Can every complex analytic space be covered by Stein spaces of finite embedding dimension?

I am almost sure that ought to be true. Here the definition of embedding dimension I have in mind is $$ \operatorname{embdim}_x(X)=\dim_\mathbb C \mathfrak m_x/\mathfrak m_x^2. $$ I am aware that every complex analytic space has a Stein cover, as well as that there exist complex spaces (Stein no less!) of infinite embedding dimension, though I do not know the explicit construtions (those too would be welcome...). My naive intuition would suggest that embedding dimension is finite locally though, hence the first question.

I would also be happy with the answer to the corresponding algebraic case, complex analytic (resp. Stein) spaces being replaced with schemes (resp. affine schemes).

Now embedding dimension is a sensible concept for any locally ringed spaces over some field $k$ (that is, a morphism of locally ringed spaces $X\to\operatorname{Spec}k$). Supposing the affirmative to the above question (in both the analytic and algebraic setting), we could imagine embedding dimension at a point $x$ to be the lowest dimension of a Stein space (resp. affine scheme) in which a neighbourhood of $x$ could be embedded.

Is there any similar "geometric" picture for locally ringed spaces over a field (or maybe just for certain classes of them)?

From what little I understand of "functor of points" theories (that is where "spaces" are not defined as locally ringed spaces, but rather as functors from some category of rings or similar things to say sets; algebraic spaces, stacks etc.), embedding dimension makes sense there as well, because the Zariski tangent space over a field $k$ to a functor $F$ can often sensibly be definied as $F(k[\varepsilon]/(\varepsilon^2))$ or something similar.

Can the intuition of embedding dimension as related to embeddings be recovered there?

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    $\begingroup$ I recommend to forget about the functor-of-points stuff and locally ringed space generalities here; it does not provide any real illumination for this question. Anyway, every point has an open neighborhood that is an closed analytic subspace of a polydisc, the dimension of which bounds the embedding dimension at all points on that open neighborhood (tangent spaces lie in those of the polydisc). That gives the local finiteness you want. By "fattening up" a discrete subset pointwise in the complex plane we get a 1-dimensional Stein space with infinite global "embedding dimension". $\endgroup$ – user27920 Jul 24 '14 at 7:06
  • $\begingroup$ I see. Thank you for the answer to the admittedly trivial first question. A shame if nothing interesting can be said about the other two, alas it is also expected. $\endgroup$ – A Rock and a Hard Place Jul 24 '14 at 14:39

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