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$s_1: A \backslash B = \emptyset$ and $B \backslash A= \emptyset$
$s_2: A \cup B = A \cap B$.

I'm not sure if what I have is a correct way to show this proof, but I have the following so far:

Suppose $A \backslash B = \emptyset$ and $B \backslash A = \emptyset$. We want to show that $A \cup B = A \cap B$.
Take any $x \in A$. If $A \cap B$, then $x \in A$ and $x \in B$.
Since $x \in A$ and $x \in B$, then $A \cap B \subseteq A \cup B$.

I'm not really sure if this is correct or how to progress.

Not2Scary
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  • If you're allowed to do proofs by illustration, I do recommend drawing the Venn diagram. (Even if you're not allowed, it's helpful to see it.) – NoName Oct 28 '20 at 02:41
  • I can seer how $s_2$ is true with a Venn diagram, but I don't know how to write the proof when I take into account $s_1$. – Not2Scary Oct 28 '20 at 02:50
  • I suggest a bridge. Prove that $s_1$ and $s_2$ are both equivalent to $A = B$. In particular, prove that $s_2 \implies (A=B).$ – user2661923 Oct 28 '20 at 02:56
  • That's actually my next step. For the assignment, I have to prove $A = B \Rightarrow s_1$ and then $s_1 \Rightarrow s_2$ and then $s_2 \Rightarrow A = B$. I've done $A = B \Rightarrow s_1$. – Not2Scary Oct 28 '20 at 03:00

2 Answers2

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Take any $x \in A$. If $A \cap B$, then $x \in A$ and $x \in B$.
Since $x \in A$ and $x \in B$, then $A \cap B \subseteq A \cup B$.

This is correct, and in fact, it doesn't depend on the assumptions in $s_1$. It's always true that $A \cap B$ is a subset of $A \cup B$.

So now you want to show that $A \cap B \supseteq A \cup B$. Well, consider a Venn diagram with sets A and B. If you draw it, you'll see that there are three parts (not counting the outside): an element is either in $A$ but not $B$, or it's in their intersection, or it's in $B$ but not $A$. So if $x$ is in $A \cup B$, and $A \backslash B$ and $B \backslash A$ are empty, then where specifically does $x$ have to be?

To complete the proof, you have to assume $s_2$ and show that it implies $s_1$. You can do this in a similar way: take an element $x \in A \cup B$. Based on where $x$ has to be according to $s_2$, you can identify parts of the Venn diagram that must be empty.

NoName
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  • Oh, okay. I think I understand. I'll have to work on it a little more, but thanks for your answer. I'll give it a shot! – Not2Scary Oct 28 '20 at 03:04
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$A-B=\phi$ means $A\subseteq B$ while $B-A=\phi$ means $B\subseteq A$ Therefore $B=A=B\cap A=B\cup A$.

herb steinberg
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