \subseteq X : |X \setminus U| < \infty\,\}.$

This is a topology: the empty set and $X$ are in $\tau$, arbitrary unions of cofinite sets are cofinite (their complement is an intersection of finite sets, which is finite), and finite intersections of cofinite sets are cofinite (their complement is a finite union of finite sets, which is finite). So $(X, \tau)$ is a valid topological space.

But $(X, \tau)$ is not metrizable: there is no metric that generates this topology. Here is one quick reason. In the cofinite topology, any two non-empty open sets must intersect (the union of their complements is finite, so the intersection of the sets themselves is cofinite, hence non-empty). So for any two distinct points $x, y \in X$, every neighborhood of $x$ intersects every neighborhood of $y$. In a metric space we can always separate distinct points by disjoint balls $B(x, d(x,y)/3)$ and $B(y, d(x,y)/3)$ — a property called the Hausdorff property. The cofinite topology on an uncountable set fails this, so no metric can generate it.

A function $f : (X, \tau_X) \to (Y, \tau_Y)$ between topological spaces is continuous if for every open set $V \in \tau_Y$, the preimage $f^{-1}(V) \in \tau_X$.

A homeomorphism between topological spaces $(X, \tau_X)$ and $(Y, \tau_Y)$ is a bijection $f : X \to Y$ such that both $f$ and $f^{-1}$ are continuous. Two spaces related by a homeomorphism are topologically equivalent — they have the same open sets up to relabeling, the same convergence, the same compactness properties, and the same connectedness properties.

Topic 15 mentioned simply connected domains and fundamental groups as the setting for its discussion of conservative vector fields and path independence. These concepts live at the general-topology level: a domain is simply connected if every closed loop in it can be continuously contracted to a point, a property defined entirely in terms of continuous maps (homotopies of loops), and the fundamental group $\pi_1(X, x_0)$ classifies loops up to continuous deformation. Fundamental groups make sense in any topological space and are preserved by homeomorphism — they are a topological invariant. Metric spaces inherit these invariants whenever they carry the metric-induced topology, which is almost always in analysis. The Urysohn Metrization Theorem gives a clean sufficient condition: a topological space is metrizable if (and essentially only if) it is second-countable, regular, and Hausdorff. Most spaces that appear in analysis satisfy these conditions, which is why metric-space theory covers so much ground. But the topological vocabulary — connectedness, path-connectedness, fundamental groups, homotopy — belongs to the general theory, not to metric spaces in particular.

A family $\mathcal{F} \subseteq C([a, b])$ is equicontinuous if for every $\varepsilon > 0$ there exists a single $\delta > 0$ — depending on $\varepsilon$ but not on $f$ — such that for every $f \in \mathcal{F}$ and every $x, y \in [a, b]$ with $|x - y| < \delta$:

$|f(x) - f(y)| < \varepsilon.$

A family $\mathcal{F} \subseteq C([a, b])$ is uniformly bounded if there exists $M > 0$ such that

$|f(x)| \leq M \quad \text{for every } f \in \mathcal{F} \text{ and every } x \in [a, b].$

Equivalently, the family fits inside a single closed ball of the sup-norm: $\|f\|_\infty \leq M$ for all $f \in \mathcal{F}$.

A subset $\mathcal{F} \subseteq C([a, b])$ (with the sup-norm metric) has compact closure if and only if $\mathcal{F}$ is uniformly bounded and equicontinuous.

Equivalently: every uniformly bounded and equicontinuous sequence $(f_n)$ in $C([a, b])$ has a subsequence that converges uniformly.

We prove both directions. The forward direction is the routine one; the reverse direction is the workhorse “diagonal subsequence” argument.

($\Leftarrow$) Uniformly bounded + equicontinuous ⟹ sequentially compact closure. Let $(f_n)$ be a sequence in $\mathcal{F}$. We will extract a subsequence that is Cauchy in the sup-norm; by completeness of $C([a, b])$ (Theorem 2) it then converges, proving sequential compactness.

Step 1: pointwise convergence on a countable dense set. Enumerate the rationals in $[a, b]$ as $r_1, r_2, r_3, \ldots$ — a countable dense set. At $r_1$, the sequence of real numbers $(f_n(r_1))$ is bounded (by uniform boundedness of $\mathcal{F}$), so by the Bolzano-Weierstrass theorem it has a convergent subsequence, which we index as $f_{n_k^{(1)}}$. Next, the sequence $(f_{n_k^{(1)}}(r_2))$ is bounded in $\mathbb{R}$, so we extract a further subsequence $f_{n_k^{(2)}}$ converging at $r_2$ — and still converging at $r_1$, since subsequences of convergent sequences converge to the same limit. Iterate this procedure: at step $j$ we have a subsequence $f_{n_k^{(j)}}$ that converges at each of $r_1, r_2, \ldots, r_j$.