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From wikipedia:

Equivalently, a set P is a partition of X if, and only if, it does not contain the empty set and:

  1. The union of the elements of P is equal to X. (The elements of P are said to cover X.)
  2. The intersection of any two distinct elements of P is empty. (We say the elements of P are pairwise disjoint.)

I clearly understand that the intersection between partition is empty (point 2), but how can the union of a partition can be the all elements in the set?

If it is a partition, shouldnt they be just a part?

I imagine a set divided in 3 and the elements in the first part are not all the elements of the second part.

How do you explain this?

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4 Answers 4

up vote 6 down vote accepted

The idea of a partition is that you take a whole (the set $X$) and you divide it to parts.

Now if I cut off an apple into slices (and one core) I have several pairwise disjoint parts of the apple, but if I reassemble the parts I get a whole apple again.

Similarly we require this from a partition of a set. We want that the union of all the parts give us the entire set we partitioned.

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Is the term exhaustive in exhaustive partition redundant? –  Kaz Oct 23 '12 at 4:44
@Kaz: I never heard this term before... –  Asaf Karagila Oct 23 '12 at 18:58

I think the word partition explain itself. it is something to do with parts. so if P are all the parts of X then definitely when you put all the P together it will form X. SO THE UNION OF ALL P IS EQUAL TO X.

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The examples will help. Examples of partitions of $ \{1,2,3\} $ are \begin{equation} \{1\}, \{2\}, \{3\} \end{equation} \begin{equation} \{1,2\},\{3\} \end{equation} \begin{equation} \{1\},\{2,3\} \end{equation} \begin{equation} \{1,2,3\} \end{equation} \begin{equation} \{2\},\{1,3\} \end{equation}

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I gave some examples. But I add this if you want. –  user29999 Oct 22 '12 at 23:36

The union of all parts gives you the whole set. So if you partition a set $X$ in three parts $P_1$, $P_2$, $P_3$, then $P_1\cup P_2\cup P_3=X$.

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