0

Prove: For all sets A, B, C, if B ∩ C ⊆ A, then (C - A) ∩ (B - A) = ∅.

Since B ∩ C is a subset of A, B is a subset of A.

Since B ∩ C is a subset of A, C is a subset of A.

Since B is a subset of A, subtracting A from B will result in the empty set.

Since C is a subset of A, subtracting A from C will result in the empty set.

The empty set intersected with the empty set equals the empty set.

NEW:

Let x belong to (C - A) ∩ (B - A)

C - A = C ∩ A^c

B - A = B ∩ A^c

x belongs to (C ∩ A^c) ∩ (B ∩ A^c)

x belongs to C ∩ A^c ∩ B ∩ A^c

x belongs to B ∩ C ∩ A^c ∩ A^c

x belongs to B ∩ C ∩ A^c

Since B ∩ C is a subset of A, B ∩ C ∩ A^c is a subset of A ∩ A^c (the empty set)

Since x belongs to B ∩ C ∩ A^c, x belongs to the empty set.

Trivial: The empty set is a subset of (C - A) ∩ (B - A)

(C - A) ∩ (B - A) = ∅

QED

Hello
  • 509

2 Answers2

1

The first bolded sentence is not a valid conclusion: for example, consider $A = \{ 1, 2 \}$, $B = \{ 1, 3 \}$, $C = \{ 1, 4 \}$. Then $B \cap C = \{ 1 \} \subseteq A$ but $B$ is not a subset of $A$.

Daniel Schepler
  • 20,364
  • I knew it seemed too simple. Any tips on how to start this then? – Hello May 09 '17 at 00:34
  • $\emptyset \subseteq (C - A) \cap (B - A)$ is trivial. For the other inclusion, start by assuming $x \in (C - A) \cap (B - A)$... – Daniel Schepler May 09 '17 at 00:35
  • Before I try anything quick question: if A ⊆ B ∩ C (switching the order), could I say that A ⊆ B and A ⊆ C? – Hello May 09 '17 at 00:36
  • @Hello Yes (and in fact, $A \subseteq B \cap C$ if and only if $A \subseteq B$ and $A \subseteq C$). – Daniel Schepler May 09 '17 at 00:38
  • Is this element argument you are referring to? Yes I agree proving the empty set is a subset is trivial, so now I should assume x is in (C - A) ∩ (B - A) and then show that x belongs to the empty set to prove equality? – Hello May 09 '17 at 00:43
  • will update the original post with my new approach..... – Hello May 09 '17 at 00:47
  • see the original post! thoughts? – Hello May 09 '17 at 00:51
  • What I was thinking was something along the lines: assume $x \in (C - A) \cap (B - A)$. Then $x \in C - A$ so $x \in C$ but $x \not\in A$; and similarly, $x \in B - A$ so $x \in B$ but $x \not\in A$. Since $x \in B$ and $x \in C$, then $x \in B \cap C$, so $x \in A$. We now have a contradiction since also $x \not\in A$. So, since a contradiction implies anything, $x \in \emptyset$. – Daniel Schepler May 09 '17 at 00:57
0

Firstly notice that your statement of the following statement is NOT true: "Since B ∩ C is a subset of A, B is a subset of A."

Notice $C \setminus A= C \cap A^c$

Thus $(C \setminus A) \cap (B \setminus A) = C \cap B \cap A^c$

Since $B \cap C \subset A$, then $B \cap C \cap A^c = \emptyset$

Jay Zha
  • 7,792