Isometry, isometric embedding, and the subspace metric on a subset

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric).

Isometric embedding and isometry. A function $f:X\to Y$ is an isometric embedding if

$d_Y(f(x),f(x^{\mathrm{\prime }}))=d_X(x,x^{\mathrm{\prime }})\text{for all }x,x^{\mathrm{\prime }}\in X,$

and an isometry if it is in addition bijective (Injection, surjection, bijection). Two metric spaces are isometric if some isometry between them exists.

Subspace metric. Let $A\subseteq X$ and let

$d_A:=d_X\upharpoonright (A\times A)$

be the restriction of $d_X$ to pairs from $A$. Then $d_A$ is a metric on $A$: the three axioms (M1), (M2), (M3) of Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric are conditions on triples of points, and each holds for points of $A$ because it holds for points of $X$. The pair $(A,d_A)$ is the metric subspace $A$ of $X$, and the inclusion $A\to X$ is an isometric embedding by construction. The metric topology of $d_A$ (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) is the subspace topology of $A$.

Balls of a subspace are traces of balls of the ambient space. For $a\in A$ and $r>0$,

$B_A(a,r)=B_X(a,r)\cap A,$

directly from the definitions: a point $z$ lies in the left side exactly when $z\in A$ and $d_A(a,z)=d_X(a,z)<r$ (Open ball, closed ball and sphere in a metric space). This is why the ambient space is always written into the ball notation, and it is the source of every apparent paradox about balls in subspaces.

• An isometric embedding is automatically injective, and it identifies $X$ with the subspace $f[X]$ of $Y$, topology and all; that is An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image. The word embedding is therefore justified rather than merely suggestive.
• A bijective isometric embedding has an isometric inverse. If $f$ is an isometry then $f^{-1}$ satisfies $d_X(f^{-1}(y),f^{-1}(y^{\mathrm{\prime }}))=d_Y(y,y^{\mathrm{\prime }})$, because writing $y=f(x)$ and $y^{\mathrm{\prime }}=f(x^{\mathrm{\prime }})$ turns that into the defining identity of $f$. So "isometric" is a symmetric relation between metric spaces, and it is transitive because a composite of isometries is one.
• Isometry is much finer than having the same topology. Isometric spaces are homeomorphic, but $\mathbb{R}$ with $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ and $\mathbb{R}$ with $\min \{\mathrm{\mid }x-y\mathrm{\mid },1\}$ have the same topology and are not isometric, the second being bounded and the first not ($\min (d,1)$ and $d\mathrm{/}(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded).