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


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Something is "closed under fill in the blank" if applying fill in the blank to elements of something yields elements of something. – G. Sassatelli Mar 1 at 19:22
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. – anonymouse Mar 1 at 19:26
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. – Steven Gregory Mar 1 at 19:27
We say something is closed under operation x if applying operation x to a set of elements y yields elements in y. – Senpai Mar 1 at 19:27
@SteveKass Well perhaps infinite sums should be considered a different operation altogether. Is such thing as applying an operation infinitely many times well defined? – Senpai Mar 2 at 1:39

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.

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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? – Marc van Leeuwen Mar 2 at 13:05
@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. – Lovsovs Mar 2 at 15:22

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

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

Which means: $ \star : S \times S \to S $ or in words: You may pick any two elements of $S$, by the operation $ \star$ and them can be assigned a new value in $S$.

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 generally.

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