# Why is a set not a partition of itself?

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?

• 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 $S$. May 12, 2019 at 17:56
• (Your conditions were incorrect - I have edited them.) May 12, 2019 at 17:57

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 $$\{1,2,3,4\}$$.

• Another instructive reply is that $\{\{1\},\{2\},\{3\},\{4\}\}$ is a partition. May 12, 2019 at 17:10
• Which part of the definition above breaks down here?*
– GooJ
May 12, 2019 at 17:10
• $S$ 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 $S$. May 12, 2019 at 17:13

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 $$\{1,2,3,4\}$$?

• 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.) May 12, 2019 at 19:09
• @DerekElkins Interesting.. May 12, 2019 at 23:32
• @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. May 13, 2019 at 4:04
• @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. May 13, 2019 at 4:19

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.