Encountered a question on set operations and am kind of lost...

$(D \bigcup \{A\})\oplus A$

$A: \{a,b,c,d\}, D: \{b,d\}$

Does $\{A\}$ mean $\{\{a,b,c,d\}\}$, making it a set of a set? Would set D union set of set A (?) be the same as set D union set A, or are they different?

I would understand how to list the elements if it were $(D \bigcup A)\oplus A$, but the $\{A\}$ is what is confusing me.

  • $\begingroup$ What does $\oplus$ mean in this context? Exclusive-or, perhaps? Or does it mean disjoint union? $\endgroup$ – goblin Jan 28 '19 at 7:44
  • $\begingroup$ Symmetric difference, so if A⊕B, elements that are in A or B but not both $\endgroup$ – Bandolero Jan 28 '19 at 7:55
  • $\begingroup$ @Bandolero the usual notation for symmetric difference is $AΔB$ $\endgroup$ – ℋolo Jan 28 '19 at 11:44

Good question!

However, you should write $$A = \{a,b,c,d\},D = \{b,d\},$$ because after all, $D$ is being defined to equal the set $\{b,d\}$, so why use a symbol other than the equality symbol to denote equality in this case?

You're correct that $\{A\}$ means $\{\{a,b,c,d\}\}$. Hence $$D \cup \{A\} = \{\{a,b,c,d\},b,d\},$$ which has three elements, assuming $b$ and $d$ are distinct. I'm sure you can take it from there.

  • $\begingroup$ Ok, so from what I understand, it should be $\{\{a,b,c,d\}b,d\}\oplus\{a,b,c,d\}$. still not sure about the answer tho, but I'm guessing it would be between $\{\{a,b,c,d\}a,c\}$ or $\{a,c\}$ $\endgroup$ – Bandolero Jan 28 '19 at 8:13
  • $\begingroup$ @Bandolero, to find $P \oplus Q$, the idea is to list all elements that are in precisely one of $P$ or $Q$, but not both. It's like the union of $P$ and $Q$, but you don't include the items that are in both sets. $\endgroup$ – goblin Jan 28 '19 at 8:20
  • $\begingroup$ I got that part, but my question is would say $\{P\}$'s elements be considered seperate from $P$'s elements; i.e. that I could list all the elements in $P$ or $Q$ (but not both) but keep $\{P\}$ untouched? $\endgroup$ – Bandolero Jan 28 '19 at 8:25
  • $\begingroup$ @Bandolero, I'm not sure I understand, but $\{P\}$ only has one element, so unless $P$ is an element of itself, you can assume that $P$ and $\{P\}$ are disjoint. Look up "urelement" for more information; the important fact is that they're disallowed in standard treatments of set theory. $\endgroup$ – goblin Jan 30 '19 at 11:58

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.