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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.

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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".

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  • $\begingroup$ 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. $\endgroup$ – JMoravitz Jul 19 at 18:23
  • $\begingroup$ That is part of the reason I suggested defining an universal set, $\Omega$. $\endgroup$ – Viktor Glombik Jul 19 at 18:26
  • $\begingroup$ 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. $\endgroup$ – Nelver Jul 19 at 18:44
  • $\begingroup$ Or will there be any ambiguity? $\endgroup$ – Nelver Jul 19 at 18:45
  • $\begingroup$ @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. $\endgroup$ – Viktor Glombik Jul 19 at 18:45
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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 $B$

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