LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), with total boundedness as in Finite $ε$ -net and totally bounded metric space and boundedness as in Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space. Then:

If $(X, d)$ is totally bounded, it is bounded.
If $(X, d)$ is totally bounded and $A \subseteq X$ , then the metric subspace $(A, d_{A})$ is totally bounded (Isometry, isometric embedding, and the subspace metric on a subset).
If $A \subseteq X$ is totally bounded, so is its closure $\overline{A}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

No choice principle is used. The one selection made is over a finite index set, which Every natural-number-indexed list of nonempty sets has a choice function on its family of values supplies in ZF.

Facts & Assumptions

Given: A metric space $(X, d)$ and a subset $A \subseteq X$ , with $(A, d_{A})$ the metric subspace and $\overline{A}$ the closure of $A$ in $X$ .

[L1]

$(X, d)$ is totally bounded exactly when for every real $ε > 0$ there is a finite $F \subseteq X$ , empty or listable as ${y_{0}, \dots, y_{m}}$ , with $X = ⋃_{y \in F} B (y, ε)$ (Finite $ε$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space).

[L2]

A subset is bounded when it is empty or contained in a ball $B (x_{0}, r)$ with $x_{0}$ in the space and $r > 0$ real (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

A metric satisfies $d (x, z) \leq d (x, y) + d (y, z)$ and $d (x, y) = d (y, x)$ , and $d (y, y) = 0$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

A nonempty finite set of reals has a maximum, which is one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

Balls of a subspace are traces: $B_{A} (a, r) = B_{X} (a, r) \cap A$ for $a \in A$ , $r > 0$ (Isometry, isometric embedding, and the subspace metric on a subset).

[L6]

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).

[L7]

$x \in \overline{A}$ exactly when $B (x, r) \cap A \neq \emptyset$ for every real $r > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

Proof

technique · direct

1.1

If $X = \emptyset$ then $X$ is bounded, emptiness being one of the two cases of the definition.

1.2

Suppose instead $X \neq \emptyset$ , and let $F$ be a finite $1$ -net for $(X, d)$ ; then $F \neq \emptyset$ , since a union over an empty family of balls is empty while $X$ is not, so $F = {y_{0}, \dots, y_{m}}$ for some $m \in N$ .

2.1

The set ${d (y_{0}, y_{j}) : j \leq m}$ is a nonempty finite set of reals, so it has a maximum $R$ , and $R \geq d (y_{0}, y_{0}) = 0$ .

L3L4step 1.2

3.1

Every $x \in X$ lies in $B (y_{j}, 1)$ for some $j \leq m$ , whence $d (y_{0}, x) \leq d (y_{0}, y_{j}) + d (y_{j}, x) < R + 1$ ; so $X \subseteq B (y_{0}, R + 1)$ with $R + 1 > 0$ , and $X$ is bounded.

L1L2L3step 2.1

4.1

Claim 1 is proved, by step 1.1 in the empty case and by step 3.1 otherwise.

step 1.1step 3.1

5.1

Claim 1 being settled, take up claim 2: assume $(X, d)$ totally bounded, let $A \subseteq X$ , let $ε > 0$ be real, fix a finite $(ε / 2)$ -net $F = {y_{0}, \dots, y_{m}}$ for $(X, d)$ , and put $J : = {j \leq m : B (y_{j}, ε / 2) \cap A \neq \emptyset}$ .

step 4.1L1

6.1

If $A = \emptyset$ then the empty set is a finite $ε$ -net for $(A, d_{A})$ ; otherwise fix $a^{*} \in A$ , put $S_{j} : = B (y_{j}, ε / 2) \cap A$ for $j \in J$ and $S_{j} : = A$ for $j \leq m$ with $j \notin J$ , all nonempty, and apply finite choice to the function $j \mapsto S_{j}$ on $σ (m)$ to obtain $a_{0}, \dots, a_{m} \in A$ with $a_{j} \in B (y_{j}, ε / 2)$ for every $j \in J$ .

L6step 5.1

7.1

Put $G : = {a_{0}, \dots, a_{m}} \subseteq A$ , a finite set; given $a \in A$ there is $j \leq m$ with $a \in B (y_{j}, ε / 2)$ , so $j \in J$ and $d (a, a_{j}) \leq d (a, y_{j}) + d (y_{j}, a_{j}) < ε / 2 + ε / 2 = ε$ , that is $a \in B_{A} (a_{j}, ε)$ .

L3L5step 6.1

8.1

So $G$ is a finite $ε$ -net for $(A, d_{A})$ , and since $ε > 0$ was arbitrary the subspace $(A, d_{A})$ is totally bounded: claim 2 is proved.

L1step 7.1

9.1

Claim 2 being settled, take up claim 3: assume $A \subseteq X$ totally bounded, let $ε > 0$ be real and fix a finite $(ε / 2)$ -net $F = {b_{0}, \dots, b_{p}} \subseteq A$ for $(A, d_{A})$ , or $F = \emptyset$ when $A = \emptyset$ .

step 8.1L1

10.1

Let $x \in \overline{A}$ ; then $B (x, ε / 2)$ meets $A$ , so there is $a \in A$ with $d (x, a) < ε / 2$ , and $a \in B_{A} (b_{i}, ε / 2)$ for some $i \leq p$ , whence $d (x, b_{i}) \leq d (x, a) + d (a, b_{i}) < ε$ .

L3L5L7step 9.1

11.1

Hence $\overline{A} = ⋃_{i \leq p} B_{\overline{A}} (b_{i}, ε)$ with ${b_{0}, \dots, b_{p}} \subseteq A \subseteq \overline{A}$ finite, so that set is a finite $ε$ -net for the subspace $\overline{A}$ ; as $ε > 0$ was arbitrary, $\overline{A}$ is totally bounded and claim 3 is proved.

L1L5step 10.1

12.1

Claims 1, 2 and 3 hold, by steps 4.1, 8.1 and 11.1 respectively.

step 4.1step 8.1step 11.1∎

Remarks

Where the halving is needed. In claim 2 the net of the subspace has to consist of points of $A$ , and a point $y_{j}$ of a net for $X$ need not lie in $A$ ; moving from $y_{j}$ to a point of $A$ within $ε / 2$ of it costs the other half of $ε$ . The same halving appears in claim 3, where the point being approximated lies in the closure rather than in $A$ .

Claim 1 does not reverse. A bounded metric space need not be totally bounded; FALSE: a bounded metric space is totally bounded states the false converse and $N$ with the discrete metric is bounded and is not totally bounded ↗ exhibits the witness.

Depends on

Used by

ℕ with the discrete metric is bounded and is not totally bounded Counterexample
FALSE: a bounded metric space is totally bounded False statement

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Totally bounded space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)