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Let $A = \{3,4\}$ be a subset of $S = \{1,2,\ldots,6\}$. Or $A \subseteq S$

and $n \in A$, what is $n \notin A$?

Would $n \notin A$ be $\{1,2,5,6\}$?

Does that question even makes sense? Help!

Trying to solve a proof question but I'm confused with the contrapositive of $n \in A$

Marconius
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misheekoh
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1 Answers1

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$n \notin A$ would be all the numbers other than $3$ and $4$. All the numbers that are not in $A$. The set of all elements not in $A$ is $A^c$

So you are not correct. $3$ is an element of $A$. However, $\{1,5,6\} \subset A^c$.

In my solution I am assuming the universe of discourse is $\mathbb{Z}$. If the universe of discourse is $A$ then $A^c$ simply contains no elements. If the universe of discourse becomes more dense then so does $A^c$.

Aleksandar
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    What about real numbers, or complex numbers, or quaternions, or polynomials? To know which elements are "not in $A$", we have to know the "universe of discourse", which may, in fact, be $S$. – David Wheeler Oct 04 '15 at 00:37
  • @ Aleksandar, sorry! It wasn't suppose to be 3 in {1,3,5,6}. It's {1,2,5,6} – misheekoh Oct 04 '15 at 00:41
  • @DavidWheeler Indeed. I assume we are working in the integers. Upvote to your comment and edit. – Aleksandar Oct 04 '15 at 00:42