A “dimension” for Tychonoff spaces

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?

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How can the last inequality fail? $\;$ (I see how it can be strict, but that's not failing.) $\hspace{1.5 in}$ –  Ricky Demer Oct 1 '11 at 6:26
Heh, thanks. Fixed. –  Adrian Keet Oct 1 '11 at 9:39
The Sorgenfrey line, or the uncountable product of the two-point discrete space, are both zero dimensional and yet satisfy D(X)=c. So this operation does not behave very nicely with non metrizable spaces. –  mathahada Oct 1 '11 at 12:16
Euclidean shape 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

It is a classic result that a separable metrizable space $X$ can be embeded into $\mathbb{R}^{2n+1}$, where $n=\dim X$; thus $D(X)\le2\dim X+1$. For the universal spaces of Menger and Nöbeling this is optimal and since they are locally homeomorphic to themselves we find that $D(X)=2\dim X+1$ is possible.

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In the uncountable, the least cardinal $k$ such that $X$ embeds into $I^k$ is simply the weight of the space $X$, i.e., the least size of a basis of the topology. A collection $\mathcal B$ is a basis for a topology $\tau$ if $\mathbb B\subseteq\tau$ and every set in $\tau$ is the union of members of $\mathcal B$.

In the light of mathahada's comment, it is interesting to note that this notion of dimension actually makes sense for zero-dimensional spaces. In the case of 0-dim spaces you might want to consider embedding in to $2^k$ rather than $I^k$, though. Now, zero-dimensional spaces are called zero-dimensional for a reason, but I don't think that giving $2^{k}$ dimension $k$ is necessarily pathological.

In any case, for spaces that don't embed into $I^{\aleph_0}$, i.e., for spaces of uncountable weight, you have defined something like a local weight that I haven't seen before. (Not that this would mean anything.)

Edit: Note that in the compact case, this local weight is just the weight: for each point choose an open neighborhood of minimal weight and cover the whole space by finitely many of these. Now weight of the whole space is the maximum of the weights of the finitely many open sets. (This argument assumes that we are dealing with infinite spaces.)

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Defining a special dimension for zero-dimensional spaces might make sense, but it's clearly not the same thing as the dimension that made them zero-dimensional, which is zero. –  Will Sawin Oct 1 '11 at 19:51
Yes, I completely agree, it is not the same dimension that makes it 0-dimensional. –  Stefan Geschke Oct 1 '11 at 20:19
Just recently we discussed with colleagues some zero-dimensional (even compact) spaces which have high Krull dimension - that is, long chains of prime ideals/filters in the lattices of their open sets. Algebraically speaking, although $\mathop{Idl}B$ cannot have long chains of primes for any Boolean algebra $B$, nevertheless $\mathop{Idl}\mathop{Idl}B$ can... –  მამუკა ჯიბლაძე Sep 2 '14 at 4:46