Question about Lebesgue Covering Dimension Suppose we have a metric space equipped with two different metrics: $(X,d), (X, d')$. What must be true of the metrics: $d, d'$ in order for $X$ to have the same Lebesgue covering dimension? A sufficient condition is that the metrics would need to be compatible, in that they induce the same topology, is this also a necessary condition? 
Update: 
To make things more explicit, my question goes as follows:
Suppose we have a metrizable space $X$ equipped with two different metrics: $d, d'$ and under these metrics, $X$ has the same Lebesgue Covering Dimension. Is it the case that $(X,d)$ is topologically equivalent to $(X,d')$? If so, great. If not, what is a counterexample? Futhermore, if the answer to my question is "No", what other assumptions are needed for the spaces to be topologically equivalent? Are there examples of metrizable spaces $X$ such that this holds? 
 A: The Lebesgue covering dimension only depends on the topology of the space, not its metric.
And being zero-dimensional is not very special:
So we can start with any set, say of size continuum $\mathfrak{c}$ (like $\mathbb{R}$) and give it any metric that makes it homeomorphic to any of the $\mathfrak{c}$ many non-homeomorphic separable metric zero-dimensional spaces that exist and have lots of examples. A concrete one: take the irrationals in their usual topology, and in the discrete metric, or even the rationals, if you prefer.
There are some theorems, that characterise zero-dimensional metric spaces:

*

*If $X$ is compact, metrisable and zero-dimensional and has no isolated points, it is topologically equivalent to the Cantor set.

*If $X$ is completely metrisable, separable, zero-dimensional and no compact set has non-empty interior (i.e. $X$ is nowhere locally compact) then $X$ is topologically equivalent to the irrationals.

*If $X$ is countable, metrisable (this already implies zero-dimensional) and has no isolated points, it is topologically equivalent to the rationals.

etc etc. So knowing a space is zero-dimensional and having some special properties can characterise spaces, sometimes. But in general, being zero-dimensional and metric is not very special.
