So I was reviewing this question and Im lost on how to do this question, and Ive seen to of misplaced the notes. The question is as follows: if (A ∩ C) ⊆ (B ∩ C) and (A ∩ C̅) ⊆ (B ∩ C̅)

then A ⊆ B

My attempt so far:

(x∈A ∩ x∈C) ⊆ (x∈B ∩ x∈C)  
(x∈A ∩ x∉C) ⊆ (x∈B ∩ x∉C)

Since x∉C and x∈C => ∅ 

(x∈A ∩ ∅) ⊆ (x∈B ∩ ∅)
(x∈A) ⊆ (x∈B ∩ ∅)

I think this is correct though Im a bit rusty and not sure if this is correct

  • $\begingroup$ Draw a Venn diagram to help you understand the problem. $\endgroup$ – avs May 10 at 21:38
  • 1
    $\begingroup$ $x \in A$ (and the like) are assertions, not sets. I'm not familiar with how you're combining these assertions with set intersection $\cap$. $\endgroup$ – Brian Tung May 10 at 21:40

Take $a\in A$. Then either $a\in C$ or $a\in C^\complement$. If $a\in C$, then $a\in A\cap C$ and therefore $a\in B\cap C$; in particular, $a\in B$. And if $a\in C^\complement$, then $a\in A\cap C^\complement$ and therefore $a\in B\cap C^\complement$; in particular, $a\in B$, again.

Concerning your proof, I don't understand the sentence “Since $x\notin C$ and $x\in C\implies\emptyset$”.

  • $\begingroup$ so since a∈C or a∈ not C we can cancel the C portions out? Still lost how do we get to the final answer $\endgroup$ – joshau May 10 at 21:34
  • $\begingroup$ Either $a\in C$ or $a\in C^\complement$ and, in both cases, $a\in B$. If there's a passage that you don't understand, please tell me which one. $\endgroup$ – José Carlos Santos May 10 at 21:41
  • $\begingroup$ I understand it a bit more in either case we have a∈B regardless of if a is in C, thus the C portion is ignored. Is this correct $\endgroup$ – joshau May 10 at 21:43
  • $\begingroup$ I did not ignore it. I proved that if $a\in C$ then $a\in B$ and if $a\in C^\complement$, then $a\in B$ too. $\endgroup$ – José Carlos Santos May 10 at 21:47
  • $\begingroup$ Yes I see and understand that, with that in mind what happens to the a∈C and a∈ not C part? Is it this: (x∈A ∩ x∉C) ⊆ (x∈B ∩ x∉C) and since Either a∈C or a∈ not C, in both cases a∈B and a∈A thus (x∈A) ⊆ (x∈B) $\endgroup$ – joshau May 10 at 21:49

I don't think I understand your attempt. Let me outline a simple proof for you:

Let $x\in A$. Then, we know that either $x\in A\cap C$ or $x\in A\cap\overline{C}$. Using this fact, the hypothesis give us that either $x\in B\cap C$ or $x\in B\cap \overline{C}$. Therefore, $x\in B$ and hence $A\subseteq B$.

  • $\begingroup$ I get confused near the end, how do you know after applying the hypothesis x∈B and thus A⊆B. $\endgroup$ – joshau May 10 at 21:40
  • $\begingroup$ Because we conclude that either $x\in B\cap C$ or $x\in B\cap\overline{C}$ and in both cases we have $x\in B$. $\endgroup$ – Ariel Serranoni May 10 at 22:37

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