1. ## Closure

hi
would you help me find out if those sets are closed under addition and scalar product or not.
$\displaystyle A_{1}=\left \{ (cosx,sinx)/x\in \mathbb{R} \right \}$
$\displaystyle A_{2}=\left \{ (x,y)/x\in \mathbb{Z},y\in \mathbb{Z} \right \}$
for the second i think $\displaystyle \mathbb{Z}$ is closed under addition,$\displaystyle \left (\forall x\in \mathbb{Z} \right ),\left ( \forall y\in \mathbb{Z} \right ); \left (x+y \right )\in \mathbb{Z}$.am i right ?
thanks for helping me.

2. Originally Posted by Raoh
hi
would you help me find out if those sets are closed under addition and scalar product or not.
$\displaystyle A_{1}=\left \{ (cosx,sinx)/x\in \mathbb{R} \right \}$
$\displaystyle A_{2}=\left \{ (x,y)/x\in \mathbb{Z},y\in \mathbb{Z} \right \}$
for the second i think $\displaystyle \mathbb{Z}$ is closed under addition,$\displaystyle \left (\forall x\in \mathbb{Z} \right ),\left ( \forall y\in \mathbb{Z} \right ); \left (x+y \right )\in \mathbb{Z}$.am i right ?
thanks for helping me.
Well $\displaystyle A_1$ is clearly not closed under multiplication by scalar: just remember that $\displaystyle |\cos{x}|\leq{1}$
Try to do the second one by yourself

Tonio

3. for the second one, $\displaystyle \mathbb{Z}$ isn't closed under scalar product,but it's closed under addition. $\displaystyle \left (\forall a\in \mathbb{Z} \right ),\left (\exists \lambda \in \mathbb{R} \right );\lambda a\notin \mathbb{Z}$,is it okay ?
thanks a lot.

4. Originally Posted by Raoh
for the second one, $\displaystyle \mathbb{Z}$ isn't closed under scalar product,but it's closed under addition. $\displaystyle \left (\forall a\in \mathbb{Z} \right ),\left (\exists \lambda \in \mathbb{R} \right );\lambda a\notin \mathbb{Z}$,is it okay ?
thanks a lot.

Exactly. Anyway, none of them is a subspace.

Tonio

5. thanks a lot.