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I have a homework assignment where, given some definitions, I need to prove that every set has a partition. There were some very elaborate ideas on how to prove this, but I realized that the definitions we were given allow a set to be a partition of itself. And if that is the case, then there clearly exists a partition of every set.

I'm not sure if I'm taking advantage of the simplicity of the definitions we are given. I tried to search if a set is a partition of itself, and not much if anything came up. The claim "every set is a partition of itself" seems to violate our english definition of partition, but be consistent with the mathematical definitions I have found.

For the sake of this discussion, I would think it be best to only consider non-empty sets because there seems to be some variety in how partitions are defined that prohibit the empty set from having partitions.

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Yes, you´re right , but for the empty set itself you will have just the empty set. –  user40276 Apr 7 at 19:56

1 Answer 1

No, a set is not a partition of itself. But rather the singleton $\{X\}$ is a partition of $X$ whenever $X$ is non-empty.

In the other end of the spectrum, $\{\{x\}\mid x\in X\}$ is a partition of $X$, again when $X$ is non-empty. And if $X$ has at least two elements, then the two suggestions are indeed distinct.

Your reasoning for excluding the empty set is correct.

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Actually I'd say that the empty set is a partition of itself: $\varnothing$ is a collection of disjoint nonempty sets whose union is $\varnothing$. –  Henning Makholm Apr 7 at 20:30
    
The reason I did not include the empty set is because some definitions state that the disjoint sets be non-empty. The definition given to me in my class did not have this constraint, and our Professor stated that by our definition, the empty set does have a partition for the exact reason you stated. –  Reck Apr 7 at 20:41
    
@Henning: That may depend on the definitions. When I'm teaching this topic, we always require that the set partitioned is non-empty. I don't find it satisfying to think of the empty set as a partition of itself very much; even if formally you might be correct. –  Asaf Karagila Apr 7 at 20:42
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@Reck: As Henning points out, one can use the empty set as a partition of itself. But I find it philosophically unsatisfying. If you definitions permit this, then you're about done (and you can note that the singletons partition gives $\varnothing$ when $X=\varnothing$). –  Asaf Karagila Apr 7 at 20:44
    
@AsafKaragila I agree with your view. My class is introductory, which is why the definitions are a bit less strict, but the definitions I've seen support your stance. –  Reck Apr 7 at 20:58

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