Symmetric relation confusion

Question

My textbook definition on symmetric relation:

A relation R on a set A is called symmetric if(b, a) ∈ R whenever(a, b) ∈ R, for all a, b ∈ A.

Giving a set $A= \{ {1,2,3,4}\}$

and a relation $R1 = \{{(1,1) (1,2)(2,1)}\}.$

$R2 = \{{(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)}\}$

How come both R1 and R2 are symmetric since they haven't considered other relations such as (1,3), (2,3)etc. Isn't for all a, b ∈ A in the definition means you have to consider every single element in set A?

Answer 1

Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)\in R_1$, then $(b,a)\in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.

In particular, the fact that $(1,3)\notin R_1$ is not a problem. On the other hand, if $(1,3)\in R_1$ but $(3,1)\notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.

Answer 2

You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b \in A$".

The right way to say this is

... reflexive if and only if $(a,b) \in R$ whenever $(b,a) \in R$ .

That does not say $(a,b)$ is in $R$ for all $a$ and $b$.

Answer 3

$\forall x~(A(x)\to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true.   No promise is about what may happen when $A(x)$ is false.   It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.

So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$\forall x\forall y~((x,y)\in R~\to~ (y,x)\in R)$$

Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$\forall x\forall y~((y,x)\notin R~\to~ (x,y)\notin R)$$ The claims are equivalent.

Answer 4

Also for e.g. $(1,3)$ it is true that $(1,3)\in R_1$ implies that $(3,1)\in R_1$.

This because $(1,3)\in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).

In short: if $p$ is false then $p\implies q$ (or equivalently $\neg p\vee q$) is true).

Especially a look on $\neg p\vee q$ makes things more clear. If $p$ is false then of course $\neg p$ is true and consequently also $\neg p\vee q$ is true.