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OK, so I have a question from Steven J. Leon's Linear Algebra with Applications, and an answer to said question. Unfortunately, I seem to be stumbling on understanding the answer and I was hoping that some kind SE comrade might help set me straight. So here is the question, answer, and what seems to be bothering me.


Prove that if S is a subspace of $ℝ^{1}$, then either S = {0} or S = $ℝ^{1}$.


Let S $\neq$ {0} be a subspace of $ℝ^{1}$ and let a be an arbitrary element of $ℝ^{1}$. If s is a non-zero element of S, then we can define the scalar $\alpha$ to be the real number a / s. Since S is a subspace it follows that

$\alpha$*s* = $\frac{a}{s}$*s* = a

is an element of S. Therefore, S = $ℝ^{1}$.

I get that I'm probably not seeing something rather simple, but here is what is hanging me up. When I see "$ℝ^{1}$", I'm thinking the set of all real numbers from $-\infty$ to $\infty$. However, can't a subset of $ℝ^{1}$ be something like {5.4, 2.009, 3.0}? The answer just seems to posit that S is a spanning set (I think). Using the technique shown in the answer S can be made to equal any number in $ℝ^{1}$, but I would think that S $\neq ℝ^{1}$.

Does this make any sense? Any help would be most appreciated!

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A subspace is by definition a vector space. – cats Feb 19 '13 at 1:00
up vote 2 down vote accepted

Recall that a subspace is not just any subset, it's also closed under scalar multiplication and addition.

If it's $\{0\}$ then we are done; otherwise it contains some $x\neq 0$, so $\frac1x\cdot x$ is there, which is $1$, and now it's easy to show that every real numbers is in our subspace by the same trick.

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That makes sense. Thank you! – BrotherJack Feb 19 '13 at 1:17

A vector (sub)space $S$ is closed under scalar multiplication with any element of the underlying field (in this case, $\mathbb{R}$ as field $F$, as well as $\mathbb{R}$ the vector space itself). So if $v \in S \subset \mathbb{R}$, then $\alpha v \in S$ for any $\alpha \in F$. This pretty much tells you right away that it has to be the entire real line, if the subspace isn't $\{0\}$.

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