3

If $A$ and $B$ are two sets, prove that $\{ A \cap B, A\setminus B \}$ is a partition of $A$.

I think this is a problem more of not knowing how to construct proofs than not comprehending the material (although, honestly, I'm finding myself to be quite weak in the Linear Algebra course I'm taking). As an aside to the question, I've acquired Velleman's How to Prove It, and I would not mind opinions as to whether that's a good start to learning how to answer questions such as this without looking at solutions.

Moving on to the question. From the way the set is defined, if $A$ is disjoint from $B$, is their intersection not the empty set?

Furthermore, I know that my proof needs to apply the definition of partition:

  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.)

But I'm not sure how to do this: I feel like if $A$ is disjoint from $B$ and my feeling that the intersection is the empty set is true, then I am in fact dealing with $A$, and that the singleton $\{A\}$ is a partition. But that's about as far as I get.

Any explanations that help me further my understanding of the material are extremely appreciated.

Peter Taylor
  • 13,425
Jake
  • 23
  • 5

2 Answers2

4

  1. Do the union of the 2 elements of your set (i.e prove that $(A \cup B) \cup (A \setminus B) = A$).

  2. Do the intersection of the 2 elements of your set (i.e prove that $(A \cup B) \cap (A \setminus B) = \varnothing$) .

Use the distribution properties of $\cap$ and $\cup$ as well as the definition of $(A \setminus B)$ to work throught the equations

Frostic
  • 1,262
  • Thanks for your answer! But the set is A ∩ B, A \ B. Applying your logic I end up reaching A ∪ (B ∩ (A/B)). I feel like perhaps I've an error along the way, though. – Jake Apr 05 '18 at 22:51
  • It’s pretty good. Keep going what is $B \cap (A/B)$? – Frostic Apr 05 '18 at 23:00
  • Oh. I understand that this must equal A, but there was a logical error in my way of thinking. I felt like B∪(A\B) includes B and B complement. Thank you. – Jake Apr 05 '18 at 23:07
  • $B\cap (A/B)$ ? (Sorry, I corrected my previous comment) – Frostic Apr 06 '18 at 18:24
  • Note that $/$ does not denote set difference. Please use $\setminus$. – Christoph Apr 10 '18 at 09:59
1

(A $\cap$ B) = A $\setminus$ (A $\setminus$ B)
So we have then:
(A $\cap$ B) $\cup$ (A $\setminus$ B ) = (A $\setminus$ (A $\setminus$ B)) $\cup$ (A $\setminus$ B ) = A
On the other hand we have:
(A $\cap$ B) $\cap$ (A $\setminus$ B ) = (A $\setminus$ (A $\setminus$ B)) $\cap$ (A $\setminus$ B ) = $\varnothing$
So it is proved that it is a partition.

Rai
  • 376
  • Thank you, I was having problems with basic set theory. This bodes ill! Haha. Much appreciated :) – Jake Apr 05 '18 at 23:08