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Let $H_0$ be $\log_2(|A|)$, where $A$ is a set.

Let $C$ be a compressor $C\colon A \to \{1,0\}^l \cup \bot$.

This is a silly question, because intuitively it seems obvious.

How can I prove that $l$ can't be less than $H_0$?

I can prove that $l = H_0$ is sufficient, but how can I prove that it can't be less.

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  • $\begingroup$ Unless you put some conditions on $C$, you can even choose $l = 0$. $\endgroup$ Apr 9, 2021 at 11:38

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The function wouldn't be one-to-one otherwise.

The size of the set in the left is $|A|$, and its image would be of size at most $2^l<2^{H_0}=|A|$, therefore, the function wouldn't be injective and thus you wont be able to de-compress some of the data.

Therefore it will be a lossy-compression for any $l<H_0$.

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