Suppose $(E,d_E)$ is a metric space that is isometric to $\mathbb R^3$ with Euclidean distance. I have some impedent :
(1) I want to state that $E$ is a metric space of dimension $n$, but as far as I know, we only have definition for dimension of vector space and topological manifold. So how can I say that $E$ is a space of $n-$dimension in this case ?
(2) I wonder with the assumption above, then should $E$ be a $n-$vector space over $\mathbb R$ ? Or is there any counter example for this ?
I hope someone would provide me with a helpful explaination. Thanks.
There are in fact definitions of 'dimension' available for general topological spaces, as the Lebesgue dimension and the inductive dimensions, that are topological invariants: $(E, d)$, being isometric to some $\mathbb{R}^n$, is of course homeomorphic to it,so that $dim(E) = dim(\mathbb{R}^n) = n$ [as these dimensions are such that they agree on $dim(\mathbb{R}^n) = n$]
Yes, it can be, but 'the specifics' may differ based on which isometry is chosen to 'witness/encode' the algebraic information: given two isometries it may well be that they pick out different elements to 'be' the zero vector
In fact, even more is generally true: if $E$ is isometric to some Banach/Hilbert space, then an isometry induces the relevant structure upon $E$ in such a way that it (the isometry) becomes an isomorphism in the appropriate sense
edit; to spell it out a bit more: for a given isometry $f:E \rightarrow V$, where $V$ is the vector space of interest, define
$x +_{E,f} y := f^{-1}(f(x) +_V f(y))$
$\forall \alpha \in \mathbb{K}, \ \alpha \cdot_{E,f} x := f^{-1}( \alpha \cdot_V f(x))$
$||x||_{E,f} := ||f(x)||_V$
$\langle x, y \rangle_{E, f} := \langle f(x), f(y) \rangle_V$
etc., etc., and check that this indeed satisfies the relevant properties