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^ That's the problem, but I'd also like to know why...

I have:

a. True

b. True

c. False, because you'd need to have a set with the null set in it on the right hand side.

d. True, seems like what was missing from c

e. True

f. True

g. True


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Everything is correct. – Alex Becker Jan 25 '13 at 4:31
up vote 6 down vote accepted


  • When $a\in A$ then you can find easily $a$ among the other elements of the set $A$.

So, for example $a$ and $b$ are true because you can easily find the element $\emptyset$ in both $\{\emptyset\},\{\emptyset,\{\emptyset\}\}$. While $c$ is wrong, since $\{\emptyset\}$ is not an element of itself.

  • When $a\in A$ then we always have $\{a\}\subset A$

So, for example, $f$ is correct. Because in it we see that $\{\emptyset\}\in\{\emptyset,\{\emptyset\}\}$ and so we have $$\{\{\emptyset\}\}\subset\{\emptyset,\{\emptyset\}\}$$

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nice addition! + 1 – amWhy Jan 25 '13 at 4:45
@amWhy: It is kind of you that you take time for my answer. – Babak S. Jan 25 '13 at 4:49

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