# Linear independence over $\mathbb Q$ and $\mathbb R$ of subsets of $2^{\mathbb N}$

I have the following doubt:

Suppose $f_1,\ldots,f_n\in 2^{\mathbb N}$ are such that $\{f_1,\ldots,f_n\}$ is linearly independent in the $\mathbb Q$-vector space $\mathbb{Q^N}$. Does this set remain linearly independent in the $\mathbb R$-vector space $\mathbb{R^N}$?

Here $2=\{0,1\}$. I would like hints, not full answers.

Thanks

Edit: I have shown that if there is some $I\subseteq\Bbb N$ such that $f_1\upharpoonright I,\ldots,f_n\upharpoonright I$ is linearly independent over $\Bbb Q$ with $|I|\geq n$,then we are done, however I can't see why such $I$ should exist.

Suppose $\lambda_1 f_1 + \cdots + \lambda_n f_n = 0$, where $\lambda_1,\dots,\lambda_n\in\mathbb{R}$. Try picking a basis for the $\mathbb{Q}$-vector space spanned by $\lambda_1,\dots,\lambda_n$.

I don't think either of these hints are really helpful.

Hint: Yes. Note for instance that you can check linear independence at the non-vanishing of some determinant.

Better hint: Call $V_Q = \mathbb{Q}^{\mathbb{N}}$, $V_R = \mathbb{R}^{\mathbb{N}}$, let $i : V_Q \to V_R$ the natural inclusion (where I am thinking of $V_R$ as a $\mathbb{Q}$ vector space). Since the $f_i$ are linearly independent, they can be extended to a basis. There is an invertible $\mathbb{Q}$ linear transformation therefore taking them to the functions $\delta_i$: $T : V_Q \to V_Q$. There is an extension of $T$ to some $Q$-linear invertible $T' : V_R \to V_R$, such that $T' \circ i = i \circ T$.

This (I think) reduces your problem to the question of showing that the $\delta_i$ remain independent in $V_R$, which should be more or less clear (for example by looking at the projection onto coordinates function).

The main point of this: All independent sets in a vector space can be extended to a basis, and all basis in a vector space look the same. But some basis are better than others.

(Note that the $\delta_i$ do not form a basis.)

• I think more needs to be said - this answer seems to suggest that a linear independence is "upwards absolute" (if $A\subseteq V$ is linearly independent when $V$ is viewed as a $k$-vector space, it stays so even if we embed $V$ in a larger $K$-vector space for $k\subset K$), which is clearly false - $\{\pi, 2\}$ is linearly independent in $\mathbb{R}$ as a $\mathbb{Q}$-vector space, but not as an $\mathbb{R}$-vector space. – Noah Schweber Feb 21 '16 at 19:37
• @NoahSchweber What I am asserting is that if $V$ is a $K$ vector space, and $S$ some finite $K$ independent set, then $S$ is independent still in $V \otimes_K L$, for $L$ any field extension of $K$. The situation here is that some $K$ basis for $V$ becomes an $L$ basis for $V \otimes_K L$. So in particular (after reducing to some finite subspace containing $S$) the matrix corresponding to some vectors in $S$ in $V$ did not change, hence neither did their determinant. In our situation, $Q^N \otimes_Q R$ is a (proper) subspace of $R^N$. – Lorenzo Feb 21 '16 at 19:46
• @NoahSchweber But we can just work in some finite subspace containing $S = (f_1, \ldots, f_n)$. Of course I agree that $\{\pi,2\}$ is independent over $Q$ but not $R$. But there is no way to write $\pi$ as a $\mathbb{Q} = Q$ linear combination of the basis element $1$ in $\mathbb{R} = R$, so it is not exactly the same situation. – Lorenzo Feb 21 '16 at 19:51
• @NoahSchweber Actually now I think my answer isn't very good, because asserting that they remain independent because a basis remains independent is somewhat circular and not very helpful. – Lorenzo Feb 21 '16 at 19:56
• @AreaMan, perhaps my edit can clarify what I want, and even it's related to your answer. Thanks for your time. – Родион Раскольников Feb 21 '16 at 20:02

Hint: For each $s: 2^n \to 2$ choose some $i_s = i$ such that $s = \langle f_j(i) : j < n \rangle$ (if these is no such $i$ choose zero). Let $I$ be the set of these $i_s$'s. Now you can restrict your $f_j$'s to $I$.

If we look at this question from a categorical view point then the main point is the question whether the canonical map

$$R \otimes \lim Q \to \lim ( R \otimes Q) \tag{*}$$

is injective. (For vector spaces this should work, for modules you may need something like flatness)

How does this relate to the question? Well $\Bbb R \otimes \prod \Bbb Q$ has the "same" basis as $\prod \Bbb Q$ and asking if any linear independent family in there is linear independent in $\prod \Bbb R$ is equivalent to ask if the map

$$\Bbb R \otimes \prod \Bbb Q \to \prod ( \Bbb R \otimes \Bbb Q),$$

induced by the universal property of the product, is injective.

So we are taking the old basis and adding even more basis elements when going from $\prod \Bbb Q$ to $\prod \Bbb R$. In contrast to this, if we look at ${\Bbb R}_{\Bbb Q} \to \Bbb R$ this would result in a map $\bigoplus_{\Bbb R / \Bbb Q} \Bbb R \to \Bbb R$ which can not be injective.

Edit: A few comments why this map is injective. The tensor product is a left adjoint so it commutes with arbitrary coproducts, in particular with direct sums. But finite sums are isomorphic to finite product. So if the index set is finite the map is even an isomorphism. The general case can be done by transfinite induction, since the map can be viewed as an inductive limit over smaller cardinalities and inductive limits preserve monomorphisms.

Also a notational remark: I want the products to be indexed over $\Bbb N$. I switched to the product notion instead of the function space ${\Bbb Q}^{\Bbb N}$ since I am using the properties of the product.

Last but not least: For the general limit case as in the map $(*)$ one can use the fact that limits are subspaces of products and $R$ is flat.

• I don't get the conclusion of this answer: is the map $\Bbb R \otimes \prod \Bbb Q \to \prod ( \Bbb R \otimes \Bbb Q)$ injective or not? (Btw, $\Bbb R \otimes \Bbb Q\simeq\Bbb R$, right?) – user26857 Feb 24 '16 at 14:47
• @user26857 yes you can show that this map is injective. The conclusion should be that you can simplify this problem to a problem where you have to check if a map is a monomorphism of vectorspaces. I think it is handier if both VS are over the same field. – user60589 Feb 24 '16 at 15:15
• Could you give some more details about the proof of injectivity of that map? – user26857 Feb 25 '16 at 12:12
• Well what is $\Bbb R \otimes \prod \Bbb Q$? It is the set of functions from $\Bbb N$ to $\Bbb R$ which are real multiples of functions from $\Bbb N$ to $\Bbb Q$. Further we know that $\prod (\Bbb R \otimes \Bbb Q) \cong \prod \Bbb R$ which is the set of functions from $\Bbb N$ to $\Bbb R$. The map then is just the inclusion of a subset which is injective. – user60589 Feb 25 '16 at 14:30
• "It is the set of functions from $\mathbb N$ to $\mathbb R$ which are real multiples of functions from $\mathbb N$ to $\mathbb Q$." I don't get this: how is a tensor product an usual product? – user26857 Feb 26 '16 at 23:30