Can the Chinese Remainder Theorem extend to an infinite number of moduli? I've been trying to find info on this and have come up lacking. The CRT says that a system of congruences with coprime moduli always has a unique answer (modulo the product of the original moduli). And the generalizations I've seen defined say you can use residues $\{a_1, a_2,..., a_n\}$ and moduli $\{m_1, m_2,..., m_n\}$ to find a unique $X$ mod $M$ (with $M = m_1 \cdot \ m_2 \cdot ... \cdot m_n$), so long as the set of moduli are all coprime.
Can this be extended for an infinite set of residues and moduli? It seems to me you could choose $n$ to be as large as you desire, but I feel like it's unclear. Thoughts? It feels vaguely like Euclid's proof of infinitude of primes, but "feels" isn't really well-defined...
 A: For example, suppose you want $x \equiv 0\bmod 2$ and $x \equiv 1 \bmod p_i$ for all $i > 1$, where $p_i$ is the $i$'th prime.
Since $x \equiv 1 \bmod p_i$ but $x \ne 1$, $|x - 1| \ge p_i$.  But then there is no $x$ that works for infinitely many $i$.
A: No. The lcm of an infinite number of moduli cannot be a finite number.
A: As Robert Israel's answer shows, we do not in general have existence. Regarding uniqueness: if the moduli are unbounded, solutions are unique. If $x,y$ are solutions then $m_i|x-y$ for all $i$. Assuming $x\ne y$, this gives $|x-y|>|m_i|$ for all $i$, i.e. bounded moduli.
Say that the residues $a_i$ are eventually constant if there is some $N$ with $N<i$ implies $a_N=a_i$. When the moduli are unbounded, a necessary condition for existence is that the residues are eventually constant. For some $N$, $N<i$ impies $x<m_i$ as the moduli are unbounded. So the requirement $x\equiv a_i \bmod m_i$ implies $x=a_i$ for $i>N$. So if a solution exists, the residues are eventually constant.
Finally, if the residues are eventually constant, the necessary and sufficient condition for existence is that if $a_N$ is the 'eventual value' of the residues, then $a_N\equiv a_i\mod m_i$ for all $i\le N$.
*Note, when the moduli are bounded, you only need the normal Chinese remainder theorem.
