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Does associativity of binary operation imply closure under this operation?

Sometimes definitions of semigroup, group or vector space omit axiom of closure under corresponding operations and sometimes they don't.

One of the arguments for omitting the axiom that I found is that associativity implies closure.

As a possible proof, let + be binary operation on set A. Assume a, b and c are elements of A. Also (b + c) is in set but (a + b) is not in a set.

Then a + (b + c) is well-defined (even though the result can be out of set).

However, if we assume that + is associative, we will get:

a + (b + c) = (a + b) + c

But that is not true because second "addition" in right part is not defined for (a + b) that is out of set and c.

So for associativity, (a + b) must be in set.

Does this argument make sense? Is it true?

UPDATE: In a possible proof the error is in the first assumption. If + is binary operation on A and a and b are in A then (a + b) must be in set by definition of binary operation (A x A -> A).

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    $\begingroup$ It's not associativity that implies closure, it's "operation" that implies closure. $\endgroup$
    – bof
    Jun 25, 2016 at 0:28
  • $\begingroup$ Why is b+c with b not in the set any less defined than a+b with a,b in set but a+b not in? Ivm not saying you are wrong but I don't really get it. How about real life example. S =odd integers and + is addition. Are you claiming + isn't associative because (1+3)+5 and 1+(3+5) aren't defined. $\endgroup$
    – fleablood
    Jun 25, 2016 at 0:35
  • $\begingroup$ Yes, your example is a good one. I think + is not associative if it is defined as sum of two odd integers. Could you say if it is wrong and explain why. $\endgroup$ Jun 25, 2016 at 0:45
  • $\begingroup$ I don't know if it's wrong or right. But I'm unconvinced. I think addition is associative whatever group it is on and that definition precludes associativity rather than the other way around. But I don't know for sure. $\endgroup$
    – fleablood
    Jun 25, 2016 at 2:21
  • $\begingroup$ Can a quasi binary operation and a set actually be "defined" if the operation is not closed. You claim if a+b is not in the set + can't be associative. Because (a+b)+c is undefined. That really "feels" like a hack to me. Ivm more inclined to simply declare + isn't binary at all if a+b isn't in the set. I'd say associativity means associative when defined, which addition is. But I don't know. $\endgroup$
    – fleablood
    Jun 25, 2016 at 2:28

2 Answers 2

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Consider the set $E=\{-1,0,1\}$, under the standard operation of addition on $\bf R$. Obviously it is associative, but is the group closed under addition?

Also, when you state that, “let $+$ be binary operation on set $A$”, you are already assuming that it is closed, since this binary operation is from $A\times A$ to $A$.

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    $\begingroup$ I'd like to emphasize that closure under an operation is a property of subsets in particular, not sets in general. Accordingly, being closed under an operation is an axiom for being a subgroup or a subspace (or a subwhatever), not for being a group, or a vector space, etc. Declaring "let $+$ be a binary operation on a set $A$" (tautologically) implies that $A$ is closed under the operation as a subset of itself; this is reflected in the fact that a group or a vector space is of course a subgroup or a subspace of itself. $\endgroup$ Jun 25, 2016 at 0:43
  • $\begingroup$ Why in associativity axiom we usually don't specify that elements are distinct? If we take (1+1)+1 = 1+(1+1) then as in my question i could say that something is wrong and {-1,0,1} is not associative with addition $\endgroup$ Jun 25, 2016 at 0:56
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    $\begingroup$ @littlewhywhat Associativity requires that for all elements of the group, say $a$, $b$ and $c$, one has $a\oplus(b\oplus c)=(a\oplus b)\oplus c$, including cases in which $a=b$, $c=b$, $a=c$ or $a=b=c$. $\endgroup$
    – Workaholic
    Jun 25, 2016 at 0:58
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    $\begingroup$ @littlewhywhat What operation would you be working on in $E=\{-1,0,1\}$ in the first place? You can't use standard addition because it isn't a binary addition on $E$, i.e. we don't have $E\times E\to E$. $\endgroup$
    – Workaholic
    Jun 25, 2016 at 1:08
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    $\begingroup$ @Workaholic oh, I think I see now. There are only two sets in this case: set A we are working with and set X (X x X -> X) on which binary operation is defined. So if X is a subset of A or they are equal then operation is always closed. Interesting case is only if A is a subset of X. And that's when we try to verify if A is a subspace or subgroup or smth else as Vladimir Sotirov mentioned. $\endgroup$ Jun 25, 2016 at 6:38
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Associativity does not imply closure - both are characteristics required of a set to form a group. Both (usually) need to be verified to show that a set forms a group under said binary operation, although it is the binary operation that usually implies the closure property.

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