Why is the sky dark at night?

[ZetaX]

That's the same thing.

Someone above protested they aren't and it sounded to me like one is the local, the other the global version.

Sure. But if I can pick an arbitrarily large subset, it guarantees that set itself is infinite. You cannot pick an arbitrarily large subset from a finite set. I don't see how you can talk about a finite time when I can chose a coordinate system with an arbitrary time span.

[...]

You cannot have a boundary one one coordinate (time) in one frame of reference, and not have boundaries on all other coordinates (spacial) in some other choices of frame of reference. A space without bounds requires time without bounds. Whether it's infinite or cyclical is dealer's choice.

That's where our definitions disagree: I fixed a frame of reference (say ours) and definied finiteness relative to that. And yes, it may then change for different frames of reference. But even those frame have infinite space in _some_ directions, just not all. So we should be careful wheter infinite means non-compact or boundaryless.

I have by the way problems to imagine how that "barrier" in space or time would behave. A singularity is one thing, but an entire codimension 1 submanifold where reality stops sounds... weird. Manifolds with boundary stop sounding like an easy concept if you want to imagine to live in one...

A good, simple to understand example is the Serpinski triangle. We start with a large triangle, cut away 1/4 of it. Then cut away 1/4 of the remainder, and so on. What's the total area of all points that remain in the Serpinski triangle? Well, it's 1 - 1/4 - (3/4)(1/4) - (3/4)(3/4)(1/4) - ... I can re-arrange it to read 1 - (1/4) * sum((3/4)^n) for n = 0 to inf. That sum is easily evaluated as a geometric progression, and is exactly 4. (Wolfram Alpha). So the total area of Serpinski triangle is 1 - (1/4) * 4 = 0.

Adding up the missing things makes things unnecessarily complicated. The area is factored by 3/4 each time, thus the final area is the limit of (3/4)^n, which is 0.

Yes, like that. Like I said, I do not see any reason why it would not work like that, other than maybe the dithering thing due to distribution.

If you think something else I am very curious what that is exactly, because I am pretty much stuck. An object the size of the Sun or more, made of stars, should produce the light of a Sun.

I don't understand what your (remaining¿) objection is. Can you please clarify¿

[K^2]

Yes, but again, locally the densities are rather high. Especially when looking in the direction of the galactic core of the Milky Way, you have most (since we live in the outer edges) of a full galaxy of stars in a relatively narrow band or the sky. It seems that such a high density would mean the brightness of a sun or collection of suns, but it is not the case. Or even other galaxies. Andromeda (M31) has an apparent width of a number of Moons (or Suns, as they are almost as big in the sky). Most of what we see of it are its stars. Yet it is actually very faint, and not as bright as a disc shaped Sun.

Apparently uniform distribution of visible stars has made people believe that we were at the center of the galaxy some time back. But no, we really just see stars in our immediate neighborhood. The galaxy just isn't big enough, in any direction, to bring in contribution from distant stars to the same level. It also isn't helping that there is a lot of dust between stars that absorbs light. Now, if uniform density of stars would extend further, interstellar gas/dust would heat up and glow as well. But, thankfully, our galaxy ends rather abruptly, and all of that interstellar stuff can cool down by radiating into intergalactic void.

It's pretty straight forward to run the numbers. You should take average density of stars in Milky Way and an average luminosity of a single star and estimate how big the galaxy would have to be to get, say, 50% of Sun's brightness across the sky. You'll get a VERY big number. But our galaxy is tiny, so only the most proximate stars matter, and they are pretty uniformly distributed in our immediate neighborhood.

[Camacha]

I don't understand what your (remaining¿) objection is. Can you please clarify¿

I am not sure how I can make myself more clear, but I will repeat my example: take object M31. It is mostly made of stars. Is has an apparent size of multiple Moons or Suns (which are about the same size) as shown in the image I posted. By logic, something that is larger than the Sun, and it made out of stars (the same 'material' as the Sun), should be as bright as the sun, lighting the night sky. Yet it is a very faint object, with nothing in between that galaxy and ours to reduce the light received. There is a huge discrepancy there.

The only explanation I can come up with is that the object is actually not full of stars at all, but in fact almost dark because it mostly shows the dark space behind. The stars that are there just make it seem that it is a solid-ish wall of stars.

[ZetaX]

Yes, it is as you say: not enough stars to really fill that part of the sky associated to it.