Show (assuming the other axioms of set theory) that the universal specification axiom, Axiom 3.8, is equivalent to an axiom postulating the existence of a “universal set” \(\Omega\) consisting of all objects (i.e., for all objects \(x\), we have \(x \in \Omega\)). In other words, if Axiom 3.8 is true, then a universal set exists, and conversely, if a universal set exists, then Axiom 3.8 is true. (This may explain why Axiom 3.8 is called the axiom of universal specification). Note that if a universal set \(\Omega\) existed, then we would have \(\Omega \in \Omega\) by Axiom 3.1, contradicting Exercise 3.2.2. Thus the axiom of foundation specifically rules out the axiom of universal specification.
- \(\bullet\) Show that if Axiom 3.8 is true, then a universal set exists, and conversely, if a universal set exists, then Axiom 3.8 is true
- - other axioms of set theory
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\(\Vdash\) { Prove each direction }
- \(\bullet\) Show that if Axiom 3.8 is true, then a universal set exists
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\(\Vdash\) { Define \(P(x)\) to be "\(x = x\)" }
- \(\{ x : P(x) \}\) exists and \(x \in \{ x : P(x) \}\) for all object \(x\)
- \(\square\)
- \(\bullet\) Show that if a universal set exists, then Axiom 3.8 is true
- - Let \(\Omega\) the universal set
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\(\Vdash\) { axiom-3.5 }
- \(\{ x \in \Omega : P(x) \text{ is true} \}\) exists \(y \in \{ x \in \Omega: P(x) \text{ is true} \} \Leftrightarrow P(y) \text{ is true}\)
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\(\vdash\) { Since \(x \in \Omega\), \(x\) is object }
- \(\{ x : P(x) \text{ is true} \}\) exists \(y \in \{ x : P(x) \text{ is true} \} \Leftrightarrow P(y) \text{ is true}\)
- \(\square\)