Does the interpretation of " $\forall$ " differ in these contexts?

Question

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

Reflexive Relation :

If $(a,a) \in R \ \forall a \in A$ then it's a reflexive relation.

Note : for all $a \in A$

This means, $R$ is a Reflexive Relation as long as it contains $(a,a)$ for all $a$ it doesn't matter what else is present in $R$.

Alright fine!

Symmetric Relation :

$(a,b) \in R \iff (b,a) \in R$ for all $ a,b \in A$

Again, note the for all

From here : https://en.m.wikipedia.org/wiki/Symmetric_relation

But I have learnt that, $ R = \{ (1,2), (2,1) \} $ is also a symmetric Relation. Then what happened to "for all" part?

Clearly not all elements of $A$ were used, but it's still symmetric? However, for reflexive relation for all part in the definition is a game changer?

Similarly with transitive relation, if $(a,b)$ and $(b,a) \in R \implies (a,c) \in R,$ for all $a,b,c \in A$ but we know $R= \{(1,2),(2,1),(1,1),(2,2) \}$ once again, I don't see the for all part happening here.

This is as detailed as I could've gotten with my doubt. Please help!

Answer 1

It is still "for all". The implication $$(1,3)\in R\iff (3,1)\in R$$ is true (as both sides are false), as are all other possible combinations.

Answer 2

The $\forall$ means the same in all statements.

Let's read the symmetric relation again, with some added brackets:

$\Big[ (a,b) \in R \iff (b,a) \in R\Big]$ for all $ a,b \in A$

An example, take the relation $R=\{\,(1,2), (2,1), (1,1)\,\}$:

$(1,2) \in R \Leftrightarrow (2,1) \in R$, that's true, since both sides lie in $R$.

$(1,4) \in R \Leftrightarrow (4,1) \in R$, that's true, since both sides do not lie in $R$.

Etc. We may conclude that $R$ is indeed symmetric.

For the transitive part, it is the same: The statement

$\forall a,b,c\in A: \Big[(a,b) \in R \text{ and }(b,c)\in R \Rightarrow (a,c)\in R\Big]$

means that for all $a,b,c\in A$, no matter what they are specifically, if $(a,b)$ and $(b,c)$ are part of the relation, then $(a,c$) has also be a part of the relation.

Let's take $R=\{\,(1,2),(2,3),(1,3),(1,4)\,\}$, which is indeed a transitive relation.

$(1,2)\in R$ and $(2,3)\in R$, so transitivity tells us that $(1,3)\in R$ aswell (which is the case).

Therefore $\Big[(1,2) \in R \text{ and } (2,3)\in R\Rightarrow (1,3)\in R\Big]$ holds.

$(1,2) \in R$ and $(2,4)\notin R$. $(1,4)\in R$, but that's not needed.

Therefore $\Big[(1,2) \in R \text{ and } (2,4)\in R\Rightarrow (1,4)\in R\Big]$ holds.

Answer 3

Another way to look at "for all" is by translating it to something with the words "there exists."

For example, $$ (\forall a \in A).((a,a) \in R)$$ is equivalent to $$ \lnot(\exists a \in A).\lnot((a,a) \in R).$$

You can read this as: "There does not exist any $a$ in $A$ such that $(a,a)$ is not a member of $R.$"

The symmetric property can similarly be written as $$ \lnot(\exists a,b \in A).\lnot((a,b) \in R \iff (b,a) \in R).$$

The expression $\lnot(P \iff Q)$ is a bit awkward to express in words, but an equivalent statement is, "exactly one of the statements $P$ or $Q$ is true, but not both." So the symmetric property is true unless there are elements $a$ and $b$ that have the relationship $aRb$ but do not have the relationship $bRa.$ Does your relationship provide any such pairs? No? Then your relationship is symmetric.

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By the way, I would prefer to define the symmetric property this way: $$ (\forall a,b \in A).((a,b) \in R \implies (b,a) \in R).$$ This turns out to be equivalent to the expression with $\iff,$ because anything that is true "for all $a,b\in A$" is also true "for all $b,a\in A,$" and vice versa. But I find the version with $\implies$ easier to describe and reason about. In particular, converting it into the "exists" wording, the property becomes:

There do not exist $a,b\in A$ such that $(a,b) \in R$ and $(b,a)\not\in R.$

Answer 4

To further elaborate the answer given by Martin, I would like to emphasize on the validity of the biconditional $p\leftarrow\rightarrow q$:

$(a,b)\in R\leftarrow\rightarrow (b,a)\in R$ $\forall a,b\in A$ holds when

• Either both $p$ and $q$ are true or

• both $p$ and $q$ are false

  Your confusion arises due to the unawareness of the second case in which the pair $(a,b)$ is absent in $R$ for some $a,b\in A$.