LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology

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) and define, for $x, y \in X$ ,

$d'(x,y) := \min\{\, d(x,y),\ 1 \,\}, \qquad d''(x,y) := \frac{d(x,y)}{1 + d(x,y)} .$

Both are well defined: $d(x,y) \ge 0$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), so $1 + d(x,y) > 0$ and is invertible, and the minimum of a two-element set of reals exists (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set). Then:

$d'$ and $d''$ are metrics on $X$ .
$d'(x,y) \le 1$ and $d''(x,y) < 1$ for all $x,y$ ; hence $(X,d')$ and $(X,d'')$ are bounded metric spaces (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and if $X \ne \emptyset$ then $\operatorname{diam}(X) \le 1$ for both.
$d'$ and $d''$ are each uniformly equivalent to $d$ , hence topologically equivalent to it (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence).

Consequently every metric space carries a bounded metric with exactly the same topology, so boundedness cannot be read off the topology alone.

Facts & Assumptions

Given: A metric space $(X,d)$ , points $x, y, z \in X$ , a real $\varepsilon > 0$ , and the two functions $\varphi_1(t) := \min\{t, 1\}$ and $\varphi_2(t) := t(1+t)^{-1}$ , defined for reals $t \ge 0$ , so that $d' = \varphi_1 \circ d$ and $d'' = \varphi_2 \circ d$ .

[L1]

A metric is nonnegative and satisfies (M1), (M2), (M3) (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

The minimum of a two-element set of reals exists, is one of the two elements, and is a lower bound of both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

$0 < 1$ (The multiplicative identity is positive); a sum of positives is positive and inequalities may be added, in the strict form of Order is preserved by adding a constant and by adding inequalities and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)).

[L4]

Inverses and order: $u > 0$ gives $u^{-1} > 0$ , and $0 < u < v$ gives $0 < v^{-1} < u^{-1}$ (Inverses of positives are positive, and reciprocation reverses order); Inverses of positives are positive, and reciprocation reverses order states only those strict forms, so the nonstrict version used below, that $0 < u \le v$ gives $0 < v^{-1} \le u^{-1}$ , is that statement together with the case $u = v$ , in which the two inverses are equal, the order being total (Ordered field, Complete ordered field (least-upper-bound property)). Multiplying an inequality by a positive preserves it, in the strict form of Sign rules for products and monotonicity of multiplication and, with the same equality case, in the nonstrict form; and $u u^{-1} = 1$ (Field).

[L5]

Bounded subset and diameter: $A$ is bounded when it lies in some ball, and for nonempty bounded $A$ the diameter is the least upper bound of the distances, so any upper bound of those distances bounds the diameter (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space, Suprema and infima are unique).

[L6]

Uniform and topological equivalence, and the implication between them (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence).

Proof

technique · direct

1.1

Properties of $\varphi_1$ on $t \ge 0$ : it is one of $t$ and $1$ , so $0 \le \varphi_1(t) \le t$ and $\varphi_1(t) \le 1$ ; $\varphi_1(t) = 0$ exactly when $t = 0$ , since $1 \ne 0$ ; it is nondecreasing, because for $0 \le s \le t$ the value $\varphi_1(s)$ is $s \le t$ or $1$ , and in both cases it is a lower bound of $\{t,1\}$ , hence at most $\varphi_1(t)$ ; and it is subadditive, since for $s, t \ge 0$ either one of $s, t$ is $\ge 1$ , and then $\varphi_1(s) + \varphi_1(t) \ge 1 \ge \varphi_1(s+t)$ , or both are $< 1$ , and then $\varphi_1(s) + \varphi_1(t) = s + t \ge \varphi_1(s+t)$ .

L2L3

1.2

Properties of $\varphi_2$ on $t \ge 0$ : here $1 + t \ge 1 > 0$ , so $(1+t)^{-1} > 0$ and $(1+t)^{-1} \le 1$ , whence $0 \le \varphi_2(t) \le t$ ; also $t < 1 + t$ gives $\varphi_2(t) < (1+t)(1+t)^{-1} = 1$ ; $\varphi_2(t) = 0$ exactly when $t = 0$ ; $\varphi_2$ is strictly increasing, because $\varphi_2(t) = 1 - (1+t)^{-1}$ and $0 \le s < t$ gives $0 < 1+s < 1+t$ , hence $(1+t)^{-1} < (1+s)^{-1}$ ; and it is subadditive, since for $s,t \ge 0$ one has $0 < 1+s \le 1+s+t$ and $0 < 1+t \le 1+s+t$ , so $\varphi_2(s+t) = s(1+s+t)^{-1} + t(1+s+t)^{-1} \le s(1+s)^{-1} + t(1+t)^{-1} = \varphi_2(s) + \varphi_2(t)$ .

L3L4

1.3

Both $d'$ and $d''$ are symmetric, being $\varphi_i$ applied to the symmetric function $d$ , and both vanish exactly on the diagonal, since $\varphi_i(t) = 0$ exactly when $t = 0$ and $d(x,y) = 0$ exactly when $x = y$ .

2.1

$d'$ is a metric: (M1) and (M2) are step 1.3, and (M3) follows because $d(x,z) \le d(x,y) + d(y,z)$ with $\varphi_1$ nondecreasing and subadditive on nonnegatives gives $d'(x,z) = \varphi_1(d(x,z)) \le \varphi_1(d(x,y) + d(y,z)) \le \varphi_1(d(x,y)) + \varphi_1(d(y,z)) = d'(x,y) + d'(y,z)$ .

step 1.1step 1.3L1

2.2

$d''$ is a metric: identically, using that $\varphi_2$ is increasing and subadditive on nonnegatives, $d''(x,z) \le \varphi_2(d(x,y) + d(y,z)) \le d''(x,y) + d''(y,z)$ .

step 1.2step 1.3L1

2.3

Boundedness: $d'(x,y) \le 1 < 2$ and $d''(x,y) < 1 < 2$ for all $x,y$ , so if $X \ne \emptyset$ then fixing any $x_0 \in X$ gives $X \subseteq B_{d'}(x_0,2)$ and $X \subseteq B_{d''}(x_0,2)$ , while $X = \emptyset$ is bounded outright; and $1$ is an upper bound of all the distances, so $\operatorname{diam}(X) \le 1$ in both metrics when $X \ne \emptyset$ . This is claim 2.

step 1.1step 1.2L3L5

3.1

$d'$ is uniformly equivalent to $d$ : given $\varepsilon > 0$ , take $\delta := \varepsilon$ , so that $d(x,y) < \delta$ gives $d'(x,y) \le d(x,y) < \varepsilon$ ; and take $\delta' := \min\{\varepsilon, 1\} > 0$ , so that $d'(x,y) < \delta' \le 1$ forces $\varphi_1(d(x,y)) \ne 1$ , hence $d'(x,y) = d(x,y)$ by [L2], hence $d(x,y) < \delta' \le \varepsilon$ .

step 1.1step 2.1L2L3

3.2

$d''$ is uniformly equivalent to $d$ : given $\varepsilon > 0$ , take $\delta := \varepsilon$ , so that $d(x,y) < \delta$ gives $d''(x,y) \le d(x,y) < \varepsilon$ ; and take $\delta' := \varphi_2(\varepsilon) > 0$ , so that $d''(x,y) < \delta'$ forces $d(x,y) < \varepsilon$ , since $d(x,y) \ge \varepsilon$ would give $\varphi_2(d(x,y)) \ge \varphi_2(\varepsilon) = \delta'$ by monotonicity.

step 1.2step 2.2L4

4.1

Uniform equivalence implies topological equivalence, so $d'$ and $d''$ have exactly the metric topology of $d$ ; this completes claim 3.

step 3.1step 3.2L6

5.1

Claims 1, 2 and 3 hold by steps 2.1 and 2.2, step 2.3, and steps 3.1, 3.2 and 4.1; hence every metric space carries a bounded metric inducing the same topology.

step 2.1step 2.2step 2.3step 4.1∎

Remarks

Two constructions rather than one, on purpose. $\min\{d,1\}$ is the shorter argument and is the one used by the counterexamples on the companion page; $d/(1+d)$ is strictly less than $1$ everywhere and is strictly increasing in $d$ , which makes it the better behaved of the two when the value of the metric is to be compared, and it is the form that generalises to countable products.
Neither is Lipschitz equivalent to $d$ when $d$ is unbounded. A Lipschitz bound $\alpha d \le d'$ with $\alpha > 0$ would force $d \le \alpha^{-1}$ everywhere, which fails as soon as $d$ takes arbitrarily large values; the real line is the witness (On $\mathbb{R}$ the metrics $|x-y|$ and $\min(|x-y|,1)$ are uniformly but not Lipschitz equivalent ↗).
Boundedness is therefore not a topological property, which is recorded as FALSE: boundedness of a metric space is determined by its topology with the real line as witness.
The bound $\operatorname{diam}(X) \le 1$ need not be an equality, and the two constructions differ on when it is. For a one-point space both new metrics are identically $0$ . For $d' = \min\{d,1\}$ the bound is attained as soon as $d$ takes some value $\ge 1$ , since then $d'$ takes the value $1$ itself, and the companion page computes one such case. For $d'' = d/(1+d)$ the value $1$ is never taken at all, by claim 2, so on a space where $d$ is bounded the diameter in $d''$ is strictly below $1$ : for instance on a two-point space with $d = 1$ the new distance is $1/2$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 18 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

Metric space (Wikipedia) (standard reference, not scraped)
Equivalence of metrics (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §20 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 20: The Metric Topology (East Tennessee State University) (standard reference, not scraped)

min⁡(d,1)\min(d,1)min(d,1) and d/(1+d)d/(1+d)d/(1+d) are metrics uniformly equivalent to ddd, so every metric space carries a bounded metric with the same topology