# 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.

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 $$B$$