Shannon's Perfect Security for Asymmetric Encryption

I have the following definition of Shannon's Perfect Security.

Assuming messages and keys are drawn randomly from some distribution then:

1. The probability of guessing plaintext m is not enhanced by knowing any ciphertexts, i.e. $$P(M=m| C=c) = P(M=m)$$
2. The keys are drawn uniformly at random and are only used for one encryption.

Does this apply to asymmetric encryption as well? How does statement 2 apply when we have public keys for encryption, say RSA encryption?

If Shannon's perfect security can be generalized then does this mean that RSA is perfectly secure?

• I can see you're in the mood, but please relax with the number of questions. Handling one or two at the time is much easier :) Commented May 3 at 0:52
• Perfect secrecy requires that key size > message size. This is the first problem. The second all-public key cryptosystem relies on the hardness of some mathematical problem that is not informatically proven to be secure. Getting $p$ is nearly half of the key whereas we can recover all ( get half of the OTP bits that is all you got). This prevents informational security. Besides this requirement is the opposite of the public key that one has to generate again. Commented May 3 at 11:10

Consider the following attacker for some encrypted message $$c=E(pk,m)$$, where $$m\in\{0,1\}^{n}$$:
1. Sample $$m_{0}\leftarrow \{0,1\}^{n}$$.
2. If $$E(pk,m_{0})=c$$, output $$m_{0}$$.
3. Otherwise, sample $$m_{1}\leftarrow \{0,1\}^{n}$$ and output it.
Clearly, knowing the ciphertext does enhance the probability of guessing the plaintext (by a negligible amount, of course, as we need to both guess the message and the randomness used by $$E(pk,\cdot)$$), which means the scheme is not perfectly secured.