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Take the set {1,2,3,4}, why is {1,2,3,4} not a partition of this, which condition does it not meet?

By my understanding, a partition of a finite set $S$ is any set $\{ S_{1},...S_{n} \}$ of n subsets of $S$, which satisfy,

  1. $S_{i} \ne \emptyset$ for all $1 \leq i \leq n$,
  2. $S_{i} \cap S_{j} = \emptyset$ for all $1 \leq i,j \leq n$, $i \neq j$
  3. $S_{1} \cup \cdots \cup S_{n} = S$

Seeing as $S$ is a subset of $S$, which part of the definition breaks down here?

Note I think $\{\{1,2,3,4\}\}$ is a partition of $\{1,2,3,4\}$... could this be explained to?

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    $\begingroup$ A partition of $S$ is a set of subsets of $S$ (with these three conditions.) Every element of the partition is a subset of $S$. Since $1 \in\{1,2,3,4\}$ and $1$ is not a subset of $S$, $\{1,2,3,4\}$ is not a partition of $S$. $\endgroup$ May 12, 2019 at 17:56
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    $\begingroup$ (Your conditions were incorrect - I have edited them.) $\endgroup$ May 12, 2019 at 17:57

3 Answers 3

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A partition of a set $A$ is a subset of its power set.

$\{\{1,2,3,4\}\}$ is, but $\{1,2,3,4\}$ is not, a partition of $\{1,2,3,4\}$.

$\{1,2,3,4\}$ is an element, not a subset of the power set of $\{1,2,3,4\}$.

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    $\begingroup$ Another instructive reply is that $\{\{1\},\{2\},\{3\},\{4\}\}$ is a partition. $\endgroup$
    – Lee Mosher
    May 12, 2019 at 17:10
  • $\begingroup$ Which part of the definition above breaks down here?* $\endgroup$
    – GooJ
    May 12, 2019 at 17:10
  • $\begingroup$ $S$ is a subset of $S$ and hence it is an element of the power set of $S$. As mentioned, a partition of $S$ should be a subset (not an element) of the power set of $S$. $\endgroup$
    – CY Aries
    May 12, 2019 at 17:13
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a partition of a finite set $S$ is any set $\{S_1, \dots, S_n\}$ of $n$ subsets of $S$...

You answered your own question. Which element of $\{1,2,3,4\}$ is a subset of $\{1,2,3,4\}$?

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  • $\begingroup$ With the von Neumann definition of naturals and assuming we started at $1$, it turns out all the elements would be subsets... (The joys of the non-abstractness of set theory.) $\endgroup$ May 12, 2019 at 19:09
  • $\begingroup$ @DerekElkins Interesting.. $\endgroup$ May 12, 2019 at 23:32
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    $\begingroup$ @DerekElkins: If you're going to play that game, you should not go around calling them $1, 2, 3, 4$ in contexts where they are intended to be sets. Someone else might be using a different construction of the naturals. Also, starting the von Neumann construction at 1 instead of 0 is in very poor taste. You want the von Neumann naturals to be the fixed points of the cardinality operator, but that construction leaves them offset by one. $\endgroup$
    – Kevin
    May 13, 2019 at 4:04
  • $\begingroup$ @Kevin You won't get an argument from me. I prefer foundations where this kind of thing can't happen in the first place. And I, of course, agree with starting from $0$; I'm a programmer. I did see someone recently define von Neumann naturals starting at $1$, and it was a bit nauseating. $\endgroup$ May 13, 2019 at 4:19
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The following remarks do not aim at proving as a general law that no set is a partition of itself, but at showing via couterexamples that it is not the case that any set is a partition of itself,

(1) A partition of a set S is, by definition, a family of sets.

So, if a set S is not itself a family of sets, it cannot be a partition of any set; and consequently, it cannot be a partition of itself.

(2) Lets consider a given set S that is a family of sets, say

S = { {1, 2} , {2,3} }

Now suppose that S is a partition of itself.

That would mean that : Intersection(S) is empty

( for by definition, the intersection of the elements of a partition is empty).

But that is not true, for here we have

Intersection (S) = {1,2} Inter {2,3} = {2}

(3) If S = { {1, 2} , {2,3} } were a partition of itself, then Union (S) would be equal to S ( this is also a necessary condition to be a partition).

But Union (S) = {1, 2} Union {2,3} = { 1,2,3}

and this is not equal to S.

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