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If someone could please verify if my operation table in the picture below is correct it'd be much appreciated.

The task was given:

$P_D=\{A: A \subset D\}$ and $D$ is a $3$-element set $D=\{a, b, c\}$ with the operation of symmetric difference defined as $A+B=(A-B) \cup (B-A)$. Write all the elements of $P_D$ and then write the operation table for $(P_D, +)$.

I got a little befuddled in doing this because I'm relatively new to dealing formally with sets, and I had trouble thinking of a, b, and as sets because in the previous problem it used A, B instead. Also, it was a little trickier with 3 sets and some simplified nicely while others did not...so I wasn't sure if that's okay or if I was perhaps doing something wrong or if you're only supposed to simplify to a certain point when doing an operation table (I didn't show my work here though...its on other pages and much messier).

I apologize for the small print and messy handwriting, but I think it's readable enough and I'd already re-written it several times so I thought that was good enough (I have trouble "organizing" my math).

enter image description here

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In general, in operation tables you're supposed to simplify until each entry looks like one of the elements in the group, so you're probably expected to simplify more in most of them.

You're missing the curly brackets around each element of $P_D$; the power set of $D$ contains subsets of $D$, and $a$ is not a set, but $\{a\}$ is. This might be why you're having trouble simplifying them.


I'll write three of the simplifications to illustrate:

\begin{align} \{a\} &\triangle \{b\} = (\{a\}\setminus \{b\}) \cup (\{b\} \setminus \{a\}) = \{a\}\cup \{b\}=\{a,b\} \\ \{a,b\} &\triangle \{a,c\} = (\{a,b\}\setminus \{a,c\}) \cup (\{a,c\} \setminus \{a,b\}) = \{b\}\cup \{c\}=\{b,c\} \\ \{a,b\} &\triangle \{c\} = (\{a,b\}\setminus \{c\}) \cup (\{c\} \setminus \{a,b\}) = \{a,b\}\cup \{c\}=\{a,b,c\} \end{align}

The rest are similar. (Feel free to ask about this if it's confusing.)

You could compute all the others like this, or you could realize that the symmetric difference of two sets $X$ and $Y$ is the set containing all the elements that are in $X$ or in $Y$ but not in both: $$ X \triangle Y = (X \setminus Y) \cup (Y \setminus X) = (X \cup Y) \setminus (Y \cap X) ,$$ so, since the sets are small, you could find the symmetric differences by inspection, by removing any elements that $X$ and $Y$ share and collecting the rest.


It would also save you a lot of work to show that the operation is commutative, because then you could compute only what's above the diagonal and complete the rest by symmetry.

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  • $\begingroup$ Yes, this is confusing to me, because I was not under the impression that A-B=A as you're saying...I thought "-" meant take the element out of A that are also in B. There can be overlap so you can't simply say that set is just a can you? $\endgroup$ – Liam Cooney Aug 16 '15 at 19:29
  • $\begingroup$ You're right, $A-B\neq A$ in general. If $A$ and $B$ are sets, $A-B$ means $A \cap B^c$, or the intersection between $A$ and the complement of $B$, so it consists of the elements that are in $A$ and are not in $B$. Could you rephrase your second question? I don't really understand what you mean. $\endgroup$ – coldnumber Aug 16 '15 at 19:33
  • $\begingroup$ Basically I don't see how your 3 "simplifications" are true. You say {a}+{b}={a, b} but I don't see how that's possible. My trouble is with {a}-{b}=a and {b}-{a}=b (and the same goes for the rest of the simplifications you worked out) If you have the set {a} and take all the element of {b} out of it, then unless {a} /cup {b}=\emptyset, then I don't see how that equals {a}. Is that clearer? If we're still not on the same page I can try making a Venn Diagram $\endgroup$ – Liam Cooney Aug 17 '15 at 1:58
  • $\begingroup$ Yes, that's clearer. Writing the difference as an intersection makes it easier to see: $\{a\}-\{b\}=\{a\}\cap \{b\}^c = \{a\} \cap \{a,c\} = \{a\}$ ( this is how set difference was actually defined in my class). I think it's better to think of $X-Y$ as removing anything that the sets $X$ and $Y$ have in common from $X$. If they don't have anything in common, there is nothing to remove. Does that make sense? $\endgroup$ – coldnumber Aug 17 '15 at 2:05
  • $\begingroup$ Your new defn/eqn does not make sense to me (what does the superscript c mean?...maybe if I know that the rest will fall in place), but the what follows it does ("I think it's better..."). What I'm saying is that we don't know whether or not X and Y have anything in common so how can we say that X-Y=X? $\endgroup$ – Liam Cooney Aug 17 '15 at 2:15

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