My question is exactly what is stated in the title: Does Cayley's theorem imply that all groups are countable?(adsbygoogle = window.adsbygoogle || []).push({});

I don't see how a well defined transitive action of an uncountable group on itself. How could you possibly find a set of permutations sending a single element to every other element in it, much less each element of a group to every other element. Such a set of permutations would have to be uncountably uncountable, if that makes any sense. I don't see how such a set could have any well defined "order" (I don't know if this is the correct term), not in the sense of magnitude but in the sense of structure imposed.

In other words, what I mean by "order" is that in mathematics, we deal spaces, or sets on which we have added some structure by way of axioms which I see as increasing the "order". We add "order" to arbitrary sets to distinguish them up to isomorphism and also to in some sense approximate what may be true in sets with no order. However, in these "unordered" sets, no structure is imposed so that the tools of conventional mathematics are not necessarily well defined becuse it is not clear as to whether or not we can distinguish between elements, whether or not individual elements even exist since such a set would transcend completeness, or whether or not either of these actually even makes any sense. So, we can't even begin to partition the set and define equivalence classes, moreover, defining relations such as inclusion may be totally ambiguous and may not even be useful or make any sense whatsoever.

In this sense of "order", I don't see how a transitive action of an uncountable group on itself can be well defined. Take for instance the real line acting on itself by left addition. We can define very simple permutations of all the elements of this group:

Fix any [itex]\epsilon > 0[/itex], and define a group by

[itex]G = \left\{\pm g|g\geq\epsilon\right\}[/itex]

and define its action on the real line [itex]\forall g\in G[/itex] by

[itex]g\cdot x=g + x [/itex]

This should define every permutation of the real line under the action right?

However, the epsilon method of analysis requires that you have fixed such an epsilon, so that in effect, what we are considering here, since epsilon is arbitrary, is the union of an arbitrary collection of actions, where each is defined to be one of an arbitrary collection uncountable groups, uniquely determined by its arbitrary but fixed minimal positive non-zero element, acting on the real line. The collection of all of such elements and zero is most definitely not the real line since the real line has no minimal element, while fixing elements element requires that a minimal group element exists. In effect, what I'm saying is that in fixing epsilon we set a precision and any number smaller in magnitude than this precision is zero, this is the the epsilon principle. Thus, by completeness,

[itex](\forall\epsilon > 0) (\exists c \in R \ni: 0< c < \epsilon)[/itex]

so that there is some c is in the kernel of the action of the real line on itself no matter how small you fix [itex]\epsilon[/itex], and so the relative complement of the intersection of all such groups determined by their respective [itex]\epsilon[/itex] in the real line is not empty and it follows that in combining all of this the epsilon delta principle implies continuity: for each x and g,

[itex](\forall\epsilon > 0) ((\exists\delta>0 \ni: 0 < |x-c| < \delta) \Rightarrow

(|g\cdot (x-c) - g\cdot x|=|(g-c)\cdot x - g\cdot x|=|((g-c)+x) - (g + x)|=c < \epsilon))[/itex]

It also follows that since on any closed and bounded subset of R the function defined by the action is uniformly continuous. So, for all x and y in some [itex]\delta[/itex]-ball dependent on the fixed [itex]\epsilon[/itex], permutations of x and y are not distinct. The inverse of the action, which should also be an action since the real line under addition is a group, is therefore ambiguous, and not well-defined.

I don't see any way around this. Can anyone help me out? Because the real line under addition should definitely be a group...

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Does Cayley's Theorem imply all groups are countable?

Loading...

Similar Threads for Does Cayley's Theorem |
---|

B How does matrix non-commutivity relate to eigenvectors? |

B Why does a matrix diagonalise in this case? |

B Why does every subfield of Complex number have a copy of Q? |

**Physics Forums | Science Articles, Homework Help, Discussion**