Recent content by wabbit

This doesn't seem quite right, unless ## A ## and ## B ## commute.

They might be dead too. Who's to say we're going to last for 3 bn years?

Sure - just staying in your case of an increasing sequence, under that assumption ##x_{k+1}-x_k\leq Ca^k## and ##x_n=x_0+\sum (x_{k+1}-x_k )\leq x_0+C\sum a^k## is bounded by the (finite since a<1) sum of a geometric series - so it must converge.

Presumably the same thing that causes the universe to collapse from now towards the big bang/bounce in reverse time - gravity. Unless I'm mistaken, models such as the LCDM bounce mentionned by Marcus are time-symmetric.

True but since we it is convergent in the present case.

Yes. In fact a set S is finite if and only if any injective map from S to itself is bijective.

If the sequence is monotonic, then there is one necessary and sufficient condition for convergence: the sequence must be bounded. Of course this can be difficult to prove, and writing it as a series ##x_n=\sum (x_{k+1}-x_k)## can help. Then you have various convergence criteria, typically...

Close - check again your statement In other words you're saying ##\sigma_g^X## is one-to-one hence it must be onto. You need to explain why this inference is true.

There are many big bounce models in quantum gravity, and there are models where the universe recollapses, but the two features are unrelated - a bounce is an alternative to a bang, it describes what may have happened ~14bn years ago and says nothing about what may happen some time in the distant...

Not sure what part 2 is, but the fact that X is finite is necessary to prove statement 1, so make sure you use it explicitly : )

Right, I should have been more specific, the map to consider is ## \sigma_g^X : X\rightarrow X, x\rightarrow gxg^{-1} ## which exists because of the assumption ##gXg^{-1}\subseteq X##. Finiteness of X is key of course.

Almost. You need to prove that if ##gXg^{-1}\subseteq X## then ##gXg^{-1}=X##. Consider the map ##\sigma_g:X\rightarrow X##. Is it injective? Surjective?

Of multiplication rather. ##x^{-1}## is the multiplicative inverse of ##x##, defined by the equation ##x^{-1}×x=1##, while ##f^{-1}## is the composition inverse of ##f##, defined by the equation ##f^{-1}\circ f=Id## (##1## and ##Id## being the identity element of the corresponding operation)...

This convention is not an oddity though, it is related to the fact that the natural, generally defined operation between functions is composition rather than multplication. ##f^{-1}## is the inverse of ##f## under the composition operation, not under multiplication, and in the same way ##f^n##...

There's a distinction between unaccelerated expansion, which as I understand it has no effect whatsoever, and a cosmological constant (leading to accelerated expansion) which is similar to a tiny repulsive force proportional to distance - the latter doesn't prevent the formation of...