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Can you give the definition of subspace and subset of $\mathbb{R}^n$ and how can I determine their dimension?

Star
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  • Why are you so confused about this? Subspace is a definition coming from vector space and I take you know what dimension of vector space means. Anything contained in $\mathbb{R}^n$ is a subset of it so notion of dimension don't really exist – Jack Yoon Jul 21 '14 at 10:59
  • @JackYoon The notion of dimension exists quite happily in $\mathbb R^n$... – 5xum Jul 21 '14 at 11:04
  • I wrongly thought that since $\mathbb{R}^n$ has dimension $n$, any subspace of it would have dimension $n$. Can you give me a counterexample of why this is not the case? Thanks – Star Jul 21 '14 at 11:04
  • @5xum so what is the notion? How would you define dimension of generic subset of $\mathbb{R}$. And in what way is it useful?

    A set {0} is a 0-dimensional vector space.

     – Jack Yoon Jul 21 '14 at 11:10
  • @Cris see the edit of my question for that. – 5xum Jul 21 '14 at 11:11
  • @Cris: $\mathbb{R}^{n-1}$ is a subspace of $\mathbb{R}^n$, and the dimension's clearly not the same. – Daniel Crane Jul 21 '14 at 11:13
  • The tag "convex-analysis" does not seem well chosen. – Vincenzo Tibullo Jul 21 '14 at 12:35

1 Answers1

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A subset of $\mathbb R^n$ is any set that contains only elements of $\mathbb R^n$. For example, $\{x_0\}$ is a subset of $\mathbb R^n$ if $x_0$ is an element of $\mathbb R^n$. Another example is the set $S=\{x\in\mathbb R^n, ||x||=1\}$.

A subspace, on the other hand, is any subset of $\mathbb R^n$ which is also a vector space over $\mathbb R$. That means that for every $x,y\in S$ and $\alpha\in\mathbb R$, $x+y$ and $\alpha\cdot x$ must also be elements of $S$ in order for $S$ to be a subspace. In our two cases above, $\{x_0\}$ is only a subspace if $x_0=0$, and $S$ is not a subspace.

EDIT:

An example of a subspace of $\mathbb R^n$ is the set $S_{x_0}=\{\alpha x_0| \alpha\in \mathbb R\}$ which is, if $x_0\neq 0,$ a one-dimensional subspace of $\mathbb R^n$ (and, if $n\neq 1$, the dimesion of $S_{x_0}$ is not $n$).

5xum
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