4
$\begingroup$

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? enter image description here

$\endgroup$
10
  • 2
    $\begingroup$ What is the question? Shall you evaluate the term in the first two lines? $\endgroup$ Jan 29 '15 at 6:57
  • 2
    $\begingroup$ computing the terms in the first two line $\endgroup$ Jan 29 '15 at 6:59
  • 2
    $\begingroup$ 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? $\endgroup$ Jan 29 '15 at 7:00
  • 2
    $\begingroup$ Then please edit your question adding the correct content! $\endgroup$ Jan 29 '15 at 7:11
  • 2
    $\begingroup$ I have edited the question. $\endgroup$ Jan 29 '15 at 7:13
12
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ would that answer be equivalent to a a blank set { } $\endgroup$ Jan 29 '15 at 7:22
  • 1
    $\begingroup$ $\emptyset=\{\}$, just another symbol for the same thing. $\endgroup$ Jan 29 '15 at 7:25
  • 1
    $\begingroup$ @melancholyx3 not sure if you will notice — I added a few things in an edit. There is also one on your own question. $\endgroup$ Jan 29 '15 at 7:47
  • 1
    $\begingroup$ so {}∪{a,b,⋯} would equal to {a,b,...}? I think I understand what you're trying to say now. $\endgroup$ Jan 29 '15 at 7:50
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 29 '15 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.