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Antisymmetric: $\forall x\forall y[ ((x,y)\in R\land (y, x) \in R) \to x= y]$ reflexive: $\forall x[x∈A\to (x, x)\in R]$

What really is the difference between the two? Wouldn't all antisymmetric relations also be reflexive?

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    $\begingroup$ E.g. "$\leq$" and "$<$" are antisymmetric and "$=$" is reflexive. $\endgroup$ Apr 27 '15 at 17:02
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Here are a few relations on subsets of $\Bbb R$, represented as subsets of $\Bbb R^2$. The dotted line represents $\{(x,y)\in\Bbb R^2\mid y = x\}$.

Symmetric, reflexive: sym_r

Symmetric, not reflexive sym_not_r

Antisymmetric, not reflexive antisym_not_r

Neither antisymmetric, nor symmetric, but reflexive not_antisym_r

Neither antisymmetric, nor symmetric, nor reflexive not_antisym_not_r

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  • $\begingroup$ This is a great visual approach to understanding the meaning of the words! $\endgroup$
    – LSpice
    Apr 27 '15 at 20:04
  • $\begingroup$ Thank you so much for making these, they're great! $\endgroup$
    – Tabrock
    Apr 29 '15 at 2:48
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No, there are plenty of anti-symmetric relations that are not reflexive.

For instance, let $R$ be the relation $R=\{(1,2)\}$ on the set $A=\{1,2,3\}$. This relation is certainly not reflexive, but it is in fact anti-symmetric. This is vacuously true, because there are no $x$ and $y$, such that $(x,y)\in R$ and $(y,x)\in R$.

Edit: Why is this anti-symmetric? Because in order for the relation to be anti-symmetric, it must be true that whenever some pair $(x,y)$ with $x\neq y$ is an element of the relation $R$, then the opposite pair $(y,x)$ cannot also be an element of $R$. This is true for our relation, since we have $(1,2)\in R$, but we don't have $(2,1)$ in $R$.

Also, the relation $R=\{(1,2),(2,3),(1,1),(2,2)\}$ on the same set $A$ is anti-symmetric, but it is not reflexive, because $(3,3)$ is missing.

As for a reflexive relation, which is not anti-symmetric, take $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$.

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  • $\begingroup$ Thank you so much for your answer, the last two parts make sense! :) I'm a little lost on the first part because the law says that if (x,y) and (y,x) then y=x. Could you elaborate a bit more on how R = {(1,2)} is anti-symmetric? $\endgroup$
    – Tabrock
    Apr 27 '15 at 17:35
  • $\begingroup$ Sorry, I think I messed up. Let me edit my post. There. Now, I have redone the last two examples, because they were wrong. I'll edit my post further to elaborate on why the first relation is in fact anti-symmetric. :)@TaylorTheDeveloper $\endgroup$
    – Mankind
    Apr 27 '15 at 17:42
  • $\begingroup$ This may sound like a naive question but would'nt this example be asymmetric also then by vacuous agument $\endgroup$ Oct 19 '20 at 11:31
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    $\begingroup$ @angshuknag Yes, the relation $R=\{(1,2)\}$ is also asymmetric. In fact, being asymmetric is equivalent to being both anti-symmetric and not reflexive. $\endgroup$
    – Mankind
    Oct 19 '20 at 15:08
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They're two different things, there isn't really a strong relationship between the two.

Based on the definitions you're using, they both give two different criteria for concluding that $(x, x) \in R$.

For any antisymmetric relation $R$, if we're given two pairs, $(x, y)$ and $(y, x)$ both belonging to $R$, then we can conclude that in fact $x = y$, so that that

$$(x, y) = (x, x) = (y, x),$$

and $(x, x) \in R$. It may really be better stated as saying that

$$\text{ If } x \neq y, \text{ then at most one of $(x, y)$ or $(y, x)$ is in $R$}.$$

That is, it may be a bit misleading to even think about $(x,y)$ and $(y, x)$ as being pairs in $R$, since antisymmetry forces them to in fact be the same pair, $(x, x)$.

Antisymmetric relations may or may not be reflexive. $<$ is antisymmetric and not reflexive, while the relation "$x$ divides $y$" is antisymmetric and reflexive, on the set of positive integers.

A reflexive relation $R$ on a set $A$, on the other hand, tells us that we always have $(x, x) \in R$; everything is related to itself. Reflexive relations may or may not be symmetric, or antisymmetric:

$\leq $ is reflexive and antisymmetric, while $=$ is reflexive and symmetric.

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  • $\begingroup$ Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! For example, the relation "$x$ divides $y$" on the set of all integers is neither symmetric nor antisymmetric (and neither reflexive, nor irreflexive). $\endgroup$
    – pjs36
    Apr 27 '15 at 17:52
  • $\begingroup$ The divisibility relation is reflexive, even on all integers. $\endgroup$
    – LSpice
    Apr 27 '15 at 20:04
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    $\begingroup$ @JadeNB Thank you, of course you're right; I'm not sure why I had decided it wasn't! $\endgroup$
    – pjs36
    Apr 27 '15 at 20:07
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    $\begingroup$ Probably the presence of 0 caused some reflexive (no pun intended!) worries. $\endgroup$
    – LSpice
    Apr 27 '15 at 20:11

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