Question with the Chinese Remainder Theorem

It seems that it wouldn't be significantly harder to show for any integer j (basically "shifting" the window you're intersecting with S).

That should then be enough to "bridge" intermediate values between m_k and m_{k+1} I think. (You can do jumps of m_k until m_{k+1} directly, and values in-between can't be different enough to break convergence to 0).

May be something more elegant, but I think that works.

Cheers, it seems this end of the question is oddly inelegant compared to the previous part.

Is the general idea that #S[N]/N = #S[j + m_k]/(j + m_k) for some k, j, then the RHS vanishes as k -> infinity?

So how I'm thinking about it is that we know that for any block [j,j+m_k) we'll have at most m_k / 2^k (I think will be the bound you have but the exact details don't matter) members of S. In particular, we can take blocks in steps of m_k to show #S[N] <=N/2^k when N is a multiple of m_k.

Now what if N>m_k *isn't* a multiple of m_k. Then we have a "partial block" of size N mod m_k at the end. But that block can't contribute more than a full block,so adds at most m_k/2^k to the total. And since N >m_k the contribution of the partial block divided by N must tend to 0 and can't mess things up.

Maybe slightly nicer is to prove that
#S(jm_k) <= j#S(m_k)
and observe #S(N) <= #S(jm_k) whenever j >= N/m_k to deal with the case where N isn't a multiple of m_k.

It's essentially the same argument but I think it's more elegant (or at least less *inelegant*!)

Seems to all check out, thanks. There was a similar idea on a previous sheet about splitting intervals into blocks so I really should've spotted the trick!

Don't think it's strictly necessary to go casewise, the fact that we can split up the interval into at most a + 1 (if we write N = am_k + b) intervals of size at most m_k/2^k seems to give a strict enough bound.