# Prove that the upper bound in the Noiseless-coding theorem is strict

Given a probability distribution $$p$$ across an alphabet, we define redundancy as:

Expected Length of codewords – entropy of p = $$\ E(S) – h(p)$$

Prove that for each $$\epsilon$$ with $$0 \le \epsilon \lt 1$$ there exists a $$p$$ such that the optimal encoding has redundancy $$\epsilon$$.

Attempts

I have tried constructing a probability distribution like $$p_o = \epsilon, p_1 = 1 – \epsilon$$ based on a previous answer, but I can’t get it to work.

Any help would be much appreciated.