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

1.1A1A2

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.

2.1step 1.1A1

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.

3.1step 2.1A1L2

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 $d_{X} (g (z), g (z^{'})) = d_{Z} (z, z^{'})$ .

3.2step 2.1A1L1

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 ${z \in Z : d_{Z} (f (x), z) < r} = B_{Z} (f (x), r)$ .

4.1step 3.1L2

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 $Z$ .

5.1step 2.1step 4.1L3

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.

6.1step 1.1step 3.2step 5.1∎

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.

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.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources