1

$\mathbb{R} \setminus \{0\}$ is not connected but $\mathbb{R^2} \setminus \{(0,0)\}$ is.

Per this post Connected spaces minus proper subspaces remain connected.

Is there a general rule behind that?

Averroes
  • 409
  • There is a topological notion of dimension: https://math.stackexchange.com/questions/2225877/dimension-of-open-subsets-of-rn My hunch for the general principle is: you can't seperate (chop up) $\mathbb{R}^n$ into two opens unless the subset has dimension at least $n-1$. – balddraz Dec 08 '17 at 16:15
  • There is no general rule here if you are working in the context of arbitrary topological spaces. One can say, however, that a closed subset of topological (say, covering) dimension $<n-1$ cannot separate a connected topological manifold of dimension $n$. Proving this is not easy; one can prove this statement using Poincare duality and Chech cohomology, but, likely, you do not know any of these. – Moishe Kohan Dec 08 '17 at 16:38
  • @ZeroXLR: That is correct if a subset is closed. I am not completely sure about general subsets. – Moishe Kohan Dec 08 '17 at 16:40
  • @MoisheCohen Thank you for the details. Do you have a reference to that result? I read cohomology 9 years ago at uni so both terms ring a bell. Do you recommend an introductory reference to the theory of cohomology (there is a hype about Hatcher but do not know if there are better references around) – Averroes Dec 08 '17 at 16:48

2 Answers2

1

The proof that I know goes through the Alexander/Poincare duality and the Chech cohomology, you can take a quick look at pages 256-257 on Hatcher's book, but for a detailed treatment you will have to look elsewhere, e.g. Eilenberg-Steenrod, Spanier, Dold, or (the most comprehensive and the least comprehensible) Massey's "Homology and Cohomology Theory" book. I will write a proof in the case of subsets of $S^n$, just for simplicity. A minor variation will apply to closed subsets of connected (possibly noncompact) manifolds.

First of all, you need to know what the Lebesgue covering dimension is and what the Chech cohomology groups are. From this, you learn that if $K$ is, say, a a locally compact metrizable space of covering dimension $\le k$, then $H^i(M)=0$ for all $i>k$. (Here and in what follows, cohomology is Chech with integer coefficients, while for homology you can use singular homology with integer coefficients.) Actually, this statement is quite easy to prove once you know the definitions involved, it follows from the fact that if $\Delta$ is a simplicial complex of dimension $\le k$ then $i$-th simplicial cohomology of $\Delta$ vanishes for all $i>k$. (Simply because the $i$-th cochain group is zero.)

The next thing to know is the Alexander duality theorem:

Let $K\subset S^n$ be a compact subset. Then $$ \tilde{H}^i(K)\cong \tilde{H}_{n-i-1}(S^n-K). $$ Here the homology/cohomology is reduced.

Now, if $dim(K)<n-1$, then for $i=n-1$, we obtain $$ 0=\tilde{H}^{n-1}(K)\cong \tilde{H}_{0}(S^n-K), $$ which means that $S^n-K$ is connected.

Moishe Kohan
  • 97,719
1

A simple, not so wide and wild, generalization is for n > 1
and countable K subset $R^n, R^n$ - K is arc connected.

Let a,b be two points on a plain and not in K. 0n that plain
draw a semicircle of radius r with ab as the diameter.
There are uncountable many such semicircles and as they are pairwise disjoint execept at the endpoints, almost all of them will miss K. Thus an arc from a to b.

William Elliot
  • 17,598