Real number system as ordered field with LUB property

At a Glance

• Frequency: 2 sub-parts across 2 of 14 years (2013, 2018)
• Priority tier: T3
• Marks (count): 14 (1), 20 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

This atom covers two fundamental properties of $\mathbb{R}$: its topology (every open set decomposes as a countable union of disjoint open intervals) and its ring-homomorphism rigidity (the only ring homomorphisms $\mathbb{R}\to\mathbb{R}$ are $0$ and the identity). Both questions are 14–20 marks and require multi-step, structured proofs — making them high-reward questions for candidates who have memorised the argument skeleton. The 2013 question tests topological understanding of open sets; the 2018 question uses the interplay of additive and multiplicative functional equations to force global dichotomy.

Minimum Theory

Ordered field with LUB. $\mathbb{R}$ is a complete ordered field: every non-empty subset bounded above has a least upper bound (supremum) in $\mathbb{R}$. The rationals $\mathbb{Q}$ are dense in $\mathbb{R}$: between any two real numbers there is a rational, and every non-empty open interval contains a rational.

Structure of open sets. An open set $U \subseteq \mathbb{R}$ is a set where every point has an open interval neighbourhood inside $U$. For each $x \in U$ one can define the maximal open interval $I_x = (\alpha_x, \beta_x) \subseteq U$ through $x$ (with $\alpha_x, \beta_x$ possibly $\pm\infty$). These maximal intervals are the connected components of $U$; any two are either equal or disjoint. Each component contains a rational, and distinct components contain distinct rationals, so there are at most countably many.

Functional equations on $\mathbb{R}$. Cauchy’s additive equation $f(x+y)=f(x)+f(y)$ alone has wild discontinuous solutions (Hamel basis maps). The multiplicative condition $f(xy)=f(x)f(y)$ forces $f(t) = f(\sqrt{t})^2 \ge 0$ for $t \ge 0$, which implies $f$ is monotone non-decreasing. A monotone additive map that agrees with the identity on $\mathbb{Q}$ must equal the identity everywhere by density and squeeze.

Question Archetypes

open-set-structure (1 question; 2013)

Recognition Cues — The question says “open subset of $\mathbb{R}$” and asks you to show it is a “countable disjoint union of open intervals.” The word disjoint (not just union) signals that you need to construct maximal connected components, not merely cover by neighbourhoods.

Solution Template

1. Define $I_x$ for each $x \in U$. Set $\alpha_x = \inf\{a : (a,x] \subseteq U\}$ and $\beta_x = \sup\{b : [x,b) \subseteq U\}$; let $I_x = (\alpha_x, \beta_x)$.
2. Show $I_x \subseteq U$ and $x \in I_x$. Use openness to get $\varepsilon$-neighbourhood $\subseteq U$, giving $\alpha_x \le x - \varepsilon < x$ and $\beta_x \ge x + \varepsilon > x$.
3. Show $I_x$ is maximal: if $J \subseteq U$ is an open interval with $x \in J$, then $J \subseteq I_x$ by definition.
4. Prove distinctness implies disjointness. If $I_x \cap I_y \ne \emptyset$, their union is an open interval in $U$ containing both $x$ and $y$, so $I_x = I_y$ by maximality.
5. Count via rationals. Each $I_x$ is a non-empty open interval $\Rightarrow$ contains a rational $q_{I_x}$. Distinct components have distinct rationals (they are disjoint). The map $I \mapsto q_I$ is injective into $\mathbb{Q}$, so there are countably many components.
6. Conclude $U = \bigsqcup_{n \ge 1} I_n$, a countable disjoint union of open intervals.

Worked Example

2013 Paper 2, 2013-P2-Q2d (14 marks)

Show that every open subset of $\mathbb{R}$ is a countable union of disjoint open intervals.

Let $U \subseteq \mathbb{R}$ be open. We construct the maximal interval through each point.

Step 1 — Maximal interval $I_x$.

Fix $x \in U$. Define

$\alpha_x = \inf\{a \in \mathbb{R} \cup \{-\infty\} : (a, x] \subseteq U\}, \quad \beta_x = \sup\{b \in \mathbb{R} \cup \{+\infty\} : [x, b) \subseteq U\}.$

Since $U$ is open, there exists $\varepsilon > 0$ with $(x-\varepsilon, x+\varepsilon) \subseteq U$, so $\alpha_x \le x - \varepsilon < x < x + \varepsilon \le \beta_x$. Set $I_x = (\alpha_x, \beta_x)$.

For any $y \in I_x$ with $y > x$: by definition of $\beta_x$ there exists $b > y$ with $[x, b) \subseteq U$, so $y \in U$. Similarly for $y < x$. Hence $I_x \subseteq U$. Conversely, any open interval $J \ni x$ contained in $U$ satisfies $J \subseteq I_x$ by definition of $\alpha_x, \beta_x$, so $I_x$ is the largest such interval.

Step 2 — Distinct components are disjoint.

Suppose $I_x \cap I_y \ne \emptyset$. Then $I_x \cup I_y$ is an open interval containing both $x$ and $y$, and $I_x \cup I_y \subseteq U$. By maximality $I_x \cup I_y \subseteq I_x$ (since $I_x$ is the largest interval through $x$ in $U$), and similarly $\subseteq I_y$. Therefore $I_x = I_y$.

Step 3 — Countably many components.

The components $\{I_x : x \in U\}$ (each distinct $I_x$ taken once) partition $U$. Each component is a non-empty open interval, so it contains a rational $q_I \in \mathbb{Q}$. Distinct components are disjoint, hence the map $I \mapsto q_I$ is injective into $\mathbb{Q}$. Since $\mathbb{Q}$ is countable, there are at most countably many components.

$\boxed{U = \bigsqcup_{n \ge 1} I_n, \quad \text{a countable disjoint union of open intervals.}}$

Common Traps

• Partition into maximal connected components, not arbitrary open balls. Without maximality the intervals need not be disjoint.
• The intervals can be unbounded; for example $U = (-\infty, 0) \cup (0, \infty)$ gives two (possibly unbounded) intervals.
• The proof that distinct components are disjoint requires showing $I_x \cap I_y \ne \emptyset \Rightarrow I_x = I_y$; do not claim disjointness without this argument.
• Countability via rationals: each component gets a different rational because the components are disjoint; the injectivity step must be explicit.

functional-equation (1 question; 2018)

Recognition Cues — The problem gives two functional equations: $f(x+y)=f(x)+f(y)$ (Cauchy additive) AND $f(xy)=f(x)f(y)$ (multiplicative). The goal is to show only $f \equiv 0$ or $f = \mathrm{id}$ are possible globally (not just pointwise).

Solution Template

1. Evaluate at $x=y=1$ in the multiplicative equation to get $f(1) \in \{0, 1\}$.
2. If $f(1)=0$: show $f \equiv 0$ using $f(x) = f(x \cdot 1) = f(x)f(1) = 0$.
3. Assume $f(1)=1$. Use additivity to show $f(n) = n$ for $n \in \mathbb{Z}$, then $f(q) = q$ for $q \in \mathbb{Q}$.
4. Derive monotonicity from the multiplicative law: $f(t) = f(\sqrt{t})^2 \ge 0$ for $t \ge 0$; use additivity to deduce $f$ is non-decreasing.
5. Squeeze with rationals: for any $x \in \mathbb{R}$ choose $q_1 \le x \le q_2$ in $\mathbb{Q}$; monotonicity and Step 3 give $q_1 \le f(x) \le q_2$; density of $\mathbb{Q}$ forces $f(x) = x$.
6. Conclude: $f \equiv 0$ or $f = \mathrm{id}_{\mathbb{R}}$.

Worked Example

2018 Paper 2, 2018-P2-Q4a (20 marks)

Suppose $f : \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x)+f(y)$ and $f(xy) = f(x)f(y)$ for all $x, y \in \mathbb{R}$. Show that either $f(x) = 0$ for all $x$ or $f(x) = x$ for all $x$.

Step 1 — $f(1) \in \{0,1\}$.

Set $x = y = 1$ in $f(xy) = f(x)f(y)$:

$f(1) = f(1)^2 \implies f(1)(f(1) - 1) = 0 \implies f(1) = 0 \text{ or } f(1) = 1.$

Case $f(1) = 0$: for any $x$, $f(x) = f(x \cdot 1) = f(x)f(1) = 0$, so $f \equiv 0$. Done.

Henceforth assume $f(1) = 1$.

Step 2 — $f$ fixes $\mathbb{Q}$.

From additivity: $f(0) = 0$ and $f(-x) = -f(x)$. By induction, $f(n) = nf(1) = n$ for all $n \in \mathbb{Z}$. For $q = p/n$ with $p \in \mathbb{Z}$, $n \in \mathbb{N}$: additivity gives $f(n \cdot q) = nf(q)$ and $f(p) = p$, so

$nf(q) = p \implies f(q) = \frac{p}{n} = q.$

Step 3 — Monotonicity via squares.

For $t \ge 0$, write $t = s^2$; then $f(t) = f(s)^2 \ge 0$. Hence $f$ maps non-negatives to non-negatives. If $x \le y$ then $y - x \ge 0$, so $f(y) - f(x) = f(y-x) \ge 0$, i.e., $f$ is non-decreasing.

Step 4 — Squeeze to identity.

Let $x \in \mathbb{R}$ be arbitrary. Choose rationals $q_1 \le x \le q_2$. Monotonicity and Step 2 give

$q_1 = f(q_1) \le f(x) \le f(q_2) = q_2.$

Since $\mathbb{Q}$ is dense, we may choose $q_1 \uparrow x$ and $q_2 \downarrow x$, so $f(x) = x$.

$\boxed{f \equiv 0 \quad \text{or} \quad f = \mathrm{id}_{\mathbb{R}}.}$

Common Traps

• Additivity alone has wild discontinuous solutions (Hamel-basis maps over $\mathbb{Q}$). The multiplicative law is essential — it forces non-negativity on squares, hence monotonicity.
• The conclusion is globally dichotomous: $f(1)=0$ forces $f \equiv 0$ everywhere; once $f(1)=1$, $f$ equals the identity everywhere. There is no pointwise mixture.
• The squeeze step needs both density of $\mathbb{Q}$ and monotonicity; state and use both.

Marks-Aware Writing

14-mark answer (open-set-structure): Present all three steps — construct $I_x$, prove distinct components disjoint, count via rationals — with explicit justifications for $I_x \subseteq U$ and for the injectivity $I \mapsto q_I$. Conclude with the boxed statement.

20-mark answer (functional-equation): All four steps must appear with full justification. Write out the $f(1) \in \{0,1\}$ case split explicitly. The monotonicity paragraph (Step 3) must explain why $f(s^2) = f(s)^2 \ge 0$, and the squeeze paragraph must invoke density of $\mathbb{Q}$ by name.

Practice Set

• 2013-P2-Q2d (14 m) — — Hint: construct maximal open intervals, then inject into $\mathbb{Q}$.
• 2018-P2-Q4a (20 m) — — Hint: the multiplicative law at $x=y=1$ is the first move; monotonicity follows from $f(s^2)=f(s)^2\ge0$.