# Closed Subsets

• October 7th 2009, 08:22 PM
studentmath92
Closed Subsets
Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under scalar multiplication, but not under vector addition.

Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under vector addition but not under scalar multiplication.
• October 7th 2009, 08:29 PM
tonio
Quote:

Originally Posted by studentmath92
Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under scalar multiplication, but not under vector addition.

A = { (x,y) ; |x| - |y| = 0 } . Clearly closed under scalar multiplication but (1, -1) and (1,1) belong to A, but (1,-1) + (1,1) = (2,0) doeen't.

Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under vector addition but not under scalar multiplication.

Try this one by yourself. It's easier than the above one.

Tonio
• October 7th 2009, 08:32 PM
Gamma
Quote:

Originally Posted by studentmath92
Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under scalar multiplication, but not under vector addition.

Give and prove an example of a subset A $\subseteq$ $\mathbb{R} ^{2}$ which is closed under vector addition but not under scalar multiplication.

1) $A=\{(x,0)|x\in \mathbb{R}\} \cup \{(0,y)|y\in \mathbb{R}\}$ See why its closed under scalar multiplication? but for instance $(0,1)+(1,0)=(1,1)$ is not in there.

2) Let A be the set of vectors with integer entries. So (1,1) is in there and it is clear that its closed under addition since the integers are.

But take .5(1,1) and it isnt in there anymore.