# Prove that if $A \setminus B \subseteq C$ and $x \in A \setminus C$ then $x \in B$

Suppose A, B, and C are sets, $$A \setminus B \subseteq C$$, and $$x$$ is anything at all. Prove that if $$x \in A \setminus C$$ then $$x \in B$$.

Suppose $$x \in A \setminus C$$.

Suppose $$x \notin B$$. It follows that $$x \in A\setminus C\setminus B$$, which means $$x \in A \setminus B$$. Since $$𝐴\setminus 𝐵⊆𝐶$$ it implies that $$x \in C$$, which contradicts our assumption that $$x \in A\setminus C$$. Hence if $$x \in A \setminus C$$ then $$x \in B$$.

Is it accurate?

One more question is, wouldn't it be better if we rephrased initial statement as:

Suppose $$A \setminus B \subseteq C$$. Prove that if $$x \in A \setminus C$$ then $$x \in B$$.

It feels that "A,B,C are sets" and "x is anything at all" are redundant here.

## 2 Answers

It feels that "A,B,C are sets" and "x is anything at all" are redundant here.

"x is anything at all" can be rephrased as $$x$$ is arbitrary". Still, it is important to introduce the variables you use. One way would be to make the sets subsets of an universal set $$\Omega$$. The you could reword: Let $$A,B,C \subset \Omega$$ be sets and $$x \in \Omega$$ arbitrary. Prove that ...

Is it accurate?

I don't quite know what you mean by this.

Since you added the tag proof writing one suggestion would be to write $$\color{red}{(}A \setminus C\color{red}{)} \setminus B = A \cap C^{\complement} \cap B^{\complement} = A \cap B^{\complement} \cap C^{\complement} = (A \setminus B) \setminus C$$ This makes it clear to see that $$[(A \setminus C) \setminus B] \subset [A \setminus B]$$

Otherwise your proof is perfectly fine, though I don't think you need the last sentence as it just repeats the task. An alternative would be something like "and this concludes the proof".

• I find it safer to avoid referring to set complements in the case where we don't have a universal set defined. Without a universal set $B^c$ is undefined, though $A\setminus B$ is perfectly well defined. – JMoravitz Jul 19 at 18:23
• That is part of the reason I suggested defining an universal set, $\Omega$. – Viktor Glombik Jul 19 at 18:26
• My point was that, if you omit "A,B,C are sets" and "x is anything at all", the meaning of the statement won't change, it will just get more concise. – Nelver Jul 19 at 18:44
• Or will there be any ambiguity? – Nelver Jul 19 at 18:45
• @Nelver You will also lose information. In your case it's pretty obvious that they are sets, but as a rule of thumb I'd always state what the given symbols mean. – Viktor Glombik Jul 19 at 18:45

Since $$x$$ is not in $$C$$ and $$A-B$$ is a subset of $$C$$ we concluded that $$x$$ is not in $$A-B$$

Since $$x$$ is in $$A$$ and it is not in $$A-B$$ we conclude that $$x$$ is in $$B$$