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I'm learning set theory and this is a homework question. I answered, "not necessarily because nothing is telling us whether $a$ is an element of $B$ and an element cannot be a subset". Am I right? What is the correct answer?

Thanks for your help!

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  • $\begingroup$ The answer is correct, the reasoning isn't (or worded weirdly). I would say that since $A$ isn't a subset, there is an element of $A$ that isn't in $B$, so hence, $a$ could be that element. $\endgroup$ Jul 25 '19 at 3:01
  • $\begingroup$ I think it would be good to give a specific counterexample $\endgroup$ Jul 25 '19 at 3:01
  • $\begingroup$ Try $B=\emptyset$ $\endgroup$ Jul 25 '19 at 3:05
  • $\begingroup$ @freethinker36 Given that we seem to have misinterpreted your question, it seems that the set $A$ has nothing to do with the question. Does this seem right to you? $\endgroup$
    – Sambo
    Jul 25 '19 at 3:16
  • $\begingroup$ I think it is a trick question. $\endgroup$ Jul 25 '19 at 3:19
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Usually you should come up with a counterexample rather than using ambiguous descriptions.

Counterexample: Let $A=\{0,1\}$ and $B=\{0\}$ and $a=1$, then clearly $A$ is not a subset of $B$. However, $a\in A$ but $a\notin B$.

Thanks to @Sambo, if we take $a=0$, then $a$ is both in $A$ and $B$. Therefore we conclude: If $A\not\subset B$, then “$a\in A$” has nothing to do with “$a\in B$”.

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  • $\begingroup$ In your example, $a$ is an element of $B$ $\endgroup$ Jul 25 '19 at 3:03
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    $\begingroup$ I would add a second case, $a=0$, to show that the statement "$x\in B$" could be either true or false. $\endgroup$
    – Sambo
    Jul 25 '19 at 3:04
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    $\begingroup$ @J.W.Tanner Sorry sorry. I’ve seen it and fixed it. $\endgroup$
    – Feng
    Jul 25 '19 at 3:05
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    $\begingroup$ @freethinker36 The operators $\subset$ and $\not\subset$ are only defined for sets. So if $a$ is not a set, it makes no sense to say “$a\not\subset B$”. $\endgroup$
    – Feng
    Jul 25 '19 at 3:29
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    $\begingroup$ @freethinker36 Also, if your problem is like this, then $A$ has nothing to do with $B$. That would make this problem meaningless. $\endgroup$
    – Feng
    Jul 25 '19 at 3:31
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You are correct. We know that there are elements of $A$ which are not in $B$, and there may be elements of $A$ that are in $B$. If we pick an element of $A$, it could fall into either of these two categories.

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