How big would the moon appear with no atmosphere? I know that atmospheric lensing is what causes the moon to look bigger on the horizon, but I assume this effect is still taking place when the moon is overhead too, albeit to a smaller effect. However, would the moon appear to change in size if you looked from an earth with no atmosphere? 
 A: Actually, atmospheric lensing has very little effect on the angular size of the moon.
The main reason the moon seems larger on the horizon is the fact that, on the horizon, there are usually other objects of known size.  Your brain sees a tree in the distance, sees how large the moon appears in comparison to the tree, and tells you "The moon is much larger than a tree". But if you are close to the tree, the moon appears (to your brain) to be much smaller than the tree.  When the moon is high in the sky, your brain doesn't have anything else to compare its size to, so doesn't provide a size estimate.  As soon as an distant airplane flies between you and the Moon, though, the Moon's size can appear to increase because your brain knows how large an airplane is.  These illusions become very apparent in images taken of the Moon through a telephoto lens when other objects of known size are within the field of view.  
You can test this principle by taking photos of the moon using the same camera, with the same lens settings, as it rises from the horizon to higher in the sky.  Print the photos and measure the diameter of the moon in the different photos.
A: No matter what kind of lens (natural or manmade) you surround yourself with, $4\pi$ steradians has to fit into $4\pi$ steradians, so you can't make everything bigger.
In (optical) astronomy, you can safely called $2\pi$ str of your view "ground", and the other half "sky"--so can you make just the sky half bigger?
Not with atmospheric lensing. Refraction allow us to see $d=1$ degree below the horizon, so the total (true) solid angle we can see is:
$$\Omega = 2\pi\int_0^{\frac{\pi}2 + d}{sin{\theta}d\theta} = 2\pi[1-\cos{(\frac{\pi}2 + d)}]\approx 2\pi \times 1.017 $$
where $\theta=0$ ($\pi/2$) is the zenith (true horizon).
So that 1.017 times a half-sphere is stuffed into out view of the sky--a net shrinkage of 1.7 percent.
