# Am I solving this question correctly?

How can I evaluate the following term: $$\left((\{a,b\}\cup\{b,a\})\times(\{b,a\}\cap\{a,b\})\right)\setminus \left((\{b,a\}\setminus\{a,b\})\cup(\{a,b\}\times\{b,a\})\right)$$

You can see the notes to my approach in this picture. Am I solving it correctly? • What is the question? Shall you evaluate the term in the first two lines? Jan 29 '15 at 6:57
• computing the terms in the first two line Jan 29 '15 at 6:59
• Also, for sets holds $\{a,b\}=\{b,a\}$ which makes your term really redundant — are you sure you wrote everything down correctly? which is the source of this assignment? Jan 29 '15 at 7:00
• I have edited the question. Jan 29 '15 at 7:13

Under the assumption that you wrote the term down correctly:

\begin{align} &\left((\{a,b\}\cup\{b,a\})\times(\{b,a\}\cap\{a,b\})\right)\setminus \left((\{b,a\}\setminus\{a,b\})\cup(\{a,b\}\times\{b,a\})\right)\\ &= \left((\{a,b\}\cup\{a,b\})\times(\{a,b\}\cap\{a,b\})\right)\setminus \left(\emptyset\cup(\{a,b\}\times\{a,b\})\right)\\ &= \left(\{a,b\}\times\{a,b\}\right)\setminus \left(\{a,b\}\times\{a,b\}\right)\\ &= \emptyset \end{align} I am meanly using the folloving theorems: $$\{a,b\}=\{b,a\}$$ $$A\cup A=A=A\cap A$$ $$A\setminus A = \emptyset$$

EDIT: If you really need to calculate the product, just do so in the last step: \begin{align} &(\{a,b\}\times\{a,b\})\setminus(\{a,b\}\times\{a,b\})\\ &= \{(a, a),(a,b),(b,a),(b,b)\}\setminus\{(a, a),(a,b),(b,a),(b,b)\}\\ &=\{\} \end{align}

LATER EDIT: As I understood what you did in the photo, I can say that your approach is correct except one little thing: $$\{\}\cup\{a,b,\cdots\}\neq\{\{\},a,b,\cdots\}$$ but rather $$\{\}\cup A=A$$ For all sets A. You confused this with $$\{\{\}\}\cup\{a,b,\cdots\}=\{\{\},a,b,\cdots\}$$

Which is a mistake that often occurs.

• would that answer be equivalent to a a blank set { } Jan 29 '15 at 7:22
• $\emptyset=\{\}$, just another symbol for the same thing. Jan 29 '15 at 7:25
• @melancholyx3 not sure if you will notice — I added a few things in an edit. There is also one on your own question. Jan 29 '15 at 7:47
• so {}∪{a,b,⋯} would equal to {a,b,...}? I think I understand what you're trying to say now. Jan 29 '15 at 7:50
• Exactly — the empty sets contains zero elements, and thus there are no elements added by toking the union. But $\{\{\}\}$ contains one element (even if it's onvy the empty set), which is thus added to the other set. Jan 29 '15 at 7:52