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seen Apr 9 at 2:09

Apr
6
awarded  Popular Question
Mar
30
revised Intersections of fractal sets with connected sets
added 13 characters in body
Mar
30
asked Intersections of fractal sets with connected sets
Feb
18
comment Is a sufficiently nice simple curve which is nulhomotopic the boundary of a surface?
@MichaelWeiss yes, this is correct.
Feb
15
comment Is a sufficiently nice simple curve which is nulhomotopic the boundary of a surface?
@studiosus Do you have a reference?
Feb
15
asked Is a sufficiently nice simple curve which is nulhomotopic the boundary of a surface?
Feb
15
accepted Is a simple curve which is nulhomotopic the boundary of a surface?
Feb
15
comment Is a simple curve which is nulhomotopic the boundary of a surface?
@user99680 $U$ is an open subset of $\mathbb R^3$, so by invariance of domain it cannot be homeomorphic to $\mathbb R^2$.
Feb
15
comment Is a simple curve which is nulhomotopic the boundary of a surface?
@DanielRust We don't necessarily know that a homotopy from $C$ to a point is injective.
Feb
15
asked Is a simple curve which is nulhomotopic the boundary of a surface?
Jan
27
awarded  Popular Question
Jan
22
awarded  Popular Question
Nov
24
accepted Functions and convergence in law
Nov
24
comment Functions and convergence in law
Ah, okay, I see. My mistake. The first example is fine, sorry for the confusion.
Nov
24
comment Functions and convergence in law
I still do not agree with your first construction. I think your second construction is somewhat better though. What topology are you using for X?
Nov
24
comment Functions and convergence in law
Thanks for the reply! It seems though that the function $\phi_n(X)$ is not independent of $\sigma(X_1 , ... , X_k)$ for $n >k$. Indeed, if we switch the value of one of the $X_j$'s for $j\leq k$, then $\phi_n(X)$ changes from 0 to 1 or vice versa.
Nov
24
comment Functions and convergence in law
Yes, I tried this. It reduces us to the case of showing that if $(X_n)$ all have the same law, $X_n\rightarrow X$, $\phi_n(X_n) \rightarrow Y$ a.s., then $Y$ is a.s. a function of $X$. However, I don't see a way to prove this.
Nov
23
asked Functions and convergence in law
Nov
18
accepted Immersion on a compact set
Nov
14
accepted Behavior of Hausdorff dimension under homeomorphisms