Give set $A = \{\emptyset, \{\emptyset\} , \{\{\emptyset\}\}\}$, is $\{\emptyset\} \cap A = \{\emptyset\}$ or just $\emptyset$? I understand that $\emptyset \cap A = \emptyset$ (and why this is true), but I'm thrown off by the repeated nesting of empty sets within sets.
Given that, I am under the impression that $\emptyset \cup A = A =\{\emptyset\} \cup A$. Is this also true?
Best Answer
In general,
$$\emptyset \cup S = S \not=\{\emptyset\} \cup S$$
Note that in our case, $A = \{\emptyset, \{\emptyset\} , \{\{\emptyset\}\}\}$, $A $ is a set of sets and $\emptyset $ is just a set with nothing. $B = \{\emptyset\}$ is also a set of sets, but happens to only contain the empty set. Then we have
$$A = \{\emptyset, B, \{B\}\} $$
And thus $A \cap B = B = \{\emptyset\}$ given that $B \subset A $.
Also, for the $A $ given, $A \cup B = A $ only because $B \subset A$. There are plenty of sets $S $ where $S \cup B \not= S$.