# Binary operations, subsets and closure

1. Jan 9, 2009

1. The problem statement, all variables and given/known data

1)
Let S be a set and p: SxS->S be a binary operation. If T is a subset of S, then T is closed under p if p: TxT->T. As an example let S = integers and T be even Integers, and p be ordinary addition. Under which operations +,-,*,/ is the set Q closed? Under which operations +,-,*,/ is the set of irrationals closed

2. Relevant equations

not sure

3. The attempt at a solution
for 1) the irrationals can't be closed under any operations. Q is probably closed under all of them except division. But it is the example and definition that gets me. If T is a subset of the Integers, there is no way it will be closed under addition. Just take the Sup of the set and add 1 or the inf of the set and subtract 1. So I guess my question is: is closure peculiar to the subset we have taken, and if so How can we say that an entire set, like the Integers is closed under some operation.

2. Jan 9, 2009

### Dick

Re: closure

The even integers are a subset of the integers and they are closed under addition, aren't they?

3. Jan 9, 2009

### HallsofIvy

Staff Emeritus
Re: closure

"Closed under an operation", *, means that if a and b are in the set then a*b is also in the set. That has nothing to do with sup or inf. You may be confusing this, algebraic, notion of "closed" with the topological notion of "closed set". They have nothing to do with one another.

If you add two integers the result is again an integer so the set of integers is closed under addition. If you add two even integers together, the result is again an even integer so the set of even integers, as Dick said, is closed under addition. The sum of two odd integers is even so the set of odd integers is not closed under addition.