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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!

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    $\begingroup$ The way you've written it, $S = \mathbb R^2$. $\endgroup$
    – littleO
    Aug 14, 2014 at 23:28
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    $\begingroup$ 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 ${\bf x}\neq {\bf 0})$. $\endgroup$
    – mwmjp
    Aug 14, 2014 at 23:31
  • $\begingroup$ 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 $\mathbb R^2$. $\endgroup$
    – littleO
    Aug 14, 2014 at 23:36
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    $\begingroup$ 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? $\endgroup$
    – Asaf Karagila
    Aug 15, 2014 at 0:24
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Aug 16, 2014 at 6:08

1 Answer 1

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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 $\mathbb{R}^2$.

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  • $\begingroup$ 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. $\endgroup$ Aug 14, 2014 at 23:36
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    $\begingroup$ Note that saying that $\mathbb{Z}_2$ is a subgroup of $\mathbb{Z}_6$ is slightly different, since there is a unique embedding. $\endgroup$ Aug 14, 2014 at 23:40
  • $\begingroup$ However, it is true that $\mathbb{Z}_2$ is not a subgroup of $\mathbb{Z}_6$. $\endgroup$ Aug 14, 2014 at 23:44
  • $\begingroup$ 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 $\mathbb{R}^2$.) $\endgroup$ Aug 14, 2014 at 23:49
  • $\begingroup$ 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. $\endgroup$ Aug 14, 2014 at 23:57

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