$\begingroup$ We say something is closed under operation x if applying operation x to a set of elements y yields elements in y. $\endgroup$

SteveKass Well perhaps infinite sums should be considered a different operation altogether. Is such thing as applying an operation infinitely many times well defined? $\endgroup$

A set is closed under

*addition*if you can add any two numbers in the set and still have a number in the set as a result. A set is closed under (scalar)

*multiplication*if you can multiply any two elements, and the result is still a number in the set.

For instance, the set $\{1,-1 \}$ is closed under multiplication but not addition.

You are watching: Which of the following sets is closed under division?

I generally see "closed under *some operation*" as the elements of the set not being able to "escape" the set using that operation.

Usually (not generally) it involves an operation, for example: the natural numbers are closed under addition means that if I add two natural numbers, the sum will also be a natural number. This same set is not closed under subtraction since $1-2=-1$, and $-1$ is not a natural number

Usually the blank is filled with an "operation". For example you have a set $S = \{a,b,c,d,... \} $ which is closed under some operation $ \star $

Which means: $ \star : S \times S \to S $ or in words: You may pick any two elements of $S$, apply $ \star$ on them and they can be assigned a new value in $S$. So to say: You are not leaving your set $S$ by using this operation.

However, in general, this does not have to be the case: You may pick the integers as your set $S$ and division $\star$ as your operation.

Now you have : $4 \star 2 = 2 \in S$, which is fine. However you also have: $4 \star 3 \notin S$ as $4 \star 3$ as by our definition would be the fraction $\frac{4}{3}$

Most common operations are addition, multiplication etc. for the natural numbers, integers, real numbers etc.. However you don"t have to be so specific and can define your set and your operation arbitrarily.

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edited Mar 3 "19 at 19:03

answered Mar 1 "16 at 19:38

ImagoImago

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(This question has good answers already, but I do not see the answer that I expected, so I am writing this.)

I wish to add a formal definition. Let $X$ be a set, $n\in\couchsurfingcook.combb{N}$ (BTW, $0\in\couchsurfingcook.combb{N}$). $f$ is an $n$-ary *operation on* $X$ iff $f$ is a function from $X^n$ to $X$. Let $Y$ be a subset of $X$. $Y$ is *closed under* $f$ iff for every $a\in Y^n$ $f(a)\in Y$.

Remarks. As you see, a closed set ($Y$ in this definition) is a subset of another set ($X$ in this definition), and the operation may take and give members of $X$ which are not in $Y$. Every set $Z$ is closed under every $n$-ary operation on $Z$, so the term “closed under” is useless when $Y=X$.

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answered Dec 31 "17 at 17:30

beroalberoal

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