Example of a subset of $\mathbb{R}^2$ that is closed under vector addition, but not closed under scalar multiplication?

Question

I've found several examples which are closed under scalar multiplication, but not vector addition, but I can't come up with one that is closed under vector addition, but not scalar multiplication.

$\mathbb Z^2$, for example. – Ittay Weiss Mar 01 '14 at 22:33 — Ittay Weiss, Mar 01 '14 at 22:33

score 10 · Answer 1 · answered Mar 01 '14 at 22:35

10

$S=\{(r,s)\in\mathbb{R}^2:r,s\in\mathbb{Q}\}$ works as well.

answered Mar 01 '14 at 22:35

r9m

17,938

score 10 · Accepted Answer · edited Mar 01 '14 at 22:42

10

The set $\{(x,y): x\ge0, y\ge0\}$ is closed under addition, but not under scalar multiplication, since $-1\cdot(1,1)=(-1,-1)$, for example.

edited Mar 01 '14 at 22:42

Jeroen Zuiddam

117

answered Mar 01 '14 at 22:35

user84413

27,211

1

I'm surprised how I wasn't able to think of this. This is visually appealing too. My example for satisfying scalar multiplication but not vector addition was the union of the x and y-axis, so an example of the opposite kind is just the first quadrant – Grid Mar 01 '14 at 22:43

score 6 · Answer 3 · answered Mar 01 '14 at 22:33

6

The subset $S=\{(n,0)\in\mathbb{R}^2:n\in\mathbb{Z}\}$ is closed under addition but not scalar multiplication, e.g., $\frac{1}{2}\cdot (1,0)=(\frac{1}{2},0)\notin S$.

answered Mar 01 '14 at 22:33

Zev Chonoles

129,973

score 5 · Answer 4 · answered Mar 01 '14 at 22:43

Let $\{v_1, v_2, \ldots\}$ be any countable subset of $\Bbb R^2$ whose elements are not all 0. Then $$\left\{\sum_{i=1}^\infty z_iv_i \middle\lvert z_i\in \Bbb Z\right\}$$ is closed under addition. But it is a countable set, and any set (except {0}) that is closed under scalar multiplication is uncountable.

Example of a subset of $\mathbb{R}^2$ that is closed under vector addition, but not closed under scalar multiplication?

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