# relation that is the symmetric closure of a f'

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 $$Y$$.

• 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 $\{ (1,2), (2,1), (2,3), (3,2), (3,3) \}$.
– Joe
Sep 26 '19 at 9:28
• By the way, the “full question” that you have included isn’t a question. Is there more? Does it ask you a question?
– Joe
Sep 26 '19 at 9:31
• Oops missed a bit thanks Sep 26 '19 at 10:03
• @Joe So basically (2,1) and (3,2) can be at least 2 elements of Y? Sep 26 '19 at 10:14
• @Joe Ahh so {(1,2)(2,1)(2,3)(3,2)(3,3)} is an element of Y? 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.

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