29
$\begingroup$

What exactly is meant by "closed under fill in the blank"?

Thanks.

$\endgroup$
7
  • 27
    $\begingroup$ Something is "closed under fill in the blank" if applying fill in the blank to elements of something yields elements of something. $\endgroup$
    – user228113
    Mar 1 '16 at 19:22
  • 2
    $\begingroup$ To complement the previous answer, the set of integers is closed under addition because if you take two integers and add them, you will always get another integer. The set of integers is not closed under division, because if you take two integers and divide them, you will not always get an integer. $\endgroup$
    – anonymouse
    Mar 1 '16 at 19:26
  • $\begingroup$ The set of all closed sets is closed under finite union means that, if $\{ A_i\}_{i=1}^n$ is a finite collection of closed sets, then $\cup_{i=1}^n A_i$ is a closed set. $\endgroup$ Mar 1 '16 at 19:27
  • $\begingroup$ We say something is closed under operation x if applying operation x to a set of elements y yields elements in y. $\endgroup$
    – user310648
    Mar 1 '16 at 19:27
  • 2
    $\begingroup$ @SteveKass Well perhaps infinite sums should be considered a different operation altogether. Is such thing as applying an operation infinitely many times well defined? $\endgroup$
    – user310648
    Mar 2 '16 at 1:39
39
$\begingroup$

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.

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

$\endgroup$
8
  • 1
    $\begingroup$ I find the example of scalar multiplication less fortunate, because the main instance where I would talk about scalar multiplication is in linear algebra, and there the description you give does not apply. In linear algebra a subspace is closed under (addition and) scalar multiplication, because multiplying any vector from the subspace by any scalar from the field (which is not from the subspace) returns a vector in the subspace. This is apparently not what you wanted to say, so why did you say "scalar" there? $\endgroup$ Mar 2 '16 at 13:05
  • $\begingroup$ @MarcvanLeeuwen I chose to include the "scalar" since "multiplication" can have different meanings in different contexts (inner/outer product, tensor product, etc.) and since my example was concerned with scalars (specifically integers, but the operation is still the same). I was simply trying to precisely convey the operation at hand, but apparently I failed somewhat. $\endgroup$ Mar 2 '16 at 15:22
  • $\begingroup$ @Charles Btw, if you found that any of these answers answer your question satisfactorily, you can accept one of them, get the question off of the "unanswered list" and make the author of the answer happy. Cheers. $\endgroup$ Sep 4 '16 at 19:41
  • 1
    $\begingroup$ @GENIVI-LEARNER No a set does not have to be infinite if its is to be closed under an operation, depending on that operation; see the $\{1,-1\}$ example in the answer (for the operation of multiplication). As for the general importance of the notion: in algebra one can study "structures" as soon as one has a hold of all values that can ever result from certain operations (for instance starting from the numbers between 0 and 1, one can get all real numbers by arithmetic operations, but it is important to know that one never gets anything else). $\endgroup$ Apr 2 '20 at 11:51
  • 1
    $\begingroup$ ... If a subset is closed under the operations one to be performed, it is relevant to know the results remain in the subset. Taking again the real numbers, the subset of rational numbers (fractions of integers) is closed under all arithmetic operations; so if a problem has as initial numbers rationals, and its solution can be obtained using arithmetic only, then you can be sure without doing any computation that the solution will be a rational number, no matter how many operations are needed to obtain the solution (but limits cannot be taken; the rationals are not closed under taking limits). $\endgroup$ Apr 2 '20 at 11:57
10
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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, $n\in\mathbb{N}$ (BTW, $0\in\mathbb{N}$). $f$ 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 $Y=X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy