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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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