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I have to say if {2} is an element of the given sets. I'm reading {2} as if it was a subset which would make the problem true for C, D and E only correct?

F it isn't because that is a nested subset and A/B don't contain subsets.

For each of the sets, determine whether {2} is an element of that set.
a) {x ∈ R | x is an integer greater than 1}
b) {x ∈ R | x is the square of an integer}
c) {2,{2}}
d) {{2},{{2}}}
e) {{2},{2,{2}}}
f ) {{{2}}}

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Looks like you're on the right track. – Andrew Jul 10 '12 at 20:34
up vote 3 down vote accepted

First recall that $a$ is an element of $b$ if and only if $a\in b$. On the other hand, $a$ is a subset of $b$ if and only if for every $x$ such that $x\in a$, $x\in b$.

My usual tip in this situation is to replace $\{2\}$ by $x$. Now we can think of the given sets,a,b are both sets of numbers, now we can see that $x$ is not an integer, nor a real number, so it is not in either a or b.

On the other hand, c can be written as $\{2,x\}$, which means that $x$ is an element of this set. Similarly d,e both have $x$ as an element. However what is the set f? $\{\{\{2\}\}\}=\{\{x\}\}$ and $x$ is not an element of this set, it is rather an element of an element of this set.

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Is your first sentence really what you intended to write? It is a tautology that gives no information unless the reader does not know the meaning of the symbol $\in$. – Rahul Jul 10 '12 at 21:55
@RahulNarain: There is always confusion; e.g., does "X contains 2" mean $2\in X$, or does it mean $2\subseteq X$? (Recall that $2$ is itself a subset). Perhaps Asaf is thinking along those lines, and something got lost in the translation... – Arturo Magidin Jul 10 '12 at 23:52
@Rahul: The $\in$ relation is atomic, how would you express it? It seems a bit strange to write this first sentence, but this is the same as truth definition "$\varphi\land\psi$ is true if and only if $\varphi$ and $\psi$ are true". – Asaf Karagila Jul 11 '12 at 5:52

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