Level 2: Topology – The Canvas

The Defect. Continuity is first met as an $\varepsilon$–$\delta$ manipulation on intervals of real numbers. The definition works, but it hides what continuity actually depends on: not distances, but which sets count as neighbourhoods. Topology isolates exactly that data. The reward is that the two deepest facts of elementary analysis, the intermediate value theorem and the extreme value theorem, stop being consequences of formulas and become consequences of space itself: the interval is connected, and it is compact.

The One-Line Reveal. A topology is a declaration of which sets count as “neighbourhoods”; a continuous map is one whose declared neighbourhoods pull back to declared neighbourhoods; and every theorem of this level follows from two properties a space may have: connectedness (it cannot be split) and compactness (every cover admits a finite sublist).

Scope. Level 1 supplied the language (categories, universal properties, quotients by relations). This level needs exactly two inputs from analysis, both stated explicitly and proved in the Real Analysis track:

1. The completeness axiom: every nonempty set of real numbers bounded above has a least upper bound $\sup S$, characterised by (i) $s \leq \sup S$ for all $s \in S$, and (ii) if $c < \sup S$, then some $s \in S$ has $c < s$. Its first consequence is used repeatedly below, so we prove it here. Archimedean property: for every real $c$ there is a natural number $n > c$. Proof. If $\mathbb{N}$ had an upper bound $b$, then $s = \sup \mathbb{N}$ would exist; property (ii) supplies $n \in \mathbb{N}$ with $n > s - 1$, and then $n + 1 \in \mathbb{N}$ exceeds the upper bound $s$: contradiction. $\blacksquare$
2. Elementary-calculus continuity: the standard functions ($\sin$, $\cos$, $\tan$, $\arctan$) are continuous on their domains, and their bijection properties are as trigonometry describes. Where these enter, the entry is flagged.

Everything else is proved here.

Definition. A metric on a set $X$ is a function $d: X \times X \to [0, \infty)$ with $d(x, y) = 0$ iff $x = y$, $d(x, y) = d(y, x)$, and the triangle inequality $d(x, z) \leq d(x, y) + d(y, z)$. The pair $(X, d)$ is a metric space. The open ball of radius $r > 0$ around $x$ is $B(x, r) = \{y \in X : d(x, y) < r\}$. Euclidean space $\mathbb{R}^n$ carries $d(u, v) = \sqrt{\sum_i (u_i - v_i)^2}$; the triangle inequality for this $d$ is proved in the Real Analysis track and used freely here.

Definition. A subset $U \subseteq X$ of a metric space is open if for every $x \in U$ there is $r > 0$ with $B(x, r) \subseteq U$.

Proposition 1.1. Every open ball is an open set.

Proof. Let $y \in B(x, r)$, so $d(x, y) < r$. Put $\varepsilon = r - d(x, y) > 0$. For any $z \in B(y, \varepsilon)$, the triangle inequality gives $d(x, z) \leq d(x, y) + d(y, z) < d(x, y) + \varepsilon = r$, so $z \in B(x, r)$. Hence $B(y, \varepsilon) \subseteq B(x, r)$: every point of the ball has a ball around it inside the ball. $\blacksquare$

Proposition 1.2. Arbitrary unions of open sets are open, and finite intersections of open sets are open. Both can fail for infinite intersections.

Proof. Unions. Let $U = \bigcup_\alpha U_\alpha$ with each $U_\alpha$ open, and let $x \in U$. Then $x \in U_\alpha$ for some $\alpha$, and some ball $B(x, r) \subseteq U_\alpha \subseteq U$: so $U$ is open. Finite intersections. Let $x \in U_1 \cap \cdots \cap U_n$; for each $i$ there is $r_i > 0$ with $B(x, r_i) \subseteq U_i$, and $B(x, \min_i r_i)$ is contained in every $U_i$, hence in the intersection. Failure for infinite intersections: $\bigcap_{n \geq 1} (-1/n, 1/n) = \{0\}$ in $\mathbb{R}$ (a real number in every $(-1/n, 1/n)$ satisfies $|t| < 1/n$ for all $n$, forcing $t = 0$ by the Archimedean property, proved in the Scope), and $\{0\}$ is not open: no ball around $0$ fits inside a single point. $\blacksquare$

The two properties just proved are the ones the theorems of analysis actually consume. We promote them to a definition.

Lemma 1.3 (generated topologies). For any set $X$ and any collection $\mathcal{S}$ of subsets of $X$, there is a smallest topology containing $\mathcal{S}$: the intersection of all topologies on $X$ that contain $\mathcal{S}$.

Proof. An intersection of topologies is a topology (each axiom is an “every member has…” or “any union/finite intersection of members is a member” statement, checked member-wise over the intersection). The intersection is over a nonempty family: the power set of $X$, containing $\mathcal{S}$, is itself a topology. Being the intersection of all containing topologies, it is contained in each of them, hence smallest. $\blacksquare$

Definition. A topology on a set $X$ is a collection $\mathcal{T}$ of subsets (the open sets) with $\emptyset, X \in \mathcal{T}$, closed under arbitrary unions and finite intersections. A topological space is a set with a topology. A metric space yields a topology by taking as open sets all unions of open balls (Definition above, justified by Propositions 1.1 and 1.2).

Examples.

• Metric topologies: $\mathbb{R}$, $\mathbb{R}^n$ as above.
• Discrete: every subset open. Axioms are immediate.
• Indiscrete: only $\emptyset$ and $X$ open. Immediate.
• Cofinite: a subset is open if it is empty or its complement is finite. Unions: the complement of a union is an intersection of complements; a finite intersection of finite sets is finite, so every union of cofinite sets is cofinite. Finite intersections: the complement of a finite intersection is a finite union of finite complements, finite. $\blacksquare$

Quick check.

1. What makes a set open in a metric space?
2. Are arbitrary unions of opens open? Finite intersections?
3. Give an infinite family of opens with a non-open intersection.
4. What is the cofinite topology?
5. What is a base, and what does second countable mean?

Answers: 1. Every point has a ball around it contained in the set. 2. Yes; yes. Infinite intersections can fail. 3. $\bigcap_n (-1/n, 1/n) = \{0\}$, not open. 4. Opens are the empty set and complements of finite sets. 5. A base is opens whose unions give every open set; second countable means a countable base.

Definition. For a function $f: X \to Y$ and a subset $B \subseteq Y$, the preimage is $f^{-1}(B) = \{x \in X : f(x) \in B\}$ (no inverse function is required; $f^{-1}$ here is set-valued bookkeeping). The function $f$ between topological spaces is continuous if $f^{-1}(U)$ is open in $X$ for every open $U \subseteq Y$.

Theorem 2.1 (composition). If $f: X \to Y$ and $g: Y \to Z$ are continuous, then $g \circ f: X \to Z$ is continuous.

Proof. For $U \subseteq Z$ open, $(g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$ (set-chase an element: $x$ maps into $U$ iff $f(x)$ maps into $U$). $g^{-1}(U)$ is open in $Y$ by continuity of $g$, and its preimage under $f$ is open in $X$ by continuity of $f$. $\blacksquare$

Theorem 2.2 ($\varepsilon$–$\delta$ is topology). For $f: X \to Y$ between metric spaces, the following are equivalent: (i) $f$ is continuous by the open-preimage definition; (ii) for every $x \in X$ and $\varepsilon > 0$ there is $\delta > 0$ with $d(x, x') < \delta \implies d(f(x), f(x')) < \varepsilon$.

Proof. (i) $\implies$ (ii): the ball $B(f(x), \varepsilon)$ is open (Proposition 1.1), so $f^{-1}(B(f(x), \varepsilon))$ is an open set containing $x$; being open, it contains some ball $B(x, \delta)$, and points of that ball map into $B(f(x), \varepsilon)$, which is (ii). (ii) $\implies$ (i): let $U \subseteq Y$ open and $x \in f^{-1}(U)$. Then $f(x) \in U$, and openness of $U$ gives an $\varepsilon$-ball around $f(x)$ inside $U$. Item (ii) supplies a $\delta$-ball around $x$ mapping into it, so $B(x, \delta) \subseteq f^{-1}(U)$. Every point of $f^{-1}(U)$ has a ball inside $f^{-1}(U)$: it is open. $\blacksquare$

Definition. A homeomorphism is a continuous bijection $f: X \to Y$ whose inverse $f^{-1}: Y \to X$ is also continuous. Spaces related by a homeomorphism are homeomorphic; a property preserved by homeomorphisms is a topological invariant.

Example (affine homeomorphisms). Any two nonempty open intervals $(a,b)$, $(c,d)$ of $\mathbb{R}$ are homeomorphic. The affine map $f(t) = \frac{d-c}{b-a} (t - a) + c$ is a bijection $(a, b) \to (c, d)$ (solve for $t$), and it is continuous with continuous inverse: every open subset of $\mathbb{R}$ is a union of the open intervals that openness supplies at each of its points, and the preimage of an interval $(u, v)$ under an affine map $t \mapsto \alpha t

• \beta$is the interval$\big((u - \beta)/\alpha, (v - \beta)/\alpha\big)$(endpoints swapped when$\alpha < 0$, still open), so preimages of opens are unions of intervals: open.$\blacksquare$

Example ($(0,1) \cong \mathbb{R}$). The map $t \mapsto \tan\!\big(\pi(t - \tfrac{1}{2})\big)$ is a continuous bijection $(0, 1) \to \mathbb{R}$ with continuous inverse $s \mapsto \tfrac{1}{\pi}\arctan(s) + \tfrac{1}{2}$. Continuity of $\tan$ and $\arctan$ on their domains and the bijection property are the elementary-calculus inputs from the Scope. So the whole real line is, topologically, a finite interval. $\blacksquare$

Quick check.

1. Define continuity through preimages.
2. Why does composition preserve continuity?
3. What is a homeomorphism?
4. Which pairs of intervals are provably homeomorphic on this page?
5. Which theorem makes $\varepsilon$–$\delta$ equal to the open-set definition?

Answers: 1. The preimage of every open set is open. 2. $(g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$ and preimages inherit openness step by step (Theorem 2.1). 3. A continuous bijection with continuous inverse. 4. Any two nonempty open intervals, by affine maps; also $(0,1) \cong \mathbb{R}$ via $\tan$. 5. Theorem 2.2.

Definition. A subset $F \subseteq X$ is closed if its complement $X \setminus F$ is open. The interior $A^{\circ}$ of $A$ is the union of all open sets contained in $A$; the closure $\overline{A}$ is the intersection of all closed sets containing $A$. A neighbourhood of $x$ is an open set containing $x$.

Proposition 3.1. Arbitrary intersections of closed sets are closed, and finite unions of closed sets are closed; $\overline{A}$ is the smallest closed set containing $A$, and $A^{\circ}$ is the largest open set contained in $A$.

Proof. By De Morgan’s laws, complements convert intersections of closed sets into unions of open sets and finite unions into finite intersections, so Proposition 1.2’s axioms convert directly. The closure of $A$ is a closed set containing $A$ (every set in the intersection contains $A$), and any closed $F \supseteq A$ participates in that intersection, so $\overline{A} \subseteq F$. Interior is dual. $\blacksquare$

Proposition 3.2 (closure via neighbourhoods). $x \in \overline{A}$ if and only if every neighbourhood of $x$ meets $A$.

Proof. ($\implies$) If some neighbourhood $N$ of $x$ missed $A$, then $A \subseteq X \setminus N$, and $X \setminus N$ is closed; so $\overline{A} \subseteq X \setminus N$, excluding $x \in N$: contradiction. ($\Longleftarrow$) If $x \notin \overline{A}$, then $N := X \setminus \overline{A}$ is an open neighbourhood of $x$ (Proposition 3.1, $\overline{A}$ closed) disjoint from $A \subseteq \overline{A}$. $\blacksquare$

Example. $\overline{(0, 1)} = [0, 1]$ in $\mathbb{R}$. Every point of $[0, 1]$ has every neighbourhood meeting $(0, 1)$: a neighbourhood of $0$ contains $(-\varepsilon, \varepsilon)$ for some $\varepsilon > 0$, which meets $(0, 1)$ in $(0, \min(\varepsilon, 1))$; similarly at $1$; points inside are trivial. Proposition 3.2 puts all of $[0,1]$ in the closure. A point $x \notin [0, 1]$ has a neighbourhood avoiding $(0, 1)$: if $x < 0$, the ball of radius $|x|/2$ lies below $0$; if $x > 1$, the ball of radius $(x-1)/2$ lies above $1$. $\blacksquare$

Definition. $x$ is a limit point of $A$ if every neighbourhood of $x$ meets $A$ in some point other than $x$ itself; points of $A$ that are not limit points of $A$ are isolated.

Quick check.

1. Define closed.
2. Describe $\overline{A}$ in two equivalent ways.
3. What is a limit point of $A$?
4. Compute $\overline{[0,1)}$ in $\mathbb{R}$.
5. What does a point outside $\overline{A}$ have?

Answers: 1. Its complement is open. 2. The smallest closed superset; equivalently, the set of points whose every neighbourhood meets $A$. 3. A point whose every neighbourhood meets $A$ in a point other than itself. 4. $[0, 1]$ (Proposition 3.2’s example method). 5. A neighbourhood disjoint from $A$.

Definition (subspace). For $A \subseteq X$, the subspace topology on $A$ has open sets exactly the intersections $A \cap U$ with $U$ open in $X$ (axioms: check each against the corresponding axiom in $X$).

Proposition 4.1 (characteristic property of the subspace). A map $g: Y \to A$ is continuous into $A$ if and only if the composite $Y \to A \hookrightarrow X$ is continuous into $X$.

Proof. ($\implies$) For $U$ open in $X$, the preimage of $A \cap U$ under $g$ is open in $Y$ (subspace definition, $A \cap U$ is open in $A$), and that same set is the preimage of $U$ under the composite. ($\Longleftarrow$) For $V$ open in $A$, write $V = A \cap U$ with $U$ open in $X$; then $g^{-1}(V) = g^{-1}(U)$, open by the composite’s continuity. $\blacksquare$

Definition (product). For spaces $X, Z$, a box is a subset $U \times V$ with $U$ open in $X$ and $V$ open in $Z$. Boxes are closed under finite intersections ($(U \times V) \cap (U' \times V') = (U \cap U') \times (V \cap V')$), so the unions of boxes form a topology (unions of unions are unions), and it is the smallest topology in which every box is open: that is the product topology on $X \times Z$.

Proposition 4.2 (characteristic property of the product). The projections $\pi_X: X \times Z \to X$, $\pi_Z$ are continuous, and a map $f = (f_1, f_2): Y \to X \times Z$ is continuous if and only if both components $f_1 = \pi_X \circ f$ and $f_2 = \pi_Z \circ f$ are continuous.

Proof. $\pi_X^{-1}(U) = U \times Z$ is a box, so $\pi_X$ is continuous; similarly $\pi_Z$. ($\implies$) $f_1$ and $f_2$ are compositions of continuous maps (Theorem 2.1). ($\Longleftarrow$) Let $W \subseteq X \times Z$ be open; by definition $W = \bigcup_\alpha (U_\alpha \times V_\alpha)$, and

a union of finite intersections of open sets: open. $\blacksquare$

This is a universal property in the sense of Level 1A: a map into a product is exactly a pair of maps into the factors, with continuity requiring nothing extra. The product is the categorical product in $\mathbf{Top}$.

Quick check.

1. What are the opens of a subspace $A \subseteq X$?
2. When is $g: Y \to A$ continuous?
3. What is a box in $X \times Z$?
4. What does the product topology contain at minimum?
5. State the product’s characteristic property.

Answers: 1. Intersections $A \cap U$ with $U$ open in $X$. 2. Exactly when its composite with the inclusion $A \hookrightarrow X$ is (Proposition 4.1). 3. $U \times V$ with $U$, $V$ open. 4. All unions of boxes. 5. $f$ into $X \times Z$ is continuous iff both components are (Proposition 4.2).

Definition. A space $X$ is Hausdorff if for any two distinct points $x \neq y$ there are disjoint open sets $U \ni x$ and $V \ni y$.

Proposition 5.1. Every metric space is Hausdorff, and in a Hausdorff space every singleton $\{x\}$ is closed and limits of sequences are unique.

Proof. Hausdorff: for $x \neq y$ put $r = d(x, y) / 2 > 0$. If $z$ lay in $B(x, r) \cap B(y, r)$, the triangle inequality would give $d(x, y) \leq d(x, z) + d(z, y) < r + r = d(x, y)$, a contradiction; so the balls are disjoint. Singletons closed: for $y \neq x$, the Hausdorff pair $(U_y, V_y)$ for $(y, x)$ has $x \notin U_y$, so $X \setminus \{x\} = \bigcup_{y \neq x} U_y$ is open. Limits: say $x_n \to x$ when every neighbourhood of $x$ contains all but finitely many $x_n$. If $x_n \to x$ and $x_n \to y$ with $x \neq y$, the disjoint neighbourhoods $U \ni x$, $V \ni y$ each contain a tail; but two tails of one sequence overlap, and the overlap lies in $U \cap V = \emptyset$: contradiction. $\blacksquare$

Quick check.

1. Define Hausdorff.
2. Why are metric spaces Hausdorff?
3. What does Hausdorff give for singletons?
4. What does it give for convergent sequences?
5. Is the indiscrete two-point space Hausdorff?

Answers: 1. Distinct points have disjoint neighbourhoods. 2. Balls of radius $d(x,y)/2$ separate (Proposition 5.1). 3. Closed. 4. At most one limit. 5. No: the only nonempty open set is the whole space.

Definition. A separation of a space $X$ is a pair of disjoint nonempty open sets $A, B \subseteq X$ with $A \cup B = X$. $X$ is connected if no separation exists; otherwise disconnected.

Theorem 6.1 (the interval is connected). Every closed bounded interval $[a, b]$ is connected (a one-point interval admits no separation, since its only nonempty open subset is the whole space; assume $a < b$).

Proof. Suppose $[a, b] = A \cup B$, a separation, and swap the names so $a \in A$. Write $A = [a,b] \cap A'$ with $A'$ open in $\mathbb{R}$; since $A'$ is open and contains $a$, some $\varepsilon' > 0$ has $[a, a + \varepsilon') \subseteq A$, so the set

contains points greater than $a$. $C$ is bounded above by $b$; let $s = \sup C$ (completeness axiom), so $a < s \leq b$. First, every point $t \in [a, b]$ with $t < s$ lies in $A$: the sup property (ii) gives $u \in C$ with $t < u$, and $[a, t] \subseteq [a, u] \subseteq A$. Now if $s \in B$, openness of $B$ gives $(s - \varepsilon, s] \subseteq B$ for some $\varepsilon > 0$, chosen smaller than $2(s - a)$ so that $s - \varepsilon/2 \in [a, b]$; but the point $s - \varepsilon/2 < s$ lies in $A$ by the previous paragraph, contradicting disjointness. So $s \in A$. Openness of $A$ gives $(s - \varepsilon, s + \varepsilon) \cap [a, b] \subseteq A$ for some $\varepsilon > 0$. If $s < b$, take $\varepsilon' = \min(\varepsilon/2, (b - s)/2) > 0$: then $[a, s + \varepsilon'] \subseteq A$, so $s + \varepsilon' \in C$, contradicting $s = \sup C$. Hence $s = b$, and the first paragraph with the present paragraph gives $[a, b] \subseteq A$, so $B = \emptyset$: no separation exists. $\blacksquare$

Theorem 6.2 (images of connected sets). If $f: X \to Y$ is continuous and $X$ is connected, then $f(X)$ is connected in its subspace topology.

Proof. Suppose $f(X) = A \cup B$ is a separation (both open in $f(X)$, disjoint, nonempty). Write $A = f(X) \cap A'$ with $A'$ open in $Y$. Then $f^{-1}(A) = f^{-1}(A')$ is open in $X$, and likewise $f^{-1}(B)$; they are disjoint (disjointness of $A$ and $B$), nonempty (surjectivity onto $f(X)$), and their union is $X$: a separation of the connected $X$. $\blacksquare$

Corollary 6.3 (every interval is connected). Open, half-open, bounded or unbounded intervals of $\mathbb{R}$ are connected.

Proof. Given a separation $(A, B)$ of an interval $I$, pick $a \in A$, $b \in B$, swap so $a < b$; then $[a, b] \subseteq I$, and $A \cap [a,b]$, $B \cap [a,b]$ are disjoint nonempty open (in $[a,b]$) sets covering $[a,b]$, contradicting Theorem 6.1. $\blacksquare$

Corollary 6.4 (intermediate value theorem). If $f: [a, b] \to \mathbb{R}$ is continuous and $c$ lies strictly between $f(a)$ and $f(b)$, then $f(x) = c$ for some $x \in [a, b]$.

Proof. Suppose not. Then $A = f^{-1}((-\infty, c))$ and $B = f^{-1}((c, \infty))$ are open (Theorem 2.2 renders continuity as open-preimages, and half-lines are open in $\mathbb{R}$), disjoint, and cover $[a, b]$ (no point maps to $c$); they are nonempty because $f(a) < c < f(b)$ places $a$ in $A$ and $b$ in $B$ up to swapping. That is a separation of the connected interval $[a, b]$: contradiction. $\blacksquare$

Lemma 6.5 ($\sqrt{2}$ is irrational).

Proof. Suppose $\sqrt{2} = p/q$ in lowest terms. Squaring: $p^2 = 2q^2$, so $p^2$ is even, hence $p$ is even (odd $p$ makes $p^2$ odd); write $p = 2r$. Then $2r^2 = q^2$, so $q$ is even too. Both even contradicts lowest terms. $\blacksquare$

Proposition 6.6 ($\mathbb{Q}$ is disconnected). The rationals with the subspace topology of $\mathbb{R}$ are disconnected.

Proof. Let $\alpha = \sqrt{2}$, which is not rational (Lemma 6.5) and satisfies $1 < \alpha < 2$. The sets $A = \mathbb{Q} \cap (-\infty, \alpha)$ and $B = \mathbb{Q} \cap (\alpha, \infty)$ are open in the subspace $\mathbb{Q}$; they are nonempty ($0 \in A$, $1 \in B$), disjoint, and their union is $\mathbb{Q}$ (no rational equals $\alpha$). $\blacksquare$

This is the first crack in the arithmetic line: connectedness is a property of completed space. It distinguishes $\mathbb{R}$ from $\mathbb{Q}$ topologically, before any measure or limit theory.

Quick check.

1. Define separation and connectedness.
2. Why is a continuous image of a connected space connected?
3. Why is every interval connected, given $[a,b]$ is?
4. State the intermediate value theorem’s mechanism.
5. Is $\mathbb{Q}$ connected?

Answers: 1. A separation is a cover by two disjoint nonempty opens; connected means none exists. 2. A separation of the image pulls back to one of the domain (Theorem 6.2). 3. A hypothetical separation restricts to a closed subinterval, contradicting Theorem 6.1. 4. Nonattainment of the intermediate value splits $[a,b]$ into two preimage half-spaces, a separation (Corollary 6.4). 5. No: split at $\sqrt{2}$ (Proposition 6.6).

Definition. An open cover of $X$ is a collection of open sets with $X \subseteq \bigcup_\alpha U_\alpha$. A finite subcover is a finite subcollection that still covers. $X$ is compact if every open cover has a finite subcover.

Theorem 7.1 (Heine–Borel for the interval). $[a, b]$ is compact.

Proof. Let $\mathcal{U}$ be an open cover of $[a, b]$. Say $t \in [a,b]$ is good if $[a, t]$ admits a finite subcover from $\mathcal{U}$, and let $C$ be the set of good points. $C$ contains $a$ (a member of $\mathcal{U}$ covers $a$, and $\{U\}$ covers $[a, a]$), and is bounded above by $b$; let $s = \sup C$ (completeness axiom), so $a \leq s \leq b$.

First, $s \in C$. Choose $U \in \mathcal{U}$ with $s \in U$; openness gives $(s - \varepsilon, s + \varepsilon) \subseteq U$ for some $\varepsilon > 0$. Since $s - \varepsilon/2 < s = \sup C$, the sup property supplies $t \in C$ with $t > s - \varepsilon/2$; the finite cover of $[a, t]$, plus the single set $U$, covers $[a, t] \cup (s - \varepsilon, s + \varepsilon) \supseteq [a, s]$. So $s$ is good.

Now if $s < b$, shrink $\varepsilon$ so that $s + \varepsilon/2 \leq b$; the same finite cover plus $U$ covers $[a, s] \cup (s - \varepsilon, s + \varepsilon) \supseteq [a, s + \varepsilon/2]$, making the point $s + \varepsilon/2$ good, contradicting $s = \sup C$. Hence $s = b \in C$: the finite subcover covers $[a, b]$. $\blacksquare$

Theorem 7.2 (constructions). (i) A closed subset of a compact space is compact. (ii) A continuous image of a compact space is compact. (iii) A compact subset of a Hausdorff space is closed. (iv) A compact subset of $\mathbb{R}$ is closed and bounded, and conversely every closed bounded subset of $\mathbb{R}$ is compact.

Proof. (i) Let $K \subseteq X$ be closed, $X$ compact, and $\mathcal{V}$ an open cover of $K$ by opens of $X$. Then $\mathcal{V} \cup \{X \setminus K\}$ covers $X$ ($X \setminus K$ is open); a finite subcover of $X$ minus the member $X \setminus K$ is a finite subcover of $K$.

(ii) Let $f: X \to Y$ be continuous, $X$ compact, and $\mathcal{V}$ an open cover of $f(X)$ by opens of $Y$. The preimages $f^{-1}(V)$ cover $X$ and are open (continuity); a finite subcover $f^{-1}(V_1), \dots, f^{-1}(V_n)$ makes $V_1, \dots, V_n$ cover $f(X)$ (surjectivity onto the image).

(iii) Let $K \subseteq X$ be compact, $X$ Hausdorff, $y \notin K$. For each $x \in K$, the Hausdorff property supplies disjoint open $U_x \ni y$ and $V_x \ni x$. The sets $\{V_x\}_{x \in K}$ cover $K$; compactness gives a finite sublist $V_{x_1}, \dots, V_{x_n}$. Then $U = U_{x_1} \cap \cdots \cap U_{x_n}$ is an open neighbourhood of $y$ disjoint from every $V_{x_i}$, hence from their union, which contains $K$. So every $y \notin K$ has a neighbourhood avoiding $K$: the complement of $K$ is open.

(iv) If $K \subseteq \mathbb{R}$ is compact, the cover $\{(-n, n)\}_{n \in \mathbb{N}}$ has a finite subcover, exhibiting a bound; closedness is (iii) with $\mathbb{R}$ Hausdorff (Proposition 5.1). Conversely, if $K$ is closed and bounded, then $K \subseteq [-R, R]$ for some $R > 0$, which is compact by Theorem 7.1, and (i) makes $K$ compact. $\blacksquare$

Corollary 7.3 (extreme value theorem). A continuous $f: [a, b] \to \mathbb{R}$ attains a maximum and a minimum value.

Proof. The image $f([a, b])$ is compact (Theorem 7.2 ii, Theorem 7.1), hence closed and bounded in $\mathbb{R}$ (Theorem 7.2 iv). Let $M = \sup f([a,b])$, which exists by boundedness. If $M$ were not attained, the complement of the image would be an open set containing $M$, so an $\varepsilon$-ball around $M$ would miss the image; but the sup property (ii) gives a value $v > M - \varepsilon/2$ with $v \leq M < M + \varepsilon/2$, placing a point of the image inside that ball: contradiction. So $f(x) = M$ for some $x$. For the minimum, apply the same argument to $-f$, which is continuous (the preimage of an open interval $(u, v)$ under $y \mapsto -y$ is the open interval $(-v, -u)$). $\blacksquare$

Example (a continuous bijection need not be a homeomorphism). Let $S^1 = \{(u, v) \in \mathbb{R} \times \mathbb{R} : u^2 + v^2 = 1\}$, a subspace of the product space $\mathbb{R} \times \mathbb{R}$ (Section 4), and let $w: [0, 2\pi) \to S^1$, $w(t) = (\cos t, \sin t)$. Then $w$ is a continuous bijection that is not a homeomorphism. This example is the reason the definition of homeomorphism insists on the inverse.

Proof. Continuity: $\cos$ and $\sin$ are continuous (elementary-calculus input from the Scope), so $w$ is continuous as a map into $\mathbb{R} \times \mathbb{R}$ (Proposition 4.2), and into $S^1$ by the characteristic property of the subspace (Proposition 4.1). Bijectivity: surjective by trigonometry (every point of the circle is reached at some angle in $[0, 2\pi)$), and injective because angles in $[0, 2\pi)$ with equal cosine and equal sine are equal (again trigonometry). Not a homeomorphism: the full map $W: [0, 2\pi] \to S^1$ on the closed interval is continuous (same formula, restricted) and surjective, so $S^1$ is compact (Theorems 7.1 and 7.2 ii). If the inverse of $w$ were continuous, then $[0, 2\pi)$ would be the continuous image of the compact $S^1$, hence compact (Theorem 7.2 ii). But $[0, 2\pi)$ is not compact: the collection $\{(-1,\, 2\pi - 1/n) \cap [0, 2\pi)\}_{n \in \mathbb{N}}$ covers it (for $t < 2\pi$ the number $2\pi - t$ is positive, so the Archimedean property gives $n > 1/(2\pi - t)$, i.e. $t > 2\pi - 1/n$), each member is open in the subspace, and no finite subcollection covers: its largest member is $(-1, 2\pi - 1/N)$, and the point $2\pi - 1/(2N)$ lies above $2\pi - 1/N$. So no such homeomorphism exists. $\blacksquare$

Invariants. Theorems 6.2 and 7.2(ii) say connectedness and compactness pass along continuous surjections; applying this to a homeomorphism in both directions, both properties are topological invariants. Connectedness separates $\mathbb{R}$ from $\mathbb{Q}$ (Proposition 6.6); compactness separates $[0, 2\pi]$ from $[0, 2\pi)$.

Quick check.

1. Define compactness.
2. Why is a closed subset of a compact space compact?
3. Why is a continuous image of a compact space compact?
4. Why are compact subsets of Hausdorff spaces closed?
5. What mechanism proves the extreme value theorem?
6. Why is $[0, 2\pi)$ not compact?

Answers: 1. Every open cover has a finite subcover. 2. Add the complement to the cover; a finite subcover minus the complement covers the subset (Theorem 7.2 i). 3. Pull the cover back, take a finite subcover, push it forward (Theorem 7.2 ii). 4. Each outside point separates finitely from the set (Theorem 7.2 iii). 5. The image is compact, hence closed and bounded; the supremum is then attained (Corollary 7.3). 6. The cover $\{(-1, 2\pi - 1/n)\}$ has no finite subcover (wrap-map example).

1. Prove that $\mathbb{R}$ with the cofinite topology is connected. (§1, §6 – two disjoint nonempty cofinite sets would have complements whose union is all of $\mathbb{R}$, making $\mathbb{R}$ finite.)
2. Give an infinite family of open subsets of $\mathbb{R}$ whose intersection is not open, and prove your claim. (§1 – $\bigcap_n (-1/n, 1/n) = \{0\}$, using the Archimedean property.)
3. Prove the $(ii) \implies (i)$ direction of Theorem 2.2 without looking. (§2 – the $\delta$-ball provided at each point of $f^{-1}(U)$ shows it is open.)
4. Prove that the product of two discrete spaces is discrete. (§4 – singletons $\{x\} \times \{z\}$ are boxes, and every subset is a union of singletons.)
5. Prove that $\mathbb{R}$ is not compact. (§7 – the cover $\{(-n, n)\}$ has no finite subcover: a finite subcollection has a largest $N$, and the point $N + 1$ escapes.)
6. Prove that $\{1/n : n \in \mathbb{N}\} \cup \{0\}$ is compact. (§7 – prove it is closed (for $y > 0$ off the set, the Archimedean property leaves only finitely many $1/n$ above $y/2$, so the nearest member of the set to $y$ exists, and the ball of half that distance, capped at $y/2$ to also avoid $0$, separates $y$ from the set) and bounded, then quote the converse in Theorem 7.2(iv).)
7. Prove that $S^1$ is connected. (§6, §7 – it is the continuous image of the connected interval $[0, 2\pi]$ under the wrap map $W$, by Theorems 6.2 and 7.1.)

Problems. The quick checks above tested recall. These chain several results and require full written arguments; hints are included where the jump is large.

8. Prove that the intersection of two compact subsets of a Hausdorff space is compact. (Each is closed by Theorem 7.2 iii; the intersection is a closed subset of one of them, so Theorem 7.2 i applies.)
9. Prove that a continuous map from a connected space into a discrete space is constant. (The image is connected and discrete, hence a single point.)
10. Prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic. (Compactness is a topological invariant; $[0,1]$ is compact and $(0,1)$ is not, by the cover $\{(1/n, 1)\}$.)

The entire page follows from one shift: neighbourhood structure, not distance, is what continuity and the great theorems use. Once “open” is axiomatised, the intermediate value theorem is what connectedness of the interval says, the extreme value theorem is what compactness of the interval says, and the wrap map shows precisely why the inverse in a homeomorphism is part of the definition.

• Level 3A: Smooth Manifolds – charts and atlases make the canvas locally linear; this page’s subspace, product, and Hausdorff theorems are the raw material of the manifold definition.
• Real Analysis – where the two declared inputs of this page (the completeness axiom and the continuity of the elementary functions) are actually proved.
• Topology: A Categorical Approach – Bradley, Bryson and Terilla’s book, free and open access from MIT Press; its Chapter 2 covers this page’s material with the categorical accent that Level 1A supplied.