Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
    
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
    
@AsafKaragila A partition of a set is a collection of non-empty subsets of the set (called "parts") which are exhaustive and mutually exclusive (pairwise disjoint). Ie: the union of all parts equals the set, and the intersection of any two parts is empty. – Graham Kemp Jun 10 at 1:23
    
@Kaz So yes, it is redundant, but sometimes we might only want to deal with a non-exhaustive partition - which is a subset of some (exhaustive) partition. – Graham Kemp Jun 10 at 1:25
    
@Graham: So a partition would be a partition of a subset, and an exhaustive partition will be a partition of the entire set? Sounds exhausting. – Asaf Karagila Jun 10 at 4:31

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 $P_1\cup P_2\cup P_3=X$.

share|cite|improve this answer

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}

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

I believe your confusion regarding the definition of a partition, P, of a set X may stem from conflating the elements of X with the elements of P. The elements of a partition are non-empty subsets of X. For P to be a partition of X it's elements (subsets of X) must be disjoint and cover all of X.

If you keep in mind that the elements of P are non-empty subsets of X, things should fall into place.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.