2,042 reputation
726
bio website tora.us.fm/erelsgl
location Israel
age
visits member for 2 years, 8 months
seen 7 hours ago

Ph.D. student in Israel, Bar Ilan University, computer science department and economics department. Research interests:

  • Fair Division of Land.
  • Compuatational Linguistics.
  • Negotiation Agents.

Recently I wrote a working paper in which I cited several answers that I got from Stack Exchange members. Many thanks to everyone!

Email: erelsgl@gmail.com


7h
answered How to prove this result involving the quotient maps and connectedness?
1d
comment Finite coloring of an interval
Is there a more basic property / term that describes such functions?
1d
revised Connectedness of the boundary
added a connectedness tag and a missing word
1d
comment Proving an intuitive fact about sets in the plane
Thanks. This means that I should add a certain condition to make the result true.
1d
comment Proving an intuitive fact about sets in the plane
This makes sense. I think the relevant condition on Blue is that it should be simply-connected, but I am not sure.
1d
comment Proving an intuitive fact about sets in the plane
@ClementC. The picture shows only the given facts - not the entire coloring of the plane.
1d
revised Proving an intuitive fact about sets in the plane
added 80 characters in body
1d
revised Proving an intuitive fact about sets in the plane
added 7 characters in body
1d
asked Proving an intuitive fact about sets in the plane
1d
asked Finite coloring of an interval
Dec
19
comment Is this two-dimensional version of the Intermediate Value Theorem correct?
I just found a MathImages article which uses pretty much the same graphic intuition to explain Brouwer's fixed-point theorem in two dimensions: mathforum.org/mathimages/index.php/Brouwer_Fixed_Point_Theorem
Dec
18
accepted Is this two-dimensional version of the Intermediate Value Theorem correct?
Dec
18
comment Cutting a pie with a fork
I think I understand what you mean. Let $(x,y)$ be the location of the fork center. Mark the values of the three pieces by $A(x,y)$, $B(x,y)$ and $C(x,y)$. Define the following function from the boundary of the pie to $R^2$: $f(x,y) = (A(x,y)-B(x,y), A(x,y)-C(x,y))$. Then there is a point in which A equals (almost) the entire cake, so $f(x,y)=(1,1)$; then there is a point in which B equals the entire cake so $f(x,y)=(-1,0)$; and a third point in which $f(x,y)=(0,-1)$. $f$ changes continuously between these points and so its winding number around $(0,0)$ is 1. Is this correct?
Dec
17
comment Coloring a circle
But, probably there should be a certain topological constraint on the coloring. Maybe that each color should be a closed set (and colors may overlap at their boundaries).
Dec
17
comment Coloring a circle
@MarkBennet Because of the rotating condition, all 3 colors must be present in the coloring (for example, if there is a blue point, then there must also be a red point and a green point). Moreover, the number of segments of each color should be equal.
Dec
17
revised Coloring a circle
added 116 characters in body
Dec
17
asked Coloring a circle
Dec
16
comment Cutting a pie with a fork
$A(S,x)$ and $A(T,x)$ have the same integral. So, if they are not the same function, there must be an $x_S$ such that $A(S,x_S)>A(T,x_S)$ and an $x_T$ such that $A(S,x_T)<A(T,x_T)$. By the IVT, there is an $x_1\in[x_S,x_T]$ such that $A(S,x_1)=A(T,x_1)$. This is what you proved. I add that, because of periodicity, there is also an $x_2\in[x_1,x_0+2\pi]$ such that $A(S,x_2)=A(T,x_2)$. So there are at least two different equalizing points.
Dec
16
asked Is this two-dimensional version of the Intermediate Value Theorem correct?
Dec
15
comment Are the roots of a smooth function, a smooth function?
@Lubin I see. You mean that $f=0$ will be a smooth curve, but not a function, since for some values of $x$ there is more than one value of $y$. So the answer to my question, as it is asked, is "no".