Is antisymmetric the same as reflexive?

Question

Note: The following definitions from my book, Discrete Mathematics and Its Applications [7th ed, 598].

This is my book's definition for a reflexive relation

This is my book's definition for a anti symmetric relation

Is a reflexive relation just the same as a anti symmetric relation? From what I've, the only way to meet that antisymmetric requirement is to have the same ordered pair, say an element a from Set A, (a,a). If you have anything other than the same ordered pair, (1,2) and (2,1), it will not meet the antisymmetric requirement. But the overall definition of reflexive relation is that it's the same ordered pair. Are they just two ways of saying the same thing? Is it possible to have one and not the other?

Answer 1

No, antisymmetric is not the same as reflexive.

An example of a relation that is reflexive, but not antisymmetric is the relation $$R={(1,1),(1,2),(2,2),(2,1)}$$ on $$A={1,2}.$$ It is reflexive because for all elements of $A$ (which are $1$ and $2$), $(1,1)\in R$ and $(2,2)\in R$. The relation is not anti-symmetric because $(1,2)$ and $(2,1)$ are in $R$, but $1\not=2$.

An example of a relation that is not reflexive, but is antisymmetric is the empty relation $R=\emptyset$ on $A={1}$. It doesn't have $(1,1)$ in it, but it is vacuously antisymmetric.

On a further note: reflexive is: $$\forall a\in A, (a,a)\in R.$$

Anti-symmetric is $$\forall (a,b),(b,a)\in R, a=b.$$

Note that the statements have different hypotheses and conclusions (even though they look similar).

Answer 2

They are not the same thing. For example, on $\mathbb R$ the strict inequality is anti-symmetric but not reflexive.... If you want one which is not vacuous, change it to be $\leq $ if at least one of the numbers is negative.

On $\mathbb Z$, if we define $(x,y) \in R$ if and only if $x-y$ is even, then this is reflexive but not santi-ymmetric.

Answer 3

No.

First, you can have a reflexive relation which is not antisymmetric. For example: an integer number $a$ is related to other integer $b$ if and only if $a$ and $b$ have the same parity. This is not antysymmetric, because $2R4$ and $4R2$, but $2\neq 4$. But it is reflexive, since every integer has the same parity than itself.

Second, you can have an antysymmetric relation that it is not reflexive. The empty relation on a non empty set, for example: that is, a non empty set in which no element is related to any element (not even to itself).

Answer 4

Reflexivity requires all elements to be both way related with themselves. However, this does not prohibit non-equal elements from being both way related with each other.

That is: $\qquad\forall a\in A: (a,a)\in R$ from which we can prove: $$\forall a\in A\;\forall b\in A:\Big(a=b \;\to\; (a,b)\in R \wedge (b,a)\in R\Big)$$

Antisymetricality requires any two elements which are both way related with each other to be equal. Which means it prohibits non-equal elements from being both-way related to each other; however, this does not require all elements to be both way related with themselves. $$\forall a\in A, \forall b\in A:\Big((a,b)\in R \wedge (b,a)\in R \;\to\; a=b\Big)$$ Or via contraposition: $\qquad\forall a\in A, \forall b\in A:\Big(a\neq b \;\to\; (a,b)\notin R \vee (b,a)\notin R\Big)$

Thus they are not the same thing.

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Examples are easy to generate. From the set $A=\{0, 1\}$ The following relations are:

$$\begin{array}{c|c|c} R & \text{Reflexive} & \text{Antisymmetric} \\ \hline \{(0,0), (1,1)\} & \text{Yes} & \text{Yes} \\ \{(0,0), (0,1), (1,0), (1,1)\} & \text{Yes} & \text{No} \\ \{(0,0), (0,1), (1,0)\} & \text{No} & \text{No} \\ \{(0,0), (0,1)\} & \text{No} & \text{Yes} \end{array}$$