Let

Prove or disprove the following.

- .
- .
- .
- .
- .
- .
- .
- .
- .

* False. *

These sets are not equal since they do not contain the same elements (since and ).
* False. *

Since contains an element (e.g. 1) that is not contained in , we have . (Again, this is asking us to observe that .)
* False. *

Again, but ; hence, .
* True. *

* Proof. * By definition of we have . Since , we have .∎
* False. *

Since , but , we have .
* False. *

The element is in , but is not in . Hence, .
* True. *

* Proof. * To show , we must show that every element in is also in , and that (so that is a proper subset). First, the only two elements of are the sets and . Since each of these sets is contained in by definition of , we have . Finally, since the set , but , we see that the inclusion is proper. Hence, . ∎
* False. *

Again, the problem is emphasizing the difference between an element and the set that contains that element. In this case and is not an element of (even though and are elements of , the set containing the sets and is not).
* True. *

* Proof. * By the definition of we have . Since , we have . ∎

*Related*

For part e) {1,2} is an element of {{1},{2},{1,2}}. So it should be true.

And thanks too for your work.

For\: part\: \left ( d \right ), \: \left \{ 1,2\right\left\right \} is\: an\: element\: of\: \left \{\left \{1\left\right \},\left \{2\left\right \},\left \{1,2\left\right\}\left. \right \}. So\: it\: should\: be\: true.

For part d) {1,2} is an element of {{1},{2},{1,2}}. So it should be true

My apologies and please delete above comments.

For part g, I think you accidentally wrote set C instead of set D starting halfway through the answer. Also, thanks for your work. Really well done.

Fixed now. Thanks for letting me know!

The same aplies to part h.

Argh. Thanks! One day all of the typos and errors will get found and corrected…