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It is known that by applying the universal coefficient theorem, the singular cohomology of closed manifold with coefficient $\mathbb{Z}_2$ vanishes in high dimensions. But for a metric space $M$ with hausdorff dimension not larger than $n$, where $n\in\mathbb{N}$, it seems impossible to applied the universal coefficient theorem to prove that $H^{n+k}(M;\mathbb{Z}_2)=0$, $k\geq 1$, as $M$ is not necessarily a manifold. Whether there is a way to prove it? Or is there a counterexample?

Edit: By Johannes Hahn's suggestion, I change my question a bit to make it a better one.

The metric space $M$ above with hausdorff dimension not larger than $n$ should replaced by a compact metric space $M$ with topological dimension $n$. Other words will not be changed.

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    $\begingroup$ Considering that the Hausdorff dimension is a metric invariant, but not a topological one, and considering the connection between Hausdorff and inductive dimension (according to wikipedia en.wikipedia.org/wiki/…) one should really ask this question for the inductive dimension. $\endgroup$
    – Johannes Hahn
    Commented Jan 19, 2015 at 15:20
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    $\begingroup$ I was going to suggest the Barratt-Milnor example (ams.org/journals/proc/1962-013-02/S0002-9939-1962-0137110-9) since that's a closed subspace of $\Bbb R^3$ with uncountable homology groups in all degrees. However, they detect this problem with rational homology and so it's not clear to me if the same method works for mod-2 homology. $\endgroup$
    – Tyler Lawson
    Commented Jan 19, 2015 at 15:31
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    $\begingroup$ @ThiKu That's exactly the point: If the cohomology in degrees $> \dim_{ind}(X)$ vanish then they will also vanish in all degree $> \dim_{Haus}(X)$. On the other hand: Vanishing of cohomology is a topological invariant, so if one is able to prove a vanishing result w.r.t. the Hausdorff dimension, then one has also proven a vanishing result for all spaces homoemorphic to X. $\endgroup$
    – Johannes Hahn
    Commented Jan 19, 2015 at 15:42
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    $\begingroup$ @TylerLawson I think there's a follow-up paper by Barratt that might answer this. $\endgroup$
    – Jeremy Rickard
    Commented Jan 19, 2015 at 17:22
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    $\begingroup$ @TylerLawson : it seems that the short proof given by Sergei Melikhov in arxiv.org/abs/0812.1407 adapts to Z/2 coefficients (maybe even simplifies, as Steenrod's realization holds in mod 2 homology). $\endgroup$
    – BS.
    Commented Jan 19, 2015 at 17:23

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Looks like the Hatcher answer in the post dimension is worth looking at for singular cohomology, and the Dranshnikov work for Cech Cohomology.

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  • $\begingroup$ Thank you for your link. It is very helpful for my question! $\endgroup$
    – Lewis Zhang
    Commented Jan 20, 2015 at 5:05

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