Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 1227

Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.

7 votes
Accepted

Potential connected non-Lie subgroup

I think the answer is, yes, the graph can be connected. By definition, if the graph G is not connected, then we can find disjoint nonempty open sets A and B, such that G is contained in A union B. In …
Martin M. W.'s user avatar
  • 6,571
11 votes

Foliation with leaves which are and are not dense

No, there doesn't exist such a foliation. The existence of any foliation would mean the Euler characteristic is zero, so the surface must be either a torus or a Klein bottle. Foliations for these surf …
Martin M. W.'s user avatar
  • 6,571
6 votes

Dynamical properties of injective continuous functions on $\mathbb{R}^d$

This is an observation, along with a sketch of a potential construction where (1) does not imply (2) for $\mathbb{R}^3$. Observation. I want to point out an example of a closed set $X \subset \mathbb …
Martin M. W.'s user avatar
  • 6,571
5 votes
Accepted

Can orientation preserving diffeomorphism in $\mathbb{R}^d$ be presented by flowmap of dynam...

No, this is usually not possible. There's a previous MO question discussing this, but to add to the material there: A time-one map $\phi_1$ commutes with the 1-parameter family of diffeomorphisms $\p …
Martin M. W.'s user avatar
  • 6,571
104 votes
Accepted

Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?

Yes, the complement of any countable set in $\mathbb{R}^3$ is simply connected, by the Baire category theorem. Say your set is $X = \{x_1, x_2, ... \}$, and let $y$ be any point in $\mathbb{R}^3 \set …
Martin M. W.'s user avatar
  • 6,571