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I think I've been confusing myself about the language of subspaces and so on. This is a rather basic question, so please bare with me. I'm wondering why we do not (or perhaps "we" do, and I just don't know about it) say that $\mathbb{ R } $ is a subspace of $\mathbb{ R }^2 $. It's elementary to prove that the set $$ S:= \left\{ c \cdot \mathbf{x} \mid c \in \mathbb{ R }, \mathbf{x} \in \mathbb{ R }^2 \right\}$$ is a vector subspace of $\mathbb{ R } ^2$. What is confusing me is that there seems to be an isomorphism between the set $S$ and $\mathbb{ R } $: \begin{align*} \varphi: S &\rightarrow \mathbb{ R } \\ c \cdot \mathbf{x} &\mapsto c \\ \end{align*} If this is indeed true, as I believe it is having checked that $\varphi$ gives an isomorphism, wouldn't we say that $\mathbb{ R } $ is a subspace of $\mathbb{ R } ^2$?
Any help sorting out this (language) problem will be greatly appreciated!

aherring
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    The way you've written it, $S = \mathbb R^2$. – littleO Aug 14 '14 at 23:28
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    Perhaps you should define $S_{\bf x}:={c\cdot {\bf x}: c\in \mathbb{R}}$ for each ${\bf x}\in \mathbb{R}^2$. In this case you have many copies of $\mathbb{R}$ (whenever ${\bf x}\neq {\bf 0})$. – mwmjp Aug 14 '14 at 23:31
  • I like this question and I think that such a short, simple question should have a simple, short answer. $\mathbb R^2$ is defined to be the set of all ordered pairs $(x,y)$ where $x,y \in \mathbb R$. By this definition, $\mathbb R$ is not a subset of $\mathbb R^2$. Therefore, $\mathbb R$ is not a subspace of $\mathbb R^2$. – littleO Aug 14 '14 at 23:36
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    Didn't we have this discussion already in the form "Is $\Bbb R$ a subset of $\Bbb C$" a few times in very recent history? – Asaf Karagila Aug 15 '14 at 00:24
  • @AsafKaragila I think that's a different story, since there is only one possible embedding of $\mathbb R$ into $\mathbb C$ as a subfield. In contrast to the many embeddings of $\mathbb R$ into $\mathbb R^2$ as a subspace. – Christoph Aug 16 '14 at 06:04
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    @Christoph: No, there are many many many many many ways to embed $\Bbb R$ into $\Bbb C$ as a subfield. Only one way is "natural" though. – Asaf Karagila Aug 16 '14 at 06:08
  • @AsafKaragila Oh, I didn't know of other embeddings, I'm surprised! Thank you. – Christoph Aug 16 '14 at 06:15
  • @Christoph: Extend $\Bbb R$ by adding any number (between $1$ and $2^{\aleph_0}$) of transcendental elements, then take an algebraic closure of this field, then that algebraic closure is isomorphic to $\Bbb C$. So by usual argument $\Bbb R$ embeds into it, but this embedding is very different than the usual embedding since the degree of $\Bbb C$ over $\Bbb R$ now is not $2$, but infinite. And there are many many other ways of doing similar tricks. – Asaf Karagila Aug 16 '14 at 06:20

1 Answers1

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This is indeed an important question. No doubt that $\mathbb{R}$ is isomorphic to many subspaces of $\mathbb{R}^2$, or in other words, $\mathbb{R}$ can be embedded in many ways in $\mathbb{R}^2$. The thing is that there isn't any specific subspace in $\mathbb{R}^2$ which is the best one to represent $\mathbb{R}$.

If you want to treat $\mathbb{R}$ as a subspace, you need to specify the embedding $\mathbb{R}\hookrightarrow\mathbb{R}^2$ you refer to.

I would say that as long as we don't choose the embedding, $\mathbb{R}$ is not a subspace of $\mathbb{R}^2$.

Amitai Yuval
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  • I'll add that, colloquially, it's often said that $\mathbb{R}$ is a subspace of $\mathbb{R}^2$, when what is really meant is, as you say, something like $\mathbb{R} \times { 0 }$ is a subspace of $\mathbb{R}^2$. This is much the same as in group theory, where we might say $\mathbb{Z}_2$ is a subgroup of $\mathbb{Z}_6$; it arises from only really caring about algebraic structures up to isomorphism. – Clive Newstead Aug 14 '14 at 23:36
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    Note that saying that $\mathbb{Z}_2$ is a subgroup of $\mathbb{Z}_6$ is slightly different, since there is a unique embedding. – Amitai Yuval Aug 14 '14 at 23:40
  • However, it is true that $\mathbb{Z}_2$ is not a subgroup of $\mathbb{Z}_6$. – Amitai Yuval Aug 14 '14 at 23:44
  • That's a bit nitpicky, but alright... $\mathbb{Z}_2 \le \mathbb{Z}_2 \times \mathbb{Z}_2$ would be an example where there is not a unique embedding. (And yes, $\mathbb{Z}_2$ isn't a subgroup of $\mathbb{Z}_6$, but it's isomorphic to one, in much the same way that $\mathbb{R}$ isn't [but is isomorphic to] a subspace of $\mathbb{R}^2$.) – Clive Newstead Aug 14 '14 at 23:49
  • I'm not saying your answer's wrong, in fact I up-voted it, I'm just adding that what's technically correct is often not what is written by mathematicians when the technically correct translation ("is" -> "is isomorphic to") is understood from context. – Clive Newstead Aug 14 '14 at 23:57
  • I agree, it's true. I just pointed out that there are a few different kinds of all these cases. – Amitai Yuval Aug 14 '14 at 23:59
  • Thank you, that makes sense, so rather I would specify the subspace $Y$ of $\Bbb R^2$ and then say that $Y\cong \Bbb R$? – Functional Analysis Sep 09 '15 at 11:47
  • @FunctionalAnalysis Yes, that would make more sense. – Amitai Yuval Sep 09 '15 at 11:49