Take the 2-minute tour ×  ^ 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   share|improve this question
Everything is correct. –  Alex Becker Jan 25 '13 at 4:31 add
comment  1 Answer  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
$$\{\{\emptyset\}\}\subset\{\emptyset,\{\emptyset\}\}$$
share|improve this answer nice addition! + 1 –  amWhy Jan 25 '13
at 4:45 @a