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1d
revised Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
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1d
revised Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
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1d
revised Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$ implies isometry?
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1d
answered Under what conditions $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ 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 $F \cong G$?
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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
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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 research-level. Consider the group $\mathbb{Q}/\mathbb{Z}$.
Jan
22
comment Do hom-sets 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 k-basis for every set X, without AC?
This argument also works for any well-orderable $k$; 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 k-basis 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 k-basis for every set X, without AC?
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