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Please let me know how to approach this problem.

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Presumably you are considering the real numbers as a real vector space (i.e., the scalars are real numbers). Is that correct? –  Jonas Meyer Jul 11 '11 at 7:24
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Yup, I'm thinking about proving U -> R and R -> U so U = R, but what I don't get is the part given that U ≠ {0} –  123 Jul 11 '11 at 7:30
    
Larry, what do you mean by "proving U -> R and R -> U"? –  Jonas Meyer Jul 11 '11 at 7:35
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like if U≠{0} is a subspace of R then U ⊆ R, and R ⊆ U as well so U = R. I don't know. I'm just guessing –  123 Jul 11 '11 at 7:40
    
Oh, you mean $U\subseteq \mathbf R$ and $\mathbf R\subseteq U$? Yeah, $U\subseteq \mathbf R$ is part of the definition of subspace. The other containment is the point, and $U\neq\{0\}$ is required because $\{0\}$ is a subspace that is not equal to $\mathbf{R}$. –  Jonas Meyer Jul 11 '11 at 7:40

5 Answers 5

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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+1 Wow, I love your exercises. Do you invent them on your own or take them from some books? –  Listing Jul 11 '11 at 14:23
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@Listing: Dear Listing, thanks for your kind words! I mostly invent these exercises on my own but of course I sometimes get inspiration from other sources. Usually, I look at the question being asked and try to think of variants of the question; e.g., I change the hypotheses involved, think about the converse, think of more general cases, think of related themes etc. In this particular case, the theme of the question is (imprecisely) "subspace equality". The notation used for the subspace $U$ immediately got me thinking of "open sets" and this inspired Exercise 2 and Exercise 3. –  Amitesh Datta Jul 11 '11 at 14:33
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@Listing: Exercise 1 generalizes in a very natural manner to any field $\mathbb{F}$ in place of $\mathbb{C}$ and hence I included it as part of a general pattern. Exercise 5 analyzes the problem at hand in the context of a higher dimension. –  Amitesh Datta Jul 11 '11 at 14:35
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@Listing: In fact I should admit that I sometimes learn from the exercises that I give. Mathematics is an amazing subject in this regard; one can take a very simple problem such as the one asked by the OP. However, if one stares at the problem and deconstructs it, then one will arrive at a host of interesting questions. Already, Exercise 3 above leads to the notion of a closed subspace of a vector space and this is a particularly ubiquituous notion in the theory of Hilbert spaces. (In fact, it is a nice exercise in this case to prove that a finite-dimensional subspace of $H$ is closed.) –  Amitesh Datta Jul 11 '11 at 14:43
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@Listing In fact, I try to ensure that these elements are present in my Exercises; namely, the nature of my Exercises usually encourages one to explore further mathematical ideas, some of which he/she will explore later in his/her mathematical studies. –  Amitesh Datta Jul 12 '11 at 1:53

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 $1 = \frac{1}{u} \cdot u$.

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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).

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Assuming that R are the real numbers: A subspace should be closed under scalar multiplication.

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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.

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