Set $S=\{(x,y,z)| x,y,z \in \mathbb{Z}\}$ is a subset of vector space $\mathbb{R}^3$, how do I show that it is not a subspace of $\mathbb{R}^3$.

Question

So I know that set $S=\{(x,y,z)| x,y,z\in \mathbb{Z}\}$ is a subset of vector space $\mathbb{R}^3$.

Specifically, it is worded in our lecture that it is a " subset of $(\mathbb{R}^3, \oplus, \odot)$ , where $\oplus$ and $\odot$ are the usual vector addition and scalar multiplication."

My teacher has stated in our lecture that this set $S$ is not a subspace of $\mathbb{R}^3$.

But from what I can tell $S$ is:

1. Closed under addition
2. Closed under multiplication
3. Contains a zero vector $(0,0,0)$

How is it not a subspace of $\mathbb{R}^3$, what am I missing?

Answer 1

$(1,0,0)$ belongs to $S$. Scale this by the real scalar $1/2$ and obtain $(1/2,0,0)$ which is not in $S$. Hence $S$ is not a subspace.

Answer 2

It is not closed under multiplication. Take $\lambda=\sqrt{2}$ and any vector $v=(a,b,c)\in S$. Since $a,b,c\in \Bbb Z$ then $a\sqrt{2}, b\sqrt{2}, c\sqrt{2}\notin \Bbb Z$. Then $\lambda\cdot v \notin S$.

Answer 3

Edit: User Randall pointed out that I misread the question. (I assumed S was under some invalid field.)

First lets look at the definition of a subspace:

1. All products and sums composed of elements within the subspace also are in the subspace.

2. All elements in the subspace must be able to be scaled by the vector space surrounding the subspace.

3. 0 is also an element of the subspace.

You remembered rule 1 and 3 however, it's clear S is violating rule 2.

In order for S to be a subset of $R^3$, all elements within S that are scaled by any number within R are also an element of S.