# Closure

• October 18th 2009, 12:20 PM
Raoh
Closure
hi(Happy)
would you help me find out if those sets are closed under addition and scalar product or not.
$A_{1}=\left \{ (cosx,sinx)/x\in \mathbb{R} \right \}$
$A_{2}=\left \{ (x,y)/x\in \mathbb{Z},y\in \mathbb{Z} \right \}$
for the second i think $\mathbb{Z}$ is closed under addition, $\left (\forall x\in \mathbb{Z} \right ),\left ( \forall y\in \mathbb{Z} \right ); \left (x+y \right )\in \mathbb{Z}$.am i right ?
thanks for helping me.
• October 18th 2009, 03:40 PM
tonio
Quote:

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

Well $A_1$ is clearly not closed under multiplication by scalar: just remember that $|\cos{x}|\leq{1}$
Try to do the second one by yourself

Tonio
• October 19th 2009, 12:35 AM
Raoh
for the second one, $\mathbb{Z}$ isn't closed under scalar product,but it's closed under addition. $\left (\forall a\in \mathbb{Z} \right ),\left (\exists \lambda \in \mathbb{R} \right );\lambda a\notin \mathbb{Z}$,is it okay ?
thanks a lot.
• October 19th 2009, 01:55 AM
tonio
Quote:

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

Exactly. Anyway, none of them is a subspace.

Tonio
• October 19th 2009, 01:59 AM
Raoh
thanks a lot.