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If $$A = \{1,2,3,4\} $$ and $$R = \{(3,3), (4,4), (1,4)\}$$

This example is antisymmetric but not symmetric.

However, the definition of Antisymmetric taken from Merriam-Webster is this:

relating to or being a relation (as “is a subset of”) that implies equality of any two quantities for which it holds in both directions (the relation R is antisymmetric if aRb and bRa implies a = b)

Please forgive my ignorance, but I'm incredibly confused by the definition. I'm obviously wrong, but it suggests to me that there is/should be a (4,1) in R. However, if there was a (4,1) in R that would make R symmetric and not antisymmetric.

I found a previous Question that asked something similar but was answered with the use of Digraphs. It made me understand how to identify antisymmetric relations, but I still don't understand the core concept.

Thank you so much for any help!

Rumblebutt
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    Anti-symmetric means that there aren't any distinct values $a\neq b$ such that both $(a,b)$ and $(b,a)$ are in $R$. That is clearly the case for your example. – lulu May 26 '16 at 22:37
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    Can you explain why you think $(4,1)$ should be in $R$? – Eric Wofsey May 26 '16 at 22:38
  • It was a critical typo. (2,1) should have been (1,4). As lulu just mentioned, it suggests both (a,b) and (b,a) are in R but (4,1) is not in R. – Rumblebutt May 27 '16 at 02:10

1 Answers1

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You're making a really common mistake, a mistake you probably even know about in abstract terms, but it's hard to recognize it when it shows up in real life. (Especially when the wording is as convoluted as in MW; that's really shoddy work on their part imo). Trying to keep as many words from the original definition as possible, I'll rewrite it so the logic is more clear:

A relation is called antisymmetic when the following statement is true:

  • If the relation holds in both directions for two quantities, then the two quantities are equal.

Adding some symbols in there, the bullet point can be rewritten:

  • If $(a,b)\in R$ and $(b,a)\in R$, then $a=b$.

Your question amounts to: "I know $(4,1)\in R$, and I know $R$ is antisymmetric, so why isn't $(1,4)\in R$?".

Hopefully the answer is a little more clear now: the bullet point doesn't say anything at all about what happens if all you know is $(a,b)\in R$. (You haven't satisfied both parts of the condition, so the bullet point tells you nothing.)

Community
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Eric Stucky
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  • Somehow I missed this explanation before and just found it here. With this and your answer I think I understand it now. Apparently, I just didn't understand the if/then statement (i probably still don't). It sounds like it should be more clearly written that everything is antisymmetric unless (a,b)∈R and (b,a)∈R and a≠b. The definition only seems to work for a=b scenarios and doesn't include (a,b)∈R but (b,a)∉R. – Rumblebutt May 27 '16 at 03:04
  • @Barry: It may be worth noting that there is an extremely related property called "asymmetry" which has a more transparent definition: "If $(a,b)\in R$ then $(b,a)\notin R$." The problem is that this definition does not allow for reflexivity, since if $b=a$ then this pair is both in $R$ and not in $R$! The whole point of antisymmetry is to capture this idea but to also allow reflexivity. – Eric Stucky May 27 '16 at 14:17
  • For feeling like you have a more solid understanding, perhaps it may help to show that "$R$ is antisymmetric" is equivalent to the following property being true: "there is at most one element of ${(a,b),(b,a)}$ contained in $R$." (Note that if $b=a$ then the set only has one element!) This is the way I think about antisymmetric, actually, but it's not so good for new folks because it hides the genuine difficulties in the definition with some cute set theory trick. – Eric Stucky May 27 '16 at 14:19