Say I have a relation where $R = X\times X$

$X =\{1,2,3\}$

So there will be $3^3$ functions in the relation. i.e. $27$

I'm struggling to understand what the symmetric closures on the set of functions $R$ are?

I know symmetric closure is filling in all the missing symmetric cases but I am not sure what the questions is asking.

Full Question

For each function $f$ in the set of functions from $X$ to $X$, consider relation that is the symmetric closure of the function $f'$. Let us call the set of theses symmetric closures $Y$. List at least $2$ elements of $Y$.

  • $\begingroup$ The symmetric closure of a relation R is the smallest relation containing R that is symmetric. For example, if one of your functions was $R = \{ (1,2), (2,3), (3,3) \}$, then its symmetric closure would be $\{ (1,2), (2,1), (2,3), (3,2), (3,3) \}$. $\endgroup$
    – Joe
    Sep 26 '19 at 9:28
  • $\begingroup$ By the way, the “full question” that you have included isn’t a question. Is there more? Does it ask you a question? $\endgroup$
    – Joe
    Sep 26 '19 at 9:31
  • $\begingroup$ Oops missed a bit thanks $\endgroup$ Sep 26 '19 at 10:03
  • $\begingroup$ @Joe So basically (2,1) and (3,2) can be at least 2 elements of Y? $\endgroup$ Sep 26 '19 at 10:14
  • 1
    $\begingroup$ @Joe Ahh so {(1,2)(2,1)(2,3)(3,2)(3,3)} is an element of Y? $\endgroup$ Sep 26 '19 at 10:27

If $R=X\times X$, then for any $(x,y)\in R$ also $(y,x)\in R$, so $R$ is already closed under symmetry. The symmetric closure of $R$ is thus $R$ itself.

You call $R$ a set of functions, but $R$ is not a set of functions. A function is a relation $f\subset X\times Y$ such that for every $x\in X$ there is exactly one pair $(x,y)\in f$. We also write $(x,y)\in f$ as $f(x)=y$ or as $f:x\mapsto y$.

As for the Full Question, if $f$ maps some $x\in X$ to $f(x)=y\in X$, then $(x,y)\in f$, and therefore $(x,y)\in f'$, where $f'$ is the symmetric closure of $f$. Since $f'$ is symmetrically closed, we must have $(y,x)\in f'$ as well.

So $f'=\{(x,y)\in X\times X\mid f(x)=y\text{ or }f(y)=x\}$. Alternatively, $f'=\{(x,y)\in X\times X\mid (x,y)\in f\text{ or }(y,x)\in f\}$. Note that $f'$ does not have to be a function anymore, but it still is a relation $f'\subset X\times X$.

Listing two different elements of the set $Y$ of all symmetric closures of functions $f:X\to X$ is not generally possible:

  • If $X=\varnothing$, then there is only one function $f=\varnothing$, so $f'=\varnothing$, therefore $Y=\{\varnothing\}$ and listing two elements is impossible.

  • If $X=\{x\}$ is a singleton, there is also only one function $f=\{(x,x)\}$, making $Y=\{\{(x,x)\}\}$, again listing two elements is impossible.

In all other cases it is possible, though.

  • $\begingroup$ @Vsotverp but X = {1,2,3} not empty or just x $\endgroup$ Sep 26 '19 at 11:46
  • $\begingroup$ Yes, for this case it's no problem. It's just not generally true. $\endgroup$
    – Vsotvep
    Sep 26 '19 at 11:50
  • $\begingroup$ @Vsotverp so you're just showing me if X was empty or had 1 element it wouldn't have 2 possible elements in Y? $\endgroup$ Sep 26 '19 at 11:54
  • $\begingroup$ Yes, your question was not clear on whether X was the same X as before, or just a general set. $\endgroup$
    – Vsotvep
    Sep 26 '19 at 12:09
  • $\begingroup$ @Vsotverp I said X = {1,2,3} . Btw i appreciate the in depth response. $\endgroup$ Sep 26 '19 at 12:12

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