# Why is $\mathbb{R}^2$ not a subset and /or a subspace of $\mathbb{R}^3$?

One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent. For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$. Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$? As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace. I would welcome any insight readers of this query can provide.

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What is true is that $\mathbb{R}^2$ is isomorphic to a subspace of $\mathbb{R}^3$. For instance via $(x,y)\longmapsto (x,y,0)$. – 1015 Mar 5 '13 at 16:04
what kind of equivalence are you looking for ? it is certainly equivalent as a vector space (and also as a topological vector space) – magguu Mar 5 '13 at 16:05
What you wrote is in fact a subspace of $\mathbb{R}^3$, namely the $x$-$y$ plane, and it is isomorphic to $\mathbb{R}^2$ (just not equal, as Asaf points out.) – Trevor Wilson Mar 5 '13 at 16:06
Appreciate the clarifications -- implication for me is that we can arbitrarily constrain $z = 0$, even under addition. I guess that's obvious, but I wasn't sure: thanks, again. – user10756 Mar 5 '13 at 16:14

The elements of $\Bbb R^2$ are vectors of two coordinates; and the elements of $\Bbb R^3$ are vectors of three coordinates. (One can easily think of those vectors as $2$-tuples and $3$-tuples, for example.)

Assuming mathematics is consistent, $2\neq 3$. Therefore no element of $\Bbb R^2$ is an element of $\Bbb R^3$. It follows that $\Bbb R^2$ is not a subset of $\Bbb R^3$.

And in order to be a subspace, one first has to be a subset. So it's not a subspace either.

What you have defined as $X$ is isomorphic to $\Bbb R^2$, but just as well you could decide that $y$ is $0$, and the identification would still be natural. $X$ is a subset of $\Bbb R^3$ and indeed a subspace, but it is not $\Bbb R^2$ as a set, it is just isomorphic to it in a very obvious way.

While isomorphism is an equivalence relation, and we often think of it almost as identity, it is still not set equality which is a stricter notion.

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Thanks very much -- I'm beginning to see the light. – user10756 Mar 5 '13 at 16:11
Uhh, that's a train. You're standing on the tracks. :-) – Asaf Karagila Mar 5 '13 at 16:14