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✓ 3 results · all verified · 3 also independently AI-judged

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The Topology of Euclidean Space — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

The Euclidean closed ball and sphere worked through the compactness equivalence chart

Example

Assume ${A C}_{ω}$ and $D C$ , let $n \geq 1$ , $c \in R^{n}$ , and $r > 0$ . The closed ball ${\overline{B}}_{2} (c, r)$ and sphere $S_{2} (c, r)$ are compact (For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact). Therefore each is closed and bounded, pseudocompact, countably compact, sequentially compact, limit point compact, and complete and totally bounded; every continuous real-valued function on either set attains both extrema (Assuming ${A C}_{ω}$ and $D C$ , compactness, sequential compactness, countable compactness, limit point compactness, completeness and total boundedness, pseudocompactness, closedness and boundedness, and the extreme-value property are equivalent for nonempty subsets of $R^{n}$ with $n \geq 1$ ). The two sets are those of Euclidean spheres and closed balls as subspaces of $R^{n}$ .

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

The open unit ball in $R^{n}$ is bounded and not compact

Statement refuted

Refuted claim: every bounded subset of $R^{n}$ is compact.

For $n \geq 1$ , the open unit ball $B_{2} (0, 1)$ is bounded but not compact.

Facts & Assumptions

Given: $n \geq 1$ , the open unit ball $B_{2} (0, 1)$ , and a standard unit vector $e_{0}$ .

[A1]

Every bounded Euclidean subset is compact.

[L1]

Euclidean compactness is equivalent to closedness and boundedness (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[L2]

The unit vector $e_{0}$ exists and has Euclidean norm $1$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L3]

The open ball $B_{2} (0, 1)$ consists of the points of Euclidean distance less than $1$ from $0$ , and metric balls form neighbourhoods in the metric topology (Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Counterexample

technique · direct

1.1

The open unit ball is bounded, since every one of its points has distance less than $1$ , hence less than $2$ , from $0$ .

1.2

It is not closed: $e_{0} \notin B_{2} (0, 1)$ , while for every $r > 0$ the point $(1 - ε) e_{0}$ , with $0 < ε < \min (r, 1)$ , lies in both $B_{2} (0, 1)$ and $B_{2} (e_{0}, r)$ . Thus every neighbourhood of $e_{0}$ meets the open unit ball.

L2L3

2.1

By [L1], the bounded nonclosed set $B_{2} (0, 1)$ is not compact. It therefore refutes [A1].

A1L1step 1.1step 1.2∎

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

$R^{n}$ is closed and unbounded and is not compact for $n \geq 1$

Statement refuted

Refuted claim: every closed subset of $R^{n}$ is compact.

For $n \geq 1$ , $R^{n}$ is closed and unbounded, hence it is not compact.

Facts & Assumptions

Given: $n \geq 1$ , Euclidean space $R^{n}$ , and a standard unit vector $e_{0}$ .

[A1]

Every closed Euclidean subset is compact.

[L1]

[L2]

The standard vector $e_{0}$ has Euclidean norm $1$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L3]

For every real radius there is a natural number larger than it (Every complete ordered field is Archimedean).

[L4]

The empty set is open, so the whole space is closed; a metric subset is bounded exactly when it lies in some ball about some centre (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Counterexample

technique · direct

1.1

The whole space $R^{n}$ is closed, since its complement is empty and the empty set is open.

1.2

It is unbounded. Indeed, for an arbitrary centre $c \in R^{n}$ and radius $r > 0$ , choose a natural $k > r + ∥ c ∥_{2}$ by [L3]. The reverse triangle inequality gives $∥ k e_{0} - c ∥_{2} \geq k - ∥ c ∥_{2} > r,$ so $R^{n}$ is contained in no ball.

L2L3L4choose

2.1

By [L1], the closed unbounded space $R^{n}$ is not compact. Hence it refutes [A1].

A1L1step 1.1step 1.2∎

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

The straight segment between two points of an open Euclidean ball stays in the ball

Example

Let $x, y \in B_{2} (c, r) \subseteq R^{n}$ , with $0 \leq t \leq 1$ . The triangle inequality and absolute homogeneity give

$∥ (1 - t) x + t y - c ∥_{2} \leq (1 - t) ∥ x - c ∥_{2} + t ∥ y - c ∥_{2} < r .$

So the segment from $x$ to $y$ stays in the ball. By A finite concatenation of straight segments in $R^{n}$ is a continuous path it is a polygonal path, and $B_{2} (c, r)$ is polygonally connected in the sense of Polygonal paths and polygonally connected subsets of $R^{n}$ . The norm and ball conventions are those of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and Open ball, closed ball and sphere in a metric space.

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

$G L_{1} (R) = R ∖ {0}$ is disconnected, whereas $R^{2} ∖ {0}$ is polygonally connected

Example

The invertible $1 \times 1$ real matrices are the nonzero real numbers, so $G L_{1} (R) = R ∖ {0}$ . This set is disconnected: it contains $- 1$ and $1$ but not the intermediate point $0$ , so it is not order-convex and cannot be connected by The connected subspaces of $R$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $R$ ". In contrast, For $n \geq 2$ , the punctured space $R^{n} ∖ {0}$ is polygonally connected gives polygonal connectedness of $R^{2} ∖ {0}$ . Connectedness is understood through Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31Open item page →

An infinite particular-point space is pseudocompact and not compact

Statement refuted

False claim: every pseudocompact topological space is compact.

Let $X$ be an infinite set with a distinguished point $p$ , carrying the particular-point topology. Then $X$ is pseudocompact and not compact.

Facts & Assumptions

Given: An infinite set $X$ , a point $p \in X$ , and the particular-point topology, whose nonempty open sets are exactly the subsets containing $p$ .

[A1]

Every pseudocompact topological space is compact.

[L1]

The particular-point topology is a topology and has exactly the stated open sets (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L2]

The usual topology on $R$ is Hausdorff, so distinct real numbers have disjoint open neighbourhoods (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L3]

A continuous map pulls back open sets to open sets, and compactness means that every open cover has a finite subcover (Continuity of a map of topological spaces at a point and globally, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L4]

A space is pseudocompact exactly when every continuous real-valued map has bounded image (Pseudocompact space: every continuous real-valued function has bounded image).

Counterexample

technique · contradiction

1.1

Let $f : X \to R$ be continuous. If $f (x) \neq f (p)$ for some $x$ , choose disjoint open neighbourhoods $U$ of $f (x)$ and $V$ of $f (p)$ by [L2]. Then $f^{- 1} [U]$ is open, contains $x$ , and does not contain $p$ , contradicting [L1].

L1L2L3assume-contra

1.2

The family $U : = {{p, x} : x \in X ∖ {p}}$ consists of open sets and covers $X$ .

2.1

Hence every continuous $f : X \to R$ is constant, so its image is bounded. Thus $X$ is pseudocompact by [L4].

step 1.1L4

2.2

No finite subfamily covers $X$ , because its union contains $p$ and only finitely many other points, whereas $X ∖ {p}$ is infinite. Thus $X$ is not compact.

L3step 1.2

3.1

The pseudocompact noncompact space $X$ contradicts [A1], refuting the claim.

A1step 2.1step 2.2discharge-contradiction∎

The Topology of Euclidean Space — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The Euclidean closed ball and sphere worked through the compactness equivalence chart

Example

The open unit ball in RnRn is bounded and not compact

Statement refuted

Facts & Assumptions

Counterexample

RnRn is closed and unbounded and is not compact for n≥1n≥1

Statement refuted

Facts & Assumptions

Counterexample

The straight segment between two points of an open Euclidean ball stays in the ball

Example

GL1(R)=R∖{0}GL1​(R)=R∖{0} is disconnected, whereas R2∖{0}R2∖{0} is polygonally connected

Example

An infinite particular-point space is pseudocompact and not compact

Statement refuted

Facts & Assumptions

Counterexample

Sources

The open unit ball in $R^{n}$ is bounded and not compact

$R^{n}$ is closed and unbounded and is not compact for $n \geq 1$

$G L_{1} (R) = R ∖ {0}$ is disconnected, whereas $R^{2} ∖ {0}$ is polygonally connected