It's well-known that any Tychonoff space $X$ can be embedded in $[0,1]^k$ for some cardinal $k$. It's natural to ask what the smallest such $k$ is (let's call it $k(X)$). However, this probably probably doesn't deserve to be called dimension, since it fails to satisfy some desirable properties. For instance, although $k(\text{point}) = 0$, we have $k(\text{2 points}) = 1$. This leads me to consider a local version:

If $X$ is Tychonoff and $x \in X$, let $D(X, x)$ be the smallest cardinal $k$ such that some neighbourhood of $x$ can be embedded in $[0,1]^k$, and let $D(X) = \sup_{x \in X} D(X, x)$. This satisfies some obvious properties:

- If $\{U_\alpha\}$ is an open cover of $X$, then $\dim(X) = \sup D(U_\alpha)$.
- If $A$ is a subspace of $X$, then $D(A) \le D(X)$. Equality holds if, for instance, $A$ contains a neighbourhood of a point $x$ with $D(X,x) = D(X)$.
- $D(X) = n$ if $X$ is a $n$-dimensional manifold.
- $D(X \times Y) \le D(X) + D(Y)$.

This last inequality may be strict; for instance, if $X$ is the Cantor set, then $D(X) = 1$ and $X \times X \cong X$.

If $\dim$ denotes the Lebesgue covering dimension, then for $X$ compact, we have $\dim(X) \le \dim([0,1]^{D(X)}) = D(X)$. I have no idea when equality holds (it would if $\dim(X) = n$ implied that $X$ could be *locally* embedded in $\mathbb{R}^n$, but I don't know if that's true).

Is there a name for this $D$, or has such an invariant been studied before? How is this related to other notions of dimension for a topological space? In particular, are there classes of nice spaces (for instance, compact metrizable) on which they agree?

T, i.e. the set of all $\ (x\ y\ z)\ \in\ \mathbb R^3\ $ such that $\ x\cdot y=x\cdot z=y\cdot z=0\ $ and $\ 0\le x+y+z\le 1,\ $ has $\ \dim T=1\ $ but $\ T\ $ cannot be embedded into $\ \mathbb R,\ $ not even locally around the ramification point $\ (0\ 0\ 0).$ – Włodzimierz Holsztyński Sep 2 '14 at 5:32