Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
Thanks very much -- I'm beginning to see the light. –  user10756 Mar 5 '13 at 16:11
4  
Uhh, that's a train. You're standing on the tracks. :-) –  Asaf Karagila Mar 5 '13 at 16:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.