Does the FLRW model have a scale factor $0$ at the singularity, meaning that even infinite distances become $0$ there according to the mapping $x \mapsto a(t)x$, for $a(t)=0$?

I am asking this because I was arguing at a respectable forum the other day that an infinite universe cannot be infinite at the singularity.

  • $\begingroup$ The scale factor $a$ approaches 0 as $t$ approaches 0, but that doesn't necessarily mean that the size of the universe becomes zero at the instant of the Big Bang. If the universe is infinite now, it's always been infinite. At $t=0$, we get the indeterminate form $0 \times \infty$ for its size. See physics.stackexchange.com/q/136860/123208 $\endgroup$ – PM 2Ring Dec 14 '19 at 17:59
  • $\begingroup$ From a pure mathematics point of view, just because $ 0 \times \infty $ is indeterminate does not imply that there is not a limit. In this example, if every $ x $ is mapped to $ 0 \times x $, a simple induction demonstrates that $ \infty $ is mapped to $ 0 $. And there is a nice geometric representation of this. $\endgroup$ – dimachaerus Dec 14 '19 at 19:44
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    $\begingroup$ @PM2Ring’s comment is the correct answer. If the Universe is infinite (which we don’t know is true), it was born infinite. $\endgroup$ – pela Dec 14 '19 at 22:02
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    $\begingroup$ Even if a(t) diminishes exponentially or at a tetration exponentiation order giving rise to inverse Aleph-n infinities multiplied by a mere Aleph-0 size of the universe, say? I have very grave doubts as to whom is right or wrong... $\endgroup$ – dimachaerus Dec 14 '19 at 22:45
  • $\begingroup$ Eg. The projections of the points of a rotating hyperplane. $\endgroup$ – dimachaerus Dec 16 '19 at 18:21

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