ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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In any metric space the range of a convergent sequence together with its limit is compact, worked out for $\{0\} \cup \{1/(k+1) : k \in \mathbb{N}\}$ in $\mathbb{R}$

Example

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $(x_k)$ be a sequence in $X$ converging to $p \in X$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences), and put

$K \;:=\; \{\, x_k : k \in \mathbb{N} \,\} \cup \{p\}.$

Then $K$ is a compact subset of $X$ (Open cover, subcover, compact metric space, and compact subset of a metric space).

In $\mathbb{R}$ with the usual metric (The absolute value makes $\mathbb{R}$ 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) the sequence $x_k := 1/(k+1)$ converges to $0$ , so $\{0\} \cup \{\, 1/(k+1) : k \in \mathbb{N} \,\}$ is compact. The index is written $k+1$ because $\mathbb{N}$ contains $0$ and $1/0$ is undefined (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Dropping the limit destroys compactness: the range alone is not closed in $\mathbb{R}$ , and a compact subset is closed (A compact subset of a metric space is closed and bounded).

Facts & Assumptions

Given: A metric space $(X,d)$ , a sequence $(x_k)$ in $X$ with $x_k \to p \in X$ , and $K := \{x_k : k \in \mathbb{N}\} \cup \{p\}$ .

[L1]

$A \subseteq X$ is compact exactly when every family $(U_i)_{i \in I}$ of open subsets of $X$ with $A \subseteq \bigcup_{i \in I}U_i$ has finitely many members whose union contains $A$ , or $A = \emptyset$ (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L2]

$x_k \to p$ means: for every rational $\varepsilon > 0$ there is $K_0 \in \mathbb{N}$ with $d(x_k,p) < \varepsilon$ for all $k \ge K_0$ (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ ).

[L3]

$U$ is open when every point of $U$ has a ball around it inside $U$ (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, Open ball, closed ball and sphere in a metric space).

[L4]

A function with domain a natural number all of whose values are nonempty sets has a choice function, in ZF (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

[L5]

For every real $\eta > 0$ there is a natural $N \ge 1$ with $1/N < \eta$ , and $0 < 1/(k+1) \le 1/N$ for $k + 1 \ge N$ ; in $(\mathbb{R},d_{\mathbb{R}})$ the distance is $|x-y|$ (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, The absolute value makes $\mathbb{R}$ 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, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Isometry, isometric embedding, and the subspace metric on a subset).

Verification

technique · direct

1.1

Let $(U_i)_{i \in I}$ be a family of open subsets of $X$ with $K \subseteq \bigcup_{i \in I} U_i$ .

2.1

Since $p \in K$ there is $i^{\ast} \in I$ with $p \in U_{i^{\ast}}$ , and openness gives a real $r > 0$ with $B(p,r) \subseteq U_{i^{\ast}}$ .

L3step 1.1

3.1

Taking a positive rational below $r$ , for instance $1/N$ with $N \ge 1$ a natural and $1/N < r$ , convergence supplies $K_0 \in \mathbb{N}$ with $d(x_k,p) < 1/N < r$ for every $k \ge K_0$ , so $x_k \in B(p,r) \subseteq U_{i^{\ast}}$ for every $k \ge K_0$ .

L2L5step 2.1

4.1

For each $j < K_0$ the set $\{\, i \in I : x_j \in U_i \,\}$ is nonempty, since $x_j \in K$ ; finite choice applied to $j \mapsto$ that set gives indices $i_0, \dots, i_{K_0-1} \in I$ with $x_j \in U_{i_j}$ for every $j < K_0$ , this list being empty when $K_0 = 0$ .

L4step 3.1

5.1

Then $K \subseteq U_{i^{\ast}} \cup U_{i_0} \cup \dots \cup U_{i_{K_0-1}}$ : the point $p$ and every $x_k$ with $k \ge K_0$ lie in $U_{i^{\ast}}$ by steps 2.1 and 3.1, and every $x_j$ with $j < K_0$ lies in $U_{i_j}$ . So finitely many members of the family cover $K$ , and $K$ is compact.

L1step 2.1step 3.1step 4.1

6.1

For the instance in $\mathbb{R}$ , the terms $1/(k+1)$ satisfy $|1/(k+1) - 0| = 1/(k+1) < \varepsilon$ for every $k$ with $k+1 \ge N$ , where $N \ge 1$ is a natural with $1/N < \varepsilon$ ; so $1/(k+1) \to 0$ and $\{0\} \cup \{1/(k+1) : k \in \mathbb{N}\}$ is compact by the general claim.

L2L5step 5.1∎

Remarks

Finitely many exceptional terms is the whole idea. All but finitely many terms are captured by the single member containing the limit, and the remaining ones are finitely many points, each needing one member. That is why the selection at step 4.1 is over a finite index set and costs nothing (Every natural-number-indexed list of nonempty sets has a choice function on its family of values).

Without the limit point the set is not compact. In $\mathbb{R}$ the set $\{1/(k+1) : k \in \mathbb{N}\}$ has $0$ in its closure and does not contain it, so it is not closed and hence not compact (A compact subset of a metric space is closed and bounded).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 85 results over 17 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

Compact space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Example 2.21(e)) (standard reference, not scraped)

In any metric space the range of a convergent sequence together with its limit is compact, worked out for {0}∪{1/(k+1):k∈N}\{0\} \cup \{1/(k+1) : k \in \mathbb{N}\}{0}∪{1/(k+1):k∈N} in R\mathbb{R}R