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A subset of X can be viewed as injection $f :S \rightarrow X$, where $S$ is a structure whose only relation is equality.

If we allow $f$ to be an arbitrary function, we get the notion of a multiset.

If we furthermore assert that $S$ is a well-ordered set, we get the notion of a sequence in $X$ (potentially transfinite).

Question: is it possible to model the notion of a subcollection $\mathcal{K} \subseteq \mathcal P(X)$ as a function $f:S \rightarrow X$?

goblin GONE
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2 Answers2

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It may be possible, but I think any method would be highly "unnatural".

Here is an alternative that you may like, though: the power set $\mathcal{P}(X)$ of $X$ has a natural interpretation as the collection $\{0,1\}^X$ of all functions $X\to\{0,1\}$, via the correspondence $$g\in\{0,1\}^X\quad\longleftrightarrow\quad g^{-1}(0)\subseteq X,$$ and then you could say that a $\mathcal{K}\subseteq\mathcal{P}(X)$ can be interpreted as an injective map $f:S\to \{0,1\}^X$.

Zev Chonoles
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  • Yeah I thought of that. But if $g:X \rightarrow Y$ is an injection, I was hoping that $g \circ f$ would be well-defined. Eg if $f$ is a tiling of $X$ and $g$ is a translation, i want $g \circ f$ to be a translation of the tiling. – goblin GONE Mar 09 '13 at 07:52
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I don't usually answer my own questions.... but, here's an idea.

Let $S$ be a structure consisting of two "sorts," call them "elements" and "subsets." Equip $S$ with a relation $\in$ between elements (on the left) and subsets (on the right). Let us write $f : S \rightarrow X$ to mean that $f$ is a function that is defined for every element of $S$.

Then whenever $\in$ is extensional, we can say that a bijection $f : S \rightarrow X$ is a "subcollection" of $X$. If $\in$ fails to be extensional, we can still call $f$ a "multisubcollection." If the subsets of $S$ are well-ordered, call $f$ "a (potentially transfinite) sequence of subsets of $X$."

Just an idea.

goblin GONE
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