There are questions that may seem to be similar to this one, but I've yet to find an answer.

I have come to understand that a flat universe, that is to say a curverature of $k=0$ which means that $S_k(r) = r$. The FLRW metric polar coordinates: $$ds^2 = -dt^2 + a^2(t) \left[ \frac{dr^2}{1 - kr^2} + r^2d\Omega^2 \right]$$ Now, since only $r$ is being altered, $dt = 0$ and $d\Omega$ = $0$. $$ds^2 = a^2(t) \frac{dr^2}{1 - kr^2}$$ This can be integrated to: $$s(r) = \frac{\sin^{-1}(\sqrt{k}r)}{\sqrt{k}}$$ So, by definition, the maximum value is when $\sin^{-1} = 90^\circ = \frac{\pi}{2}$ which occurs when $\sqrt kr=1$ To find the highest value, we replace $\sin^{-1}(\sqrt kr)$ with $\pi \over 2$ and get: $$s(r)_{\text max} = \frac{\pi}{2\sqrt k}$$ Therefore, as $k \to 0,\space s(r)\to \infty$. Given $k=0$, there is an infinite possible distance.

Now that we have that out of the way, when physicists talk about the size of the universe, by which I mean "when the universe was the size of a grapefruit" r a similar comparisson, space must have still been infinite, so what is this a description of?

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    $\begingroup$ The phrase size of the universe is used in two ways: (1) to mean the size of the currently observable universe, or (2) to indicate that the speaker has no understanding of general relativity. $\endgroup$ – John Rennie Oct 15 '14 at 7:54
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    $\begingroup$ This question appears to be off-topic because it is asking about the meaning of a commonly misused phrase. $\endgroup$ – John Rennie Oct 15 '14 at 7:55
  • $\begingroup$ Hello! I'm a big fan of yours. Thanks for contributing. Anyways, it is the very correct 'usage' of the phrase and the meaning behind it that I am inquiring. $\endgroup$ – Goodies Oct 15 '14 at 7:59
  • $\begingroup$ "Observable universe" or "observable region" imply the possible fallacy that all matter and energy in them are necessarily "observable" by at least some member of the intra-communicating species. $\endgroup$ – Edouard Feb 3 at 15:19
  • $\begingroup$ The universe might be geometrically infinite but contain only a finite number of events, although the GR-based version of this idea (known as "primeval atom" or "cosmic egg" cosmology), favored by Eddington (who liked Aristotle's idea that time has no beginning) and developed by his student Lemaitre in 1927, has been rejected much more recently by Vilenkin, who feels that any primeval atom would've been "unstable to quantum fluctuations". $\endgroup$ – Edouard Feb 3 at 18:37

All statements like "when the universe was the size of a grapefruit" refer to the currently observable universe. As the universe has a finite age and light travels at a finite speed (and there is nothing infinite going on with expansion), the observable universe is a finite patch.

I discussed some of the different notions of horizons in answering another question. The "observable universe" is taken to extend out to the particle horizon. That is, it includes precisely the points in our current time slice whose past worldlines (assuming they simply go with the expansion of space and have no peculiar velocity with respect to our reference frame) intersect the interior of our past light cone.

If you think of galaxies as marking these points, these are precisely the galaxies that we can see assuming arbitrarily good telescopes, since the light reaching us today was emitted as the galaxy crossed our past light cone.

Galaxies that started out too far away from us in an infinite universe haven't been able to get their photons to us. And indeed expansion will prevent most of them from ever getting to us.

The scale factor $a$ when the universe was the size of a grapefruit is simply the radius of a grapefruit divided by the radius of the current observable universe (about $46\ \mathrm{Gly}$), or something like $10^{-28}$ (corresponding to a redshift of about $z = 10^{28}$). The idea is that the galaxies (or rather their precursor quantum fluctuations) inside this grapefruit-sized volume are exactly the galaxies inside our observable universe today. In comoving coordinates the grapefruit is the same $46\ \mathrm{Gly}$ in radius then as our observable universe is now.

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  • $\begingroup$ Thanks! Correct my if I'm wrong, but this would seem to indicate that the observable universe would have expanded from a singular point and would, indeed, have a center, though the Universe as a whole would not. $\endgroup$ – Goodies Oct 15 '14 at 7:58
  • $\begingroup$ @Goodies The observable universe can be made arbitrarily small as you go back closer to the big bang, but see the last few sentences I added. If you scale your coordinates with the expansion of space, there is just as much volume then as now. The "point" you envision is the whole observable universe, not a part of it. $\endgroup$ – user10851 Oct 15 '14 at 8:00

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