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I'm reading a book on basic analysis and came across the result that given a set $X$, there exists a set $\{Y : Y$ is a subset of $X\}$. The proof is left as an exercise and the hint states that one can start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with $f^{-1}(\{1\})$.

While the above route isn't difficult to understand, why can't we just define the required set as: $\{f(X) \mid f\colon X \to X\}$? I suppose this also requires the axiom of replacement (the objects in question being $X$ and $f$) and looks fine to me, but there could be a problem with this reasoning, otherwise the author would've used this definition. If so, what's the problem?

Apologies if this is a really basic question, but I'm working on improving my mathematical logic skills.

Matthew Leingang
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user9343456
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  • I'm not sure why this is marked as a duplicate. The OP is not asking "why is this a set?", but rather "can we do it this other way instead?" – Alex Provost Mar 20 '16 at 18:31

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Hint: How would you obtain the empty set as the image of a function from $X$ to $X$?

Alex Provost
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  • What is your definition of a function? Because if a function $f:X\to Y$ is a subset $f$ of $X\times Y$ such that for $\forall x\in X((x,y_1)\in f \wedge (x,y_2)\in f \rightarrow y_1=y_2)$ then $f=\emptyset$ is a perfectly reasonable function and its image is exactly the empty set. – Darío G Mar 20 '16 at 18:37
  • @Wore You forgot to include that every "$x$-coordinate" must appear once. (If you meant to write down the usual definition of a function.) Then if $X \neq \emptyset$ the empty function does not exist. – Alex Provost Mar 20 '16 at 18:51
  • https://en.wikipedia.org/wiki/Function_(mathematics)#Definition – Darío G Mar 20 '16 at 19:21
  • @Wore Did you read what I wrote? From your very own link: "every element of $X$ is the first component of one and only one ordered pair in the subset." If $X \neq \emptyset$ your "empty function" isn't actually a function according to that definition because there exists an element in $X$ that isn't the first component of any ordered pair in your function. – Alex Provost Mar 20 '16 at 19:27
  • @AlexProvost: (edit to original comment) I guess in that case the union of the set I defined and ${\emptyset}$ should suffice. – user9343456 Mar 20 '16 at 22:46
  • @ShirishKulhari Correct. Given any nonempty subset $A \subseteq X$, you may define $f:X \to X$ as the identity on $A$ and map everything outside of $A$ inside of $A$ arbitrarily. Then the image of $f$ is precisely $A$. – Alex Provost Mar 20 '16 at 22:56