Keyspace of asymmetric ciphers. (RSA) I have difficulty finding information about key space calculation of RSA. I think there is a gap in my math knowledge that prevents me from understanding the problem. 

I know that symmetric ciphers such as AES 256 have key space equal to 2^256, but that can't be the same with RSA of key length 4096 that uses the product of prime numbers, can it?
The question makes a formal confusion between the keyspace, which is a set; and the size of the keyspace, which is an integer. AES-256 has keyspace equal to the set of 256-bit bitrings, noted $\{0,1\}^{256}$, which has size $2^{256}$.

Often, equivalent keys are counted only once. That's why DES is considered to have a keyspace of size $2^{56}$ even though its key is 8 octets (the low order bit of each octet is either ignored, or checked to be the complement of the XOR of the other 7). Similarly, TEA can be considered to have a keyspace of size $2^{126}$.

For symmetric algorithms, the size of the keyspace is of direct interest, since it determines the resistance to brute force key search, which ideally is the best attack.

For asymmetric algorithms like RSA, size of the keyspace is no longer directly related to security, because the public key is assumed avaialble to attackers. For this reason, size of the keyspace is not much studied, and for RSA there is not even consensus about exactly what that is. Rules vary for what a valid $N$ is, much more for a valid $e$. And it is unclear if we count (non-equivalent) pairs $(N,e)$, or just the possible $N$. We'll do the later, which alleviates the need to define a valid $e$, and makes a lot of sense since knowing a private key for $(N,e)$ allows to find a private key for any given $(N,e')$.

One possible definition of a valid 4096-bit RSA modulus is $N=p\,q$ with $p$ and $q$ prime, and $2^{(4096-1)/2}&lt;p&lt;q&lt;2^{4096/2}$ (that's about what's FIPS&nbsp;186-4 asks). By the prime number theorem, about one out of $\log(2^{2047.75})$ integers in the interval $[2^{2047.5},2^{2048}]$ are prime, giving about $u=(2^{2048}-2^{2047.5})/\log(2^{2047.75})$ choices for $p$. That gives roughly $u^2/2\approx2^{4070.5}$ choices for $N$. That's one possible estimate of the size of the keyspace for 4096-bit RSA. Which, again, has just nothing to do with its security.