# Closure — why does it matter?

I understand the concept of a set having closure under a particular operation; that, for example, the natural numbers are not closed under subtraction. But why does closure matter? Is it necessary for an operation to be closed to be useful? If the set is not closed under an operation does that mean that the outcome of the operation is not defined? Or is it only a warning not to assume the outcome of the operation to be the same kind of number?

The term "closure" is useful for communicating an idea to someone else. If you are discussing an operation on an underlying set, you might want to say "If I take two elements from my set and operate on them, will I end up in my set." But this is wordy, so instead you say "Is the set closed under this operation?"

The context will determine if it "matters." For example in Group Theory, we care about something called "Groups." But in order to have a group (which is a set and operation following particular rules), and all the nice properties that come with them, we need to have our operation be closed.

The outcome of the operation can still be defined. In the example you gave, 3-5 is certainly defined (because you get an answer), but the natural numbers are not, as you said, closed under subtraction, because you take two elements from your set, operate on them, and do not remain inside your set.

When we talk about a set being closed under a particular operation, we are indeed talking about whether or not the operation is well-defined and will yield the same kind of thing. It's possible that the operation has a perfectly well-defined rule, but gives us a different kind of thing (subtraction of natural numbers, for example), or it may be that the operation isn't even well-defined (such as multiplicative inverse on the set of all $2\times 2$ real matrices).

• How can you tell if an operation that is not closed is nevertheless well defined? – Thurber Nov 8 '13 at 19:10
• Usually that will happen if your set is a smaller part of a larger set which is closed under the operation (such as the integers, in the case of subtraction). More generally, we say that a set $X$ is closed under an $n$-ary operation $\#$ if $\#$ is a function $X^n\to X$. In that sense, $X$ is closed under $\#$ if and only if $\#:X^n\to X$ is well-defined. – Cameron Buie Nov 8 '13 at 19:11
• Thanks. I have in mind the common practice in psychology and political science of calculating the means of ordinal scales. Ordinals are not closed under division. This seems to me to be an operation that would not be well defined? How could you tell whether it was well defined? – Thurber Nov 8 '13 at 19:12
• I would say that it is not well-defined, in the sense that the mean of a set of ordinal numbers is not necessarily an ordinal number. However, if we are dealing with finite ordinal numbers, then we can consider them as real numbers, and get a well-defined function of sorts, just not an operation that necessarily gives back ordinals. – Cameron Buie Nov 8 '13 at 19:16
• In particular, when we say "operation," we tend to mean that our output is the same kind of thing that our inputs are. Otherwise, we talk about functions. – Cameron Buie Nov 8 '13 at 19:17