# Subset proof, show A⊆B

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 ∩ ∅)
A⊆B


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

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

• 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 – joshau May 10 at 21:34
• 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. – José Carlos Santos May 10 at 21:41
• 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 – joshau May 10 at 21:43
• 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. – José Carlos Santos May 10 at 21:47
• 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) – 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 $$A\subseteq B$$.

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