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Proof: Let $A, B,$ and $C$ be sets. Prove that if $A \subseteq B \setminus C$, then $A \cap C=\varnothing$.

Question: I was marked in such a way that my proof seems to be considered incorrect. Is my proof incorrect?

My Proof:

Assume that $A \subseteq B \setminus C$. This means that every element in $A$ can also be found in $B$ and not in $C$.

Now let’s consider the intersection between $A$ and $C$, denoted as $A \cap C$. This represents the set of all elements that are both in $A$ and in $C$.

However, from our assumption we know that no element of $A$ can be in $C$. Therefore, there cannot be any elements that are in both $A$ and $C$.

Hence $A \cap C$ must be an empty set, denoted as $\varnothing$. Thus we have proven that if $A \subseteq B \setminus C$, then $A \cap C= \varnothing$. QED.

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    $\begingroup$ Argument seems fine to me $\endgroup$
    – spaceisdarkgreen
    Commented Dec 14, 2023 at 0:42
  • $\begingroup$ Maybe it was marked incorrect because too much wordy and so not a formal proof. A formal proof is as follows: assume by contradiction that there exists $x \in A \cap C$. Hence, $x \in A$ and $x \in C$. By hypothesis $A \subseteq B\setminus C$ and so, from $x \in A$, we deduce $x \in B$ and $x \notin C$. So we got $x \in C$ and $x \notin C$, a contradiction. $\endgroup$
    – Bernkastel
    Commented Dec 14, 2023 at 0:47
  • $\begingroup$ Not sure why it was marked wrong. It could be that the instructor wasn't sure what you are saying. You say "from our assumption we know that no element of A can be in C". It's not clear what assumption and when we made it. The DEFINITION of $B\setminus C$ is elements of $B$ that are not in $C$ but that is not an "assumption" and you never actually stated that no element of $A$ can be in $C$. But I wouldn't call those errors. Note a for more succinct proof could be: $A\subset B\setminus C$ so all elements of $A$ are in $B$ but not $C$. So no element of $A$ is in $C$ so $A\cap C$ is empty" $\endgroup$
    – fleablood
    Commented Dec 14, 2023 at 5:55
  • $\begingroup$ I wonder if you just have a bad teacher. Maybe they want you to prove a specific way and not by a reasoned argument (which is a proof). You may have a problem I had as well, that the definitions of set theory are so specific and clear the results seem to come directly and we don't really know what needs stating. $\endgroup$
    – fleablood
    Commented Dec 14, 2023 at 6:04

1 Answer 1

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Your proof is right. In fact,$A\subseteq{(B\setminus{C})}$, we have the $A\bigcap{C}\subseteq{(B\setminus{C})\bigcap{C}}=\emptyset$

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – CrSb0001
    Commented Dec 14, 2023 at 1:22
  • $\begingroup$ @CrSb0001 This answer is fine. $\endgroup$
    – Robert Shore
    Commented Dec 14, 2023 at 1:36
  • $\begingroup$ @RobertShore oh okay my bad $\endgroup$
    – CrSb0001
    Commented Dec 14, 2023 at 1:37

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