LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces and let $f : X \to Y$ be an isometric embedding (Isometry, isometric embedding, and the subspace metric on a subset). Write $Z : = f [X] \subseteq Y$ with its subspace metric $d_{Z}$ . Then:

$f$ is injective (Injection, surjection, bijection).
$f$ , viewed as a map $X \to Z$ , is an isometry.
$f [B_{X} (x, r)] = B_{Z} (f (x), r)$ for every $x \in X$ and $r > 0$ (Open ball, closed ball and sphere in a metric space).
A subset $U \subseteq X$ is open in $(X, d_{X})$ if and only if $f [U]$ is open in $(Z, d_{Z})$ (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). So $U \mapsto f [U]$ is a bijection from the metric topology of $X$ onto the subspace topology of $f [X]$ , and $f$ is a homeomorphism onto its image.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ , $(Y, d_{Y})$ , an isometric embedding $f : X \to Y$ , the image $Z : = f [X]$ with the subspace metric $d_{Z} = d_{Y} ↾ (Z \times Z)$ , and the map $g : Z \to X$ inverse to $f : X \to Z$ once claim 2 is available.

[A1]

Isometric embedding: $d_{Y} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ for all $x, x^{'} \in X$ ; the subspace metric on $Z$ is the restriction of $d_{Y}$ (Isometry, isometric embedding, and the subspace metric on a subset).

[A2]

Separation (M1): $d_{X} (x, x^{'}) = 0$ if and only if $x = x^{'}$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

Balls: $B_{X} (x, r) = {x^{'} : d_{X} (x, x^{'}) < r}$ , and likewise in $Z$ with $d_{Z}$ (Open ball, closed ball and sphere in a metric space).

[L2]

Continuity in the $ε$ - $δ$ form (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form), and the equivalence of continuity with "preimages of open sets are open" (Metric continuity characterisations, with countable choice for the sequential converse, 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).

[L3]

A bijection $h$ and its inverse satisfy $h [S] = (h^{- 1})^{- 1} [S]$ and $h^{- 1} [h [S]] = S$ for every subset $S$ of the domain (Injection, surjection, bijection).

Proof

technique · direct

1.1

Injectivity: if $f (x) = f (x^{'})$ then $d_{X} (x, x^{'}) = d_{Y} (f (x), f (x^{'})) = 0$ , hence $x = x^{'}$ by (M1); this is claim 1.

A1A2

2.1

As a map $X \to Z$ the function $f$ is surjective, $Z$ being its image by definition, and it is injective by step 1.1, so it is a bijection $X \to Z$ ; and $d_{Z} (f (x), f (x^{'})) = d_{Y} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ , since $d_{Z}$ is the restriction of $d_{Y}$ , so it is an isometry, which is claim 2.

step 1.1A1

3.1

Both $f : X \to Z$ and its inverse $g : Z \to X$ are continuous, with $δ : = ε$ serving at every point in both directions, because $d_{Z} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ and, writing $z = f (x)$ , $z^{'} = f (x^{'})$ , also $d_{X} (g (z), g (z^{'})) = d_{Z} (z, z^{'})$ .

step 2.1A1L2

3.2

Claim 3: $f [B_{X} (x, r)] = {f (x^{'}) : d_{X} (x, x^{'}) < r} = {f (x^{'}) : d_{Z} (f (x), f (x^{'})) < r}$ , and as $f$ is onto $Z$ the latter set is ${z \in Z : d_{Z} (f (x), z) < r} = B_{Z} (f (x), r)$ .

step 2.1A1L1

4.1

By [L2] applied to the continuous maps of step 3.1, the preimage under $f : X \to Z$ of every open subset of $Z$ is open in $X$ , and the preimage under $g$ of every open subset of $X$ is open in $Z$ .

step 3.1L2

5.1

Claim 4: for $U \subseteq X$ we have $f [U] = g^{- 1} [U]$ , so if $U$ is open in $X$ then $f [U]$ is open in $Z$ by step 4.1; conversely $U = f^{- 1} [f [U]]$ , so if $f [U]$ is open in $Z$ then $U$ is open in $X$ by step 4.1. Hence $U \mapsto f [U]$ maps the topology of $X$ into that of $Z$ , is injective because $f$ is, and is onto because any open $W \subseteq Z$ equals $f [f^{- 1} [W]]$ with $f^{- 1} [W]$ open.

step 2.1step 4.1L3

6.1

Claims 1, 2, 3 and 4 are established by steps 1.1, 2.1, 3.2 and 5.1, so an isometric embedding identifies $X$ with the metric subspace $f [X]$ of $Y$ , as a metric space and hence as a topological one.

step 1.1step 3.2step 5.1∎

Remarks

Only the image carries the right topology. Claim 4 compares the topology of $X$ with the SUBSPACE topology of $f [X]$ , not with the topology of $Y$ . The image of an open set is in general not open in $Y$ : the inclusion of $[0, 1]$ into $R$ is an isometric embedding, and $[0, 1]$ is open in itself but not in $R$ (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).
This is what licenses treating a subset of a metric space as a space in its own right, and it is used on the companion page whenever a subset of $R$ is called a metric space.

Depends on

Used by

Dependency tree · two levels

25 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

Isometry (Wikipedia) (standard reference, not scraped)
Embedding (Wikipedia) (standard reference, not scraped)
Subspace topology (Wikipedia) (standard reference, not scraped)