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A semigroup $S$ is said to be $0-$ direct union of completely $0-$ simple semigroup if $S = \cup_{i \in I} S_i$, where each $S_i$ is a completely $0-$ simple semigroup and $S_i \cap S_j = S_i S_j = \{0\}$ if $i \neq j$.

I have also searched on google but did not get any example.

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Well, just take the semigroup with zero $S = \{a, b, 0\}$ where $a^2 = a$, $b^2 = b$ and $ab = ba = 0$. Then $S_a = \{a, 0\}$ and $S_b = \{b, 0\}$ are both completely $0$−simple and satisfy $S = S_a \cup S_b$ and $S_aS_b = S_a \cap S_b = \{0\}$.