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Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).

9 votes
Accepted

Boundaries of subsets of simply-connected domains

It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also …
Saúl RM's user avatar
  • 10.6k
4 votes
Accepted

Subset in $[0,1]^k$ with positive density

Apparently not. Let $\gamma=1/3$ and choose some $\varepsilon<\frac{1}{4}$. For any $k\in\mathbb{N}$ we can consider the set $$A=\left\{(x_1,\dots,x_k)\in[0,1]^k;\;\sum_{i=1}^k\lfloor2x_i\rfloor\equiv …
Saúl RM's user avatar
  • 10.6k
7 votes
Accepted

For a closed Riemannian manifold $M$, must the set of points with non-unique closest points ...

More generally, for any closed subset $S$ of a complete manifold $M$, the set of points $x$ at whose minimal distance to $S$ is attained at more than point has measure $0$. Indeed, consider the distan …
Saúl RM's user avatar
  • 10.6k
2 votes
Accepted

Estimating the volume of a convex shape in higher dimensions based only on normal sections

Those constants don't exist for any $d\geq4$, here is an idea of why. For each $\varepsilon>0$ let $A_\varepsilon=\{(x_1,\dots,x_d)\in[-1,1]^d;\lvert (d-1)x_d-\sum_{i=1}^{d-1} x_i\rvert\leq\varepsilon …
Saúl RM's user avatar
  • 10.6k
40 votes
2 answers
2k views

Can the nth projective space be covered by n charts?

That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$? I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t b …
Saúl RM's user avatar
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11 votes
Accepted

Uncountable collections of distinct subsets of an interval (existence)

My comment reposted as an answer: If the continuum hypothesis holds, then we can give a well order $\prec$ to $\mathbb{R}$ isomorphic to the first uncountable ordinal. And then for each $j\in[-1,1]$ w …
Saúl RM's user avatar
  • 10.6k