# Property: closure under an binary operation

We have the following definition:

Definition: let $*_A:A^2 \rightarrow A$ and $B \subseteq A$, with $B \neq \emptyset$, $B$ is closed under $*_A$ if $a *_A|_Bb \in B$ $\forall a,b \in B$.

Property: let $*_A:A^2 \rightarrow A$ and $A$ is closed under $*_A$, then $a *_A b \in A$ $\forall a,b \in A$

Proof: by hypothesis $A$ is closed under $*_A$ then for all $a,b \in A$ we have $a *_A|_Ab \in B$, but $*_A|_A:A^2 \rightarrow A$ is restriction function of $*_A$ on $A$ and for all $a,b \in A$ we have $a*_A|_Ab=a *_A b$, therefore $a *_A b \in A$ by Leibniz's Law.

It is correct?

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$B \neq \emptyset$ is a nonsense assumption –  Martin Brandenburg Apr 11 '13 at 17:50

By definition, if you're given a function $*:A\times A\to A$, $*(a,b)=a*b\in A$ for any $a,b\in A$. This is simply because the function is defined to have codomain $A$, that is, any pair of elements $a,b\in A$ are by definition sent to an element $*(a,b)=a*b$ which is in $A$. This is in general written as

An operation on a set $A$ is a function $*:A\times A\to A$.

It makes sense to consider subsets of $A$, and ask whether the image of $B\times B$ under $*$ is contained in $B$, that is, if $$*\mid_{B\times B}:A\times A\to A$$ is such that $$\operatorname{Im}(*\mid_{B\times B})\subseteq B$$

And example would be the mapping $$*:\Bbb Z\times \Bbb Z\to \Bbb Z$$ that maps $$n*m\to nm+n+m$$

In this case, $*\mid_{\Bbb N}$ is an operation on $\Bbb N$ because $mn+m+n$ will be a natural for any pair of natural numbers $m,n$.

ADD Consider an example that fails, $$*:\Bbb Z\times \Bbb Z\to\Bbb Z$$ defined via $$*(a,b)=a+(-b)$$

This is not an operation in $\Bbb Z_{>0}$, for example.

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Hi @Peter Tamaroff, i don't know... I am confused, but in my case I think the restriction function like $f {\restriction_{X \times Y}}: X \to Y$ with $X \subseteq S$ and $Y \subseteq T$ and $f: S \to T$... thank you for everything! –  Гарнак Олэговытц Apr 11 '13 at 19:28
@GarnakOlegovitc You can only restrict a function $f:A\to B$ to a subset of $A$. Thus, it makes no sense to talk about the restriction of $f:X\to Y$ in $X\times Y$. My whole point is that what you should worry about is proving a subset of $A$ is closed under $A$'s operation, but by definition $A$ is closed under the operation: that is what an operation is, a function $f:A\times A\to A$. –  Pedro Tamaroff Apr 12 '13 at 2:59