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The below exercise and the axiom to which it pertains can be found in Tao's Analysis I, which I am currently self studying.

Axiom 3.11 (Power set Axiom) For any two sets $X$ and $Y$, there exists a set, denoted $Y^X$, which consists of all functions from $X$ to $Y$. Thus, $$f\in Y^X \ \iff \ (f \ \text{is a function such that} \ f:X\to Y).$$ Exercise 3.4.6 Show that the above axiom implies the existence of the set $$2^X:=\{S\mid S\subseteq X\},$$ which consists of all subsets of a set $X$. (Hint: start with the set $\{0,1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$.

As the author suggested, we start with the set $$\{0,1\}^X = \{f\mid f:X\to\{0,1\}\},$$ and replace its elements with subsets of $X$. Clearly, this procedure yields the set $2^X$ however, I am unsure how I can apply the axiom of replacement. How can I do this bit? Any hints or tips are greatly appreciated.

Edit: This question has been asked and answered elsewhere, I have just found.

Crosby
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By the powerset axiom, $Q:=\{0,1\}^X$ exists, but its elements are functions $X\to\{0,1\} $ and not subsets of $X$.

We thus replace them to the appropriate subsets, using the replacement axiom with a formula $\phi(y, z) $ that expresses that $y$ as a function maps exactly the elements of $z$ to $1$. $\phi$ needs to satisfy $\phi(y, z_1)=\phi(y, z_2)\ \to\ z_1=z_2$ and then by replacement, we obtain $P(X) =\{z:\exists y\, (y\in Q\land\phi(y, z)) \} $.

Berci
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