# Is P a partition of set A

1. Mar 24, 2012

### iHeartof12

For the given set A, determine whether P is a partition of A.

A= ℝ, P=(-∞,-1)$\cup$[-1,1]$\cup$(1,∞)

Is it correct to say that P is not partition? I don't understand why.

Thank you

2. Mar 24, 2012

### zooxanthellae

A partition, as far as I know, is just a division of the set into non-intersecting subsets s.t. the union of all subsets is the original set and all subsets are non-empty.

How can you show whether or not the sets are non-intersecting? How can you show whether or not the union of the sets covers the original set? It is pretty easy to show no subset is empty.

More simply, do any of the subsets overlap with other subsets? Is there any element of ℝ that isn't in one of the subsets? If either of these is true, then the definition of a partition fails. If both are false, then we have satisfied the definition of a partition.

Last edited: Mar 24, 2012
3. Mar 24, 2012

### SithsNGiggles

Here's the definition of partition as described by my one of my professor's notes:

Let $X \not= \emptyset$ and let each $A_\alpha$, where $\alpha \in \Omega$, be a subset of $X$. Then the family of subsets $\{ A_\alpha : \alpha \in \Omega \}$ of $X$ is a partition of $X$ if and only if

(i) $A_\alpha \not= \emptyset, \forall \alpha \in \Omega$

(ii) $\bigcup_{\alpha \in \Omega} A_\alpha = X$

(iii) $\forall \alpha, \beta \in \Omega$, either $A_\alpha = A_\beta$ or $A_\alpha \cap A_\beta = \emptyset$.​

Does this make sense?

4. Mar 24, 2012

### Fredrik

Staff Emeritus
The definition makes sense, yes. Which of these three axioms would you say are true in the problem you're considering?