Proof: The answer is yes; the concept behind constructing a counterexample is in adducing an infinite collection of sets such that the intersection of any two distinct sets results in a singleton set whose element is independently identifiable as belonging to precisely those two sets.
For all i∈ℕ+ comprehend from P(ℕ) the set Ai={{i,j} | j∈ℕ}. For distinct i,j we see Ai∩Aj={{i,j}} yet this {i,j}∉Ak for any third distinct k.
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