This should be really well known but I don't seem to find a statement about it nor a question in MO answering this. 


Consider a Compact Hausdorff topological space $X$. The [cohomological dimension](http://en.wikipedia.org/wiki/Cohomological_dimension) of $X$ is the first natural number where the cohomology vanishes. The [topological dimension](http://en.wikipedia.org/wiki/Inductive_dimension) is as defined in wikipedia (see also Hurewicz Wallman's book). 


For CW complexes, it seems easy to prove that finiteness of topological dimension implies finiteness of the other. However, for general spaces I don´t figure out how to prove (or construct a counterexample) to 


Q) Does finite topological dimension of $X$ implies finite cohomological dimension? 


Maybe there are well known inequalities also (for example, it seems that for spaces homotopy equivalent to CW complexes one should have that the topological dimension is bigger than or equal to the cohomological one). Information about this kind of inequalities would be welcome. 



Topological dimension versus cohomological dimension
Well, I think it depends on which dimension you mean and which cohomology. The best fit I think is covering dimension and Čech cohomology. The Čech cohomological dimension is indeed bounded (more or less by definition) by the covering dimension.


**Addendum**: Just to comment on BCnrd's comment. The usual definition of cohomological dimension is the vanishing of all sheaf cohomology above that dimension. I however got the impression that the OP interpreted it in the sense of all cohomology with constant coefficients vanishing. In any case the Čech cohomology of sheaves also vanish above the covering dimension so for that it doesn't matter what one means, unless sheaf cohomology is meant. However, a Hausdorff space (I am asuming that that is part of the OP's assumption) that either has finite covering dimension or is compact is paracompact and hence it doesn't matter after all.


