Please let me know how to approach this problem.
Show that if $U ≠ \{0\}$ is a subspace of $\mathbb{R}$, then $U = \mathbb{R}$
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0@Larry I have given a detailed step-by-step answer and 5 exercises (which might be useful to you since they are very pertinent to your question). – 2011-07-11
5 Answers
Let $U$ be a subspace of $\mathbb{R}$. The following steps lead to a proof that $U=\{0\}$ or $U=\mathbb{R}$:
(1) If $U=\{0\}$, then there is nothing to prove; thus we may assume that $U\neq \{0\}$ and choose a non-zero vector $u\in U$.
(2) If $v\in \mathbb{R}$, prove that there exists (a scalar, if you prefer) $\alpha\in \mathbb{R}$ such that $v=\alpha\cdot u$. (The $\cdot$ denotes multiplication of the "scalar" $\alpha$ by the "vector" $u$.)
(3) Since $u\in U$ and $U$ is a subspace of $\mathbb{R}$, $v=\alpha\cdot u\in U$. Since $v\in \mathbb{R}$ was an arbitrary element (vector, if you prefer) of $\mathbb{R}$, it follows that every element (vector, if you prefer) of $\mathbb{R}$ is an element of $U$. Therefore, $\mathbb{R}\subseteq U$.
(4) Of course, the reverse inclusion $U\subseteq \mathbb{R}$ is subsumed in the very definition of a subspace.
(5) Therefore, $U=\mathbb{R}$ and the proof is complete.
Hint for (2):
Let $\alpha=\frac{v}{u}$, a fraction of real numbers, and view $\alpha\in \mathbb{R}$ as a scalar. (We note that $\alpha$ is well-defined since $u\neq 0$ as a real number.)
The following exercises are relevant:
Exercise 1: Let $U\subseteq \mathbb{C}$ be such that $\alpha\cdot u\in U$ whenever $\alpha\in\mathbb{C}$ and $u\in U$. Prove that $U=\{0\}$ or $U=\mathbb{C}$.
Exercise 2: Let $U$ be an open subspace of the real vector space $\mathbb{R}^n$ ($n$ is a positive integer). Prove that $U=\mathbb{R}^n$.
Exercise 3: Prove that every subspace of $\mathbb{R}^n$ is closed. In fact, use this and the fact that $\mathbb{R}^n$ is connected as a topological space to give another proof of Exercise 2.
Exercise 4: Prove, without using the proof that I have given as an answer to your question, that if a subset $U\subseteq\mathbb{R}$ has the property that $\alpha\cdot u\in U$ for all $\alpha\in\mathbb{R}$ and $u\in U$, then $U$ is a subspace of $\mathbb{R}$. Use the answer to the question you have asked to deduce that $U=\{0\}$ or $U=\mathbb{R}$.
Exercise 5: If $U\subseteq \mathbb{R}^2$ and if $\alpha\cdot u\in U$ for all $\alpha\in\mathbb{R}$ and $u\in U$, then is it true that $U=\{0\}$ or $U=\mathbb{R}^2$? Prove or give a counterexample. Similarly, under these conditions, is it true that $U$ is a subspace of $\mathbb{R}^2$?
I hope this helps!
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0@AD. I think that this is known as a "spoiler" and was discussed, for example, [here](http://meta.math.stackexchange.com/questions/2443/regarding-spoiler-text) – 2011-12-07
Here is a hint. Since $U$ is $\neq \{0\}$, it contains some nonzero number $u$. let $v$ be any other real number. Can you show that $v$ is a multiple of $u$? It may be useful to observe that $v = 1 \cdot v$ and that $1 = \frac{1}{u} \cdot u$.
Would this not be correct, and simple?
Since $H$ is not the trivial subspace $\{0\}, H$ contains at least one nonzero vector, let's say $\vec \alpha$. In this problem vectors are nothing more than real numbers. So we can generate a another vector in $H$ by multiplying the vector $\vec \alpha$ by the scalar $1/\alpha$. (The unrestricted set of reals is available to us for operations on $H$.) That gives us the vector $\vec 1$. Next we can generate a still another vector in $H$ by multiplying the just-generated vector 1 by the scalar $\beta$. Thus by the scalar multiplication axiom of vector spaces, $H$ includes all the reals as its vectors. So $H = \Bbb R$ (which implies that $\Bbb R$ cannot have a proper subset).
Assuming that R are the real numbers: A subspace should be closed under scalar multiplication.
Of course the way to go is directly show that every real number must be in $U$. But alternatively, $U$ a subspace of $\mathbb{R}$ implies $0\leq \dim U\leq\dim\mathbb{R}=1$. The assumption $U\neq\{0\}$ means $\dim U\neq 0$, so $\dim U=1$. Since proper subspaces have strictly smaller dimension, we conclude $U=\mathbb{R}$.
Unfortunately, if you can prove that proper subspaces have strictly smaller dimension, you are probably capable of proving the original question directly as well.