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I'm stuck at one exercise from chapter of sets from Terence Tao's analysis book. I need to proof the lemma:

Lemma: Let $X$ be a set. Then the set $\{Y : Y \:\text{is a subset of}\: X\}$ is a set.

Note: The set $\{Y : Y \:\text{is a subset of}\: X\}$ is known as the power set of $X$, defined as $2^X $

I can understand why the lemma is true, but I have no clue how to prove it. Furthermore, the author gave a hint that I found very confusing, it follows:

Hint: start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$.

JKnecht
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user312479
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    Usually your "lemma" is counted as a set-theoretic axiom in its own right, the Axiom of Power Sets, which is a standard part of ZFC. Which axiom system are you working in here? – hmakholm left over Monica Feb 11 '16 at 00:32
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 11 '16 at 00:40
  • Well, I'm studying from Tao's book. Everything I am learning of set theory it's from there... The author stated the following axioms: 1) sets are objects; 2) empty set; 3) singleton and pair set ; 4) pair wise union ; 5)axiom of specification; 6) axiom of replacement; 7) axiom of inifinity; 8) axiom of regularity and 9) power set axiom (the one that tell us the existence of a set of all functions from X to Y. – user312479 Feb 11 '16 at 00:43
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    Changing the standard axioms is, I guess, OK. Not so sure about giving a new axiom an old name. – André Nicolas Feb 11 '16 at 00:48
  • JKnecth, the problem is that I dont know where to start to solve this exercise. I have solved all the exercises before this one (and checked my proofs with others proofs).. I think I'm not prepared to work with axioms, I thought of skipping this exercise but there are others similars to this. :/ – user312479 Feb 11 '16 at 00:48
  • @AndréNicolas Especially when the old name is a bad name for the new axiom. I wonder, though: does he call it the power set axiom?, or just the power axiom? The latter would make more sense. – BrianO Feb 11 '16 at 06:54

2 Answers2

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Going from the hint:

The set $\{{}0,1\}{}^X$ is the set of all functions $f:X\rightarrow{}\{{}0,1\}{}$. We can consider for elements $x\in{}X$ that $f(x)=0$ means that $x\notin{}Y$ and $f(x)=1$ means that $x\in{}Y$.

For a subset $Y$ of $X$, there will be a function $f_Y:X\rightarrow{}\{{}0,1\}{}$ describing the elements of $Y$ as above. Then using replacement we replace the function $f_Y$ with $f_Y^{-1}(\{{}1\}{})=Y$. Proceeding for all functions $f$, we define all subsets of $X$ and $\{{}0,1\}{}^X$ being a set implies this is a set.

This is fairly hand-wavy when it gets to replacement since I'm not sure how your book is defining it. This should be a general approach though.

user312238
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Usually this is an axiom (the Powerset Axiom). Since apparently Tao adopts a different axiom instead — existence of $Y^X$, the set of all functions $X\to Y$ for any $X,Y$ — you'll have to use that to prove the usual Powerset axiom.

You can use existence of empty set and the axiom of pairing to define a set $T = \{\emptyset, \{\emptyset\}\}$, and prove that it has two distinct elements. Every $f\in T^X$ is essentially the characteristic function of a subset of $X$: $$ S_f = \{x\in X\mid f(x) \ne \emptyset\}, $$

and every subset of $S\subseteq X$ is represented by some $f$ in $T^X$ via $$ I_S = \left\{(x,y) \in X\times T\mid y = \begin{cases} \{\emptyset\}&\text{if $x\in S$,}\\ \emptyset&\text{if $x\notin S$} \end{cases} \right\}, $$ $S_f$ and $I_S$ both exist by the Axiom of Specification, and these operations are inverses of each other.

Now you can use the Axiom of Replacement to show that the following is a set: $$ P := \{S_f\mid f\in T^X\}. $$ Using $I_S$, you can show that $P = \mathcal{P}(X)$, i.e. $P = \{S\mid S\subseteq X\}$.

BrianO
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