Prove Lemma 3.4.9. (Hint: start with the set \(\{0,1\}^X\) and apply the replacement axiom, replacing each function \(f\) with the object \(f^{-1}(\{1\})\).)
See also Exercise 3.5.11.
- \(\bullet\) Let \(X\) be a set. Then the set \(\{Y : Y\text{ is a subset of }X\}\) is a set.
- – Let set \(F = \{0,1\}^X\)
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\(\Vdash\) { replacement axiom }
- For a given \(x \in X\) and \(f \in F\), a set \(f^{-1}(\{1\}) = \{ x : f(x) = 1 \}\) exists and it a subset of \(X\).
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\(\vdash\) { replacement axiom }
- \(\{ f^{-1}(\{1\}) : f(x) = 1 \text{ for } x \in f^{-1}(\{1\}) \text{ and } f \in F \}\) exists and a set.
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\(\vdash\) { For any subset \(Y \subset X\), define \(f(x) = \text{ if } x \in Y \text{ then } 1 \text{ else } 0\), then \(S = f^{-1}(\{1\})\) and for any \(f \in F\), \(f^{-1}(\{1\})\) is a subset of X.}
- \(\{Y : Y\text{ is a subset of }X\}\) exists and a set.