bio  website  

location  
age  25  
visits  member for  5 years, 3 months 
seen  52 mins ago  
stats  profile views  12,421 
1d

revised 
Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry?
deleted 194 characters in body 
1d

revised 
Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry?
added 110 characters in body 
1d

revised 
Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry?
added 38 characters in body 
1d

answered  Under what conditions $\xy\=n\iff\f(x)f(y)\=n.$ for $n\in\mathbf{N}$ implies isometry? 
1d

comment 
Infinite graphs isomorphic to their line graph
@TimothyChow: It is not hard to see that if $G\cong L(G)$ and $G$ has a vertex of degree at least 3, then $G$ has a vertex of degree at least $n$ for all $n\in\mathbb{N}$. Basically, start with a graph with a vertex of degree 3 and one more connected edge and then iterate the operation $L$; all the graphs obtained from this must embed in $G$, but they have vertices of arbitrarily high degree. 
1d

awarded  Enlightened 
1d

awarded  Nice Answer 
2d

revised 
Given functors $F$ and $G$, does $\mathrm{Res}_F \cong \mathrm{Res}_G$ imply $F \cong G$?
added 207 characters in body 
2d

answered  Given functors $F$ and $G$, does $\mathrm{Res}_F \cong \mathrm{Res}_G$ imply $F \cong G$? 
2d

comment 
Image of poset with Hausdorff interval topology
It is not necessarily the case that $e(\downarrow \alpha)=\downarrow e(\alpha)$; only $\subseteq$ holds in general. If instead of $e(s_i)$ and $e(t_j)$ you mean things like $\downarrow e(\alpha)$, then they will still cover $Q$ and you get two disjoint open sets in $Q$, but you can't be sure that these sets contain $x'$ and $y'$. 
2d

comment 
Infinite graphs isomorphic to their line graph
bof: Thanks, corrected. @Carl: $L(G)$ is not complete, but it contains a complete subgraph of the same size (all the edges containing some particular vertex). 
2d

revised 
Infinite graphs isomorphic to their line graph
added 4 characters in body 
2d

answered  Infinite graphs isomorphic to their line graph 
2d

comment 
group homomorphisms from the real line to infinite torsion abelian groups
Unless you are interested in the dependence of this on the axiom of choice, this isn't really researchlevel. Consider the group $\mathbb{Q}/\mathbb{Z}$. 
Jan 22 
comment 
Do homsets really live in the category Set?
It sounds like you may be interested in structural set theory. 
Jan 21 
comment 
Does k(X) have a kbasis for every set X, without AC?
This argument also works for any wellorderable $k$; you just have to find $k$ copies of the regular representation in each $r(I)$. In particular, assuming choice, this shows $k(X)$ has a symmetric basis for any $k$ of characteristic zero. 
Jan 20 
comment 
Maximum matchings in infinite graphs
Your last sentence does not follow (you seem to have accidentally swapped matchings and independent sets). 
Jan 20 
comment 
Maximum matchings in infinite graphs
If $M$ is any maximal matching, it is easy to see that any other matching must be no larger than $M$ if it is infinite or finite if it is finite. 
Jan 20 
comment 
Does k(X) have a kbasis for every set X, without AC?
Essentially the same argument also gives a direct proof that if $k$ is a field of positive characteristic, then $k(x_1,x_2,\dots)$ does not have an almost symmetric basis. If $f$ is any basis element of such an almost symmetric basis, pick $p$ variables not occuring in $f$ (or the finite set your permutations must fix) and apply this argument to get a contradiction. 
Jan 20 
revised 
Does k(X) have a kbasis for every set X, without AC?
deleted 5 characters in body 