TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Rudin 4.20, the sharp converse: on a noncompact $E \subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous

Statement

Let $E \subseteq \mathbb{R}$ be nonempty and not compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset). Then:

there is a function $f : E \to \mathbb{R}$ , continuous on $E$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point), that is unbounded on $E$ ;
there is a function $g : E \to \mathbb{R}$ , continuous and bounded on $E$ , such that $\sup g[E]$ exists and is not attained; in particular $g$ has no greatest value on $E$ (Maximum and minimum of a set);
if in addition $E$ is bounded (Lower bound, bounded below, bounded set), there is a function $h : E \to \mathbb{R}$ , continuous on $E$ , that is not uniformly continuous on $E$ (Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ ).

Together with A continuous real function on a compact subset of $\mathbb{R}$ is bounded, Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value and Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness this says that compactness is exactly the hypothesis those three theorems need: on a compact set every continuous function is bounded, attains its extrema and is uniformly continuous, and on a set that is not compact each of those three conclusions fails for some continuous function.

Claim 3 carries the boundedness hypothesis because it must. On an unbounded closed set every uniformly continuous function is still uniformly continuous, and a noncompact set may well carry only uniformly continuous functions of interest; what claim 3 asserts is the sharp statement for the bounded case, which is the case Heine-Cantor leaves open. The unbounded case is covered by claims 1 and 2, which hold with no extra hypothesis.

Every witness is exhibited, not merely asserted to exist. Four functions do the work: $x$ and $-1/(1+x^{2})$ when $E$ is unbounded, and $1/(x-x_0)$ and $-|x - x_0|$ when $E$ is bounded, where $x_0$ is then a point of $\overline{E} \setminus E$ .

Facts & Assumptions

Given: A nonempty set $E \subseteq \mathbb{R}$ that is not compact.

[L1]

A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded; so $E$ is not closed or not bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Lower bound, bounded below, bounded set).

[L2]

Boundedness: $S$ is bounded when there are reals $\ell, u$ with $\ell \le s \le u$ for every $s \in S$ ; equivalently when there is a real $M \ge 0$ with $|s| \le M$ for every $s \in S$ . So if $S$ is unbounded then for every real $M > 0$ some $s \in S$ has $|s| > M$ (Lower bound, bounded below, bounded set, Basic properties of the absolute value, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

Closure: $\overline{E}$ is the set of points every neighbourhood of which meets $E$ , it contains $E$ , and $E$ is closed exactly when $E = \overline{E}$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L4]

Algebra of continuous functions: constants, the identity and polynomial functions are continuous on any subset of $\mathbb{R}$ ; sums, scalar multiples, products and absolute values of continuous functions are continuous; and if $q$ is continuous on $S$ and $q(x) \ne 0$ for every $x \in S$ , then $p/q$ is continuous on $S$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, Integer powers $a^m$ ).

[L5]

Suprema: a nonempty subset of $\mathbb{R}$ bounded above has a least upper bound (Complete ordered field (least-upper-bound property)), and for $u = \sup S$ every real $\varepsilon > 0$ admits $s \in S$ with $u - \varepsilon < s$ (Epsilon characterisation of the supremum).

[L6]

Archimedean property in reciprocal form, reciprocals, and squares: for every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ ; $0 < s < t$ implies $0 < 1/t < 1/s$ ; $0 \le a < b$ implies $a^{2} < b^{2}$ ; and $t \ge 1$ implies $t^{2} \ge t$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order, Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$ , Integer powers $a^m$ ).

[L7]

Extension theorem: a uniformly continuous real function on a nonempty $D \subseteq \mathbb{R}$ extends to a continuous function on $\overline{D}$ (A uniformly continuous real function on a subset $D \subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$ , Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ ).

[L8]

Ordered-field arithmetic in $\mathbb{R}$ : totality and trichotomy; $|u| > 0$ exactly when $u \ne 0$ ; $1 + t^{2} \ge 1 > 0$ for every real $t$ ; and the minimum of a two-element set of reals (Ordered field, Basic properties of the absolute value, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · constructive

1.1

By [L1] the set $E$ is not closed or not bounded, and these two possibilities are exhaustive: if $E$ is bounded then it is not closed. The two cases below are treated separately, and claim 3 arises only in the second.

1.2

First case: $E$ is unbounded. Claim 1. Put $f(x) := x$ , continuous on $E$ by [L4]. Given a real $M > 0$ , [L2] supplies $x \in E$ with $|x| > M$ , that is $|f(x)| > M$ ; so $f$ is unbounded on $E$ .

L2L4construct

1.3

First case, claim 2. Put $g(x) := -1/(1+x^{2})$ . The denominator is a polynomial function, continuous by [L4], and satisfies $1 + x^{2} \ge 1 > 0$ by [L8], so $g$ is continuous on $E$ by [L4]; moreover $0 < 1/(1+x^{2}) \le 1$ , so $-1 \le g(x) < 0$ for every $x \in E$ and $g$ is bounded. Hence $g[E]$ is nonempty and bounded above by $0$ , so $u := \sup g[E]$ exists by [L5] and $u \le 0$ .

L4L5L8construct

2.1

First case: the supremum is $0$ and is not attained. Let a real $\varepsilon > 0$ be given and put $M := \max\{1, 1/\varepsilon\} \ge 1$ . By [L2] there is $x \in E$ with $|x| > M$ , so $x^{2} > M^{2} \ge M \ge 1/\varepsilon$ by [L6] and [L8], hence $1 + x^{2} > 1/\varepsilon > 0$ and $1/(1+x^{2}) < \varepsilon$ by [L6], that is $g(x) > -\varepsilon$ . So no real below $0$ is an upper bound of $g[E]$ , and $0$ is one; therefore $u = 0$ . Since $g(x) < 0$ for every $x \in E$ by step 1.3, the value $0$ is not attained, and for each $x \in E$ the number $\varepsilon := -g(x) > 0$ produces by [L5] some $x' \in E$ with $g(x') > -\varepsilon = g(x)$ , so $g$ has no greatest value.

step 1.3L2L5L6L8

2.2

Second case: $E$ is bounded, hence not closed. By [L3] we have $E \subseteq \overline{E}$ and $E \ne \overline{E}$ , so there is $x_0 \in \overline{E} \setminus E$ . Every neighbourhood of $x_0$ meets $E$ by [L3]; and $x - x_0 \ne 0$ for every $x \in E$ , since $x_0 \notin E$ , so $|x - x_0| > 0$ there by [L8].

step 1.1L3L8choose

3.1

Second case, claim 1. Put $f(x) := 1/(x - x_0)$ for $x \in E$ . The denominator is a polynomial function, continuous by [L4], and does not vanish on $E$ by step 2.2, so $f$ is continuous on $E$ by [L4]. Given a real $M > 0$ , step 2.2 supplies $x \in E$ with $|x - x_0| < 1/M$ , and $|x - x_0| > 0$ , so $|f(x)| = 1/|x - x_0| > M$ by [L6]. Hence $f$ is unbounded on $E$ .

step 2.2L4L6construct

3.2

Second case, claim 2. Put $g(x) := -|x - x_0|$ for $x \in E$ , continuous on $E$ by [L4]. Since $E$ is bounded, [L2] gives a real $M \ge 0$ with $|x| \le M$ on $E$ , so $|x - x_0| \le M + |x_0|$ and $-(M + |x_0|) \le g(x) < 0$ for every $x \in E$ : $g$ is bounded, and $g[E]$ is nonempty and bounded above by $0$ . For a real $\varepsilon > 0$ , step 2.2 supplies $x \in E$ with $|x - x_0| < \varepsilon$ , that is $g(x) > -\varepsilon$ ; so $\sup g[E] = 0$ by [L5], and it is not attained because $g(x) < 0$ everywhere on $E$ . As in step 2.1, $g$ therefore has no greatest value on $E$ .

step 2.1step 2.2L2L4L5L8construct

4.1

Second case, claim 3. Put $h := f$ of step 3.1, continuous on $E$ . Suppose $h$ were uniformly continuous on $E$ . By [L7] there would be a continuous $H : \overline{E} \to \mathbb{R}$ with $H(x) = h(x)$ for $x \in E$ , and $x_0 \in \overline{E}$ . Continuity of $H$ at $x_0$ with $\varepsilon := 1$ gives a real $\delta > 0$ such that every $z \in \overline{E}$ with $|z - x_0| < \delta$ satisfies $|H(z) - H(x_0)| < 1$ , hence $|H(z)| < |H(x_0)| + 1 =: B$ , a real with $B > 0$ . Put $r := \min\{\delta, 1/B\} > 0$ ; by step 2.2 there is $x \in E$ with $|x - x_0| < r$ , and then $0 < |x - x_0| < 1/B$ gives $|h(x)| = 1/|x - x_0| > B$ by [L6], while $x \in \overline{E}$ with $|x - x_0| < \delta$ gives $|h(x)| = |H(x)| < B$ . That is impossible, so $h$ is not uniformly continuous on $E$ .

step 2.2step 3.1L6L7L8

5.1

The two cases of step 1.1 are exhaustive, and in each of them claims 1 and 2 have been established by exhibiting the functions named, while claim 3, whose hypothesis places $E$ in the second case, is step 4.1.

step 1.2step 1.3step 2.1step 3.1step 3.2step 4.1discharge-construct: the four witnesses x and -1/(1+x^2) and 1/(x-x_0) and -|x-x_0|∎

Remarks

The bounded non-closed case is where all three failures happen at once. There $x_0 \in \overline{E} \setminus E$ is a hole in the domain, and $1/(x-x_0)$ blows up at it: it is unbounded, it is not uniformly continuous, and $-|x-x_0|$ approaches its supremum $0$ without reaching it. The unbounded case needs a different witness for claim 2, because $-|x - x_0|$ need not be bounded there, and $-1/(1+x^{2})$ is the standard substitute.
Claim 3 is proved through the extension theorem rather than through sequences. The textbook route takes a sequence in $E$ converging to $x_0$ , notes that it is Cauchy, and observes that a uniformly continuous function must carry it to a Cauchy, hence bounded, sequence. Producing that sequence from adherence spends countable choice (A point lies in the closure of $A \subseteq \mathbb{R}$ iff some sequence in $A$ converges to it, so a subset of $\mathbb{R}$ is closed iff it is sequentially closed). The argument above spends none: A uniformly continuous real function on a subset $D \subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$ constructs the extension without selecting anything, and the contradiction is then a single $\varepsilon$ - $\delta$ estimate at the point $x_0$ .
What "not attained" means here, precisely. The supremum of $g[E]$ exists as a real number and equals $0$ , and no point of $E$ has $g$ -value $0$ . That is stronger than saying $g$ has no maximum: it identifies the value the function fails to reach. The companion page works both witnesses out concretely in The identity on $(0,1)$ is bounded with no greatest value, and on $[0,\infty)$ it is continuous and unbounded ↗.

Depends on

Used by

The identity on (0,1) is bounded with no greatest value, and on [0,∞) it is continuous and unbounded Counterexample
FALSE: a continuous real function on a bounded domain attains a greatest value False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 155 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (Thm 4.20) (standard reference, not scraped)
Extreme value theorem (Wikipedia) (standard reference, not scraped)
Uniform continuity (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis (standard reference, not scraped)

Rudin 4.20, the sharp converse: on a noncompact E⊆RE \subseteq \mathbb{R}E⊆R there is an unbounded continuous function and a bounded continuous function with no greatest value, and if EEE is bounded there is a continuous function on EEE that is not uniformly continuous