Why isn't $\Bbb{R}^2$ a subset of $\Bbb{R}^3$? I'm trying to learn a little about vector spaces and such before heading off to university after the Summer. My question is a misunderstanding I have over, what seems, two conflicting pieces of information:  I read that the vector space $\Bbb{R}^2$ is not a sub space of the vector space $\Bbb{R}^3$.
I don't understand why this is, given the case that if you consider the set of all vectors:   $S=\begin{bmatrix}x\\y\\0\end{bmatrix}$  which is by definition a vector space because it follows the conditions of being closed under addition and scalar multiplication. All of $S$ can be found in the vector space $\Bbb{R}^3$, and therefore it is a sub space of $\Bbb{R}^3$.   However, isn't $S$ nothing other than the real $x$-$y$ plane, which is, by definition the vector space $\Bbb{R}^2$? This would suggest that $\Bbb{R}^2$ is a sub space of $\Bbb{R}^3$. But then we look again at the first statement, which says that this cannot be the case! Please show me the flaw in my thinking.   Thank you all in advance!
 A: $T:\Bbb{R^2}\to\Bbb{R}^3$ defined by $$T(x, y) =(x, y, 0) $$
is a one-to-one linear map.
$\Bbb{R}^2\cong \text{im} T \subset \Bbb{R}^3$
I.e $\Bbb{R}^3$ contains a isomorphic copy of $\Bbb{R}^2$ as subspace. Infact that subspace is the $xy$-plane.
Not exact $\Bbb{R}^2$ but an isomorphic copy of $\Bbb{R}^2$ which have same linear structure as of $\Bbb{R}^2$ but as a set they are different. Since $(x, y) \neq (x, y, 0) $ . Infact we can't compare them as they have different length!
A: It depends on who you ask. Yes, $S=\{(x,y,0)\,|\,x,y\in\mathbb{R}\}$ is a two-dimensional subspace of $\mathbb{R}^3$. However, technically, $S\neq\mathbb{R}^2$, because the lengths of the vectors are not the same.
In $S$, you have length-3 vectors with the final component set to zero; in $\mathbb{R}^2$, we just have length-2 vectors. So, the mathematical statement $S=\mathbb{R}^2$ which means $(\forall s\in S) s\in \mathbb{R}^2$ and $(\forall z\in\mathbb{R}^2) z\in S$ does not hold, since a statement like $(1,1,0)\in\mathbb{R}^2=\{(x,y)\,|\,x,y\in\mathbb{R}\}$ is false (due to the third component in $(1,1,0)$.)
