7
$\begingroup$

And when can this subspace be chosen to be open?

As the answer to this question indicates, any manifold contains an open dense subset, which is homeomorphic to $\mathbb{R}^{n}$, and so for manifolds the stronger of the two properties holds. I wonder how widespread this phenomenon is.

Note that simply connected spaces are assumed to be connected, and so we only consider connected topological spaces. Also feel free to narrow the class of spaces further, e.g. add local path connectedness, local compactness, etc.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ A negative example for the stronger property is the Sierpinski carpet en.wikipedia.org/wiki/Sierpinski_carpet , a connected compact plane set, whose non-empty open subsets are not simply connected. $\endgroup$
    – Pietro Majer
    May 3, 2017 at 15:52
  • $\begingroup$ @PietroMajer Although it looks very intuitive, how exactly do you prove it? $\endgroup$
    – erz
    May 3, 2017 at 16:01
  • 1
    $\begingroup$ The infinite-dimensional torus $S^{\mathbb N}$ with product topology is a more obvious example perhaps of a space with no simply connected open sets. $\endgroup$
    – Christian Remling
    May 3, 2017 at 17:22
  • 1
    $\begingroup$ @erz: any non-empty open subset of the Sierpinski carpet $S$ contains the boundary of some square $Q$ disjoint from $S$ (one of the squares removed in the construction). Of course $\partial Q$ is a loop which is not contractible in $\mathbb{R}^2\setminus Q$, and a fortiori in $S$ . $\endgroup$
    – Pietro Majer
    May 3, 2017 at 17:50
  • $\begingroup$ @JeremyBrazas why cannot we just remove a point from each of the circles? $\endgroup$
    – erz
    May 4, 2017 at 14:18

2 Answers 2

Reset to default
3
$\begingroup$

Every connected topological graph has a dense, open, simply connected subspace.

Proof: Let $T$ be a spanning tree for the graph. Your open set is obtained by removing a point from every edge not in $T$.

Improve this answer
$\endgroup$
3
$\begingroup$

Any real semialgebraic set $X \subset \mathbb{R}^N$ has a dense, open subset that is a submanifold: just take the complement of its singular set. In fact, the singular set is Zariski closed in $X$, hence nowhere dense in the euclidean topology of $X$.

Using your remark about manifolds, it follows that any semialgebraic set contains an open, dense subspace homeomorphic to $\mathbb{R}^n$, with $n = \dim X$.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I believe you can say more generally that subanalytic sets have this property, for the same reason. $\endgroup$
    – Sean Lawton
    May 4, 2017 at 23:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.