3
$\begingroup$

Let $X$ be a set.

Suppose $\forall x\in X$, there exists $(y_1,...,y_n)$ such that $\phi(x,y_1,...,y_n)$.

Then, does there exist a set $Y$ such that for each $x\in X$, there exists $(y_1,...,y_n)$ such that $\phi(x,y_1,...,y_n)$ and $(x,(y_1,...,y_n))\in Y$?

Informally speaking, when $\mathscr{C}$ be a proper class and if we know that for each $x\in X$, there exists $y\in \mathscr{C}$ satisfying $\phi(x,y)$, can we construct a choice function $f:X\rightarrow \mathscr{C}$ such that $\phi(x,f(x))$?

EDIT

Here is a concrete example.

Definition

Let $X$ be a set and $x\in X$. Let's say $(U,\phi,E)$ is a chart of $X$ at $x$ when $U\subset X$ and $\phi:U\rightarrow E$ is an injection and $E$ is a real Banach space and $x\in U$.

Let $X$ be a set and $Y\subset X$. Assume that for each $x\in X$, there exists a chart $(U,\phi,E)$ of $X$ at $x$ such that there exists a closed subspace $E',V$ of $E$ such that $E=E'\oplus V$ and $\phi(U\cap Y)=\phi(U)\cap E'$.

Now, I want to extract a collection $\{(U_x,\phi_x,E_x)\}_{x\in X}$ from this definition.

I read articles about Scott's trick, but I don't get how to apply this to do so.

Question1: Does there exist the ordinal $\alpha_x$ such that it is the least rank of charts $(U,\phi,E)$'s of $X$ at $x$ such that there exists a closed subspace $E',V$ of $E$ such that $E=E'\oplus V$ and $\phi(U\cap Y)=\phi(U)\cap E'$?

Since the collection of all such $(U,\phi,E)$'s are not a set, I don't know how to extract such $\alpha_x$.

Question2. Say, we have well-defined those $\alpha_x$'s for all $x\in X$. Then, does there exist a function $f$ whose domain is $X$ and $f(x)=\alpha_x$?

$\endgroup$
2
$\begingroup$

Yes, to both.

The point of Replacement is that if we have a definable class function, applying it on a set gives us that the image is also a set. If you have a function which is a set, then Replacement is not even needed.

We can ask what is the least ordinal $\alpha_x$ such that a chart for $x$ exists inside $V_{\alpha_x}$. This is a definable function, and $X$ is a set. So the range of this function is a set of ordinals. So there is some $\alpha$ such that all the needed evidence lies inside $V_\alpha$.

Now you have a set of non-empty sets, all of which lie inside $V_\alpha$. Using the axiom of choice, you get what you were looking for.

To the second question, note that saying that $\alpha_x$ is well-defined, means that it is in fact a function. So again, by Replacement this gives us a set, since now $\{(x,\alpha_x)\mid x\in X\}$ is a subset of the product of $X$ and $\{\alpha_x\mid x\in X\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.