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 $\mathcal P(ℕ)$ the set $A_i=\{\{i,j\}~|~j∈ℕ\}$. For distinct $i,j$ we see $A_i∩A_j = \{\{i,j\}\}$ yet this $\{i,j\}∉A_k$ for any third distinct $k$.
No comments:
Post a Comment