The title basically says it all. Let $A \subset \mathbb{R}^n$ be any subset. If we impose some restrictions on $A$ (like $A$ being a neighborhood retract) one can show that $A$ cannot have homology in dimensions $\geq n$ (e.g. using barycentric subdivision). However, what can we say if we do not impose any restrictions on $A$?
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Maybe look at Milnor's paper (basically, take a certain wedge of spheres and singular homology using rational coefficients): https://www.ams.org/journals/proc/1962-013-02/S0002-9939-1962-0137110-9/S0002-9939-1962-0137110-9.pdf – Alvin Jin May 27 '22 at 23:04
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There exists a compact subset of $\mathbb{R}^3$ which has nonzero homology in infinitely many degrees. Specifically, this is true of the 2-dimensional "Hawaiian earring" which consists of infinitely many shrinking 2-spheres which intersect at a point. More generally, for any $n\geq 2$, the $n$-dimensional Hawaiian earring has nonzero homology in infinitely many degrees. This is a theorem of Barratt and Milnor in
Barratt, M. G.; Milnor, John W., An example of anomalous singular homology, Proc. Am. Math. Soc. 13, 293-297 (1962). ZBL0111.35401.
On the other hand, a subset of $\mathbb{R}^2$ always has trivial homology in degrees $\geq 2$. This is a theorem of Andreas Zastrow which as far as I can tell has never been published but appears in this unpublished preprint. (The corresponding result for $\mathbb{R}^1$ is also rather trivially true.)
Eric Wofsey
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