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The reals can be thought of as a vector space over the rationals. The properties of a vector space are that addition and "scaling" by some scalar are well defined and this certainly holds for the reals. There are many posts explaining this, examples here and here. I completely understand that the reals satisfy all the properties of vector spaces with the scalars defined as rational numbers. However, how do we know that this vector space is infinite dimensional? Isn't it just 1 dimensional? And if it is infinite dimensional, what is the first dimension? Can there not be some infinite dimensional basis for this vector space then? In the second post, @AsrafKaragila says its impossible to write such a basis by hand and we can only prove its existence. Why is it impossible?

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    $\begingroup$ Any finite-dimensional vector space over a countable field is itself countable, but $\Bbb R$ is uncountable. $\endgroup$ Sep 5, 2021 at 20:45
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    $\begingroup$ “Isn’t it just 1 dimensional?” Can you write down a basis that has a single element? $\endgroup$
    – littleO
    Sep 5, 2021 at 20:46
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    $\begingroup$ but I can express $\pi$ as an infinite sum over ${\mathbb Q}$ --- But can you express $\pi$ as a finite sum? Look at the definition of "linear combination". You'll see that only finite sums are involved. Indeed, in an arbitrary (algebraic only) vector space, there is no reasonable way to define what an infinite sum of vectors is. $\endgroup$ Sep 5, 2021 at 21:17
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    $\begingroup$ May be the following will help you understand what is going on? Consider the following subsets of the reals: $$V_2:=\Bbb{Q}(\sqrt2)=\{a+b\sqrt2\mid a,b\in\Bbb{Q}\},$$ this is a 2-dimensional subspace (over the rationals) of $\Bbb{R}$. The set $B_2:=\{1,\sqrt2\}$ is a basis for it. The set $B_2$ obviously spans $V_2$, and it is linearly independent because $\sqrt2$ is irrational. $\endgroup$ Sep 5, 2021 at 21:36
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    $\begingroup$ I think because there are only countably many "definable" numbers, but the basis would be uncountable. $\endgroup$
    – Joe
    Sep 6, 2021 at 3:08

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I think your question is best answered by considering (un)countability. Recall that we call a set $M$ countable when there is some surjective function $f : \mathbb{N} \rightarrow M$, where $\mathbb{N}$ denotes the naturals. Intuitively, we can say that $f$ enumerates all of the elements of $M$ (this includes finite sets, since $f$ need not be injective). It is a well known result (see Cantor's Diagonal Argument) that $\mathbb{R}$ is not countable.

Thus, assume we have a countable set of vectors $B$ of $\mathbb{R}$. Consider the space spanned by $B$. This consists of finite linear combinations of vectors selected from a countable set, and is thus again countable (I'm not going to prove this here, but this essentially amounts to proving that the set of all finite strings over a countable set is countable, although there are many other ways to show this).

So a countable set $B$ cannot span all of $\mathbb{R}$. Thus $B$ is uncountable, and especially not finite. This should also answer your question to why we can't "write a basis down" for this; besides the fact that we can't write down infinitely many things, we can't even have a way to enumerate all the (first $n$) elements of a basis $B$ of $\mathbb{R}$, since this would amount to a surjective function $\mathbb{N} \rightarrow B$, implying that $B$ is countable.

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  • $\begingroup$ the set $B$ could be countably infinite. Is the set of linear combinations of vectors selected from a countably infinite set also countable even if the number of vectors taken in the linear combination can be countably infinite? $\endgroup$ Sep 10, 2021 at 19:54
  • $\begingroup$ @RohitPandey If you were to allow countably infinite linear combinations, you would indeed get an uncountable set. However, vector spaces don't allow infinite linear combinations (a linear combination is by definition a finite sum), since there isn't even a valid notion of infinite sums in general; thus, $B$ being countably infinite would not suffice. When we look at infinite series in $\mathbb{R}$, we are using the topological structure intensely (see the definition of infinite series as the limits of partial sums), and not every vector space necessarily has a "nice" structure like this. $\endgroup$
    – tolUene
    Sep 11, 2021 at 0:01
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Since you've asked several questions, I will only address the one that says "Why can't we explicitly write out the basis?" In other words, assuming the family $\{e_\alpha\}_{\alpha \in I}$ to be such a basis, why can't we write a formula for $e_\alpha$?

The only reason we know that $e_\alpha$ exists to begin with is the Axiom Of Choice. The existence proof would not work without it. So, we are stuck with whatever the AOC gives us. But the AOC never gives a constructive answer. That's just how it is.

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Suppose the vector space $V=(\mathbb{R},+,\cdot)$ over $\mathbb{Q}$ is finite dimensional. Let $\dim V=n$. Then, since $V$ is a $n$ dimensional vector space over $\mathbb{Q}$, $V$ is isomorphic to the vector space $\mathbb{Q}^n$ over $\mathbb{Q}$. Hence there is a bijection $\phi:\mathbb{R}\to\mathbb{Q}^n$, but it's a contradiction since $\mathbb{Q}^n$ is countable and $\mathbb{R}$ is uncountable.

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Question: "However, how do we know that this vector space is infinite dimensional? Isn't it just 1 dimensional?"

Answer: An "elementary" proof that $dim_{\mathbb{Q}}(\mathbb{R}) \neq n$ for some integer $n\geq 1$: We "know" there are real numbers $x=\pi$ that are non-algebraic ($\pi$ is "known" to be non-algebraic). If $\mathbb{Q} \subseteq \mathbb{R}$ was a finite extension of fields it would follow that any real number is algebraic:

Given any $x\in \mathbb{R}$ it would follow the set

$$x,x^2,x^3,..,x^n,..$$

is a linearly dependent set of vectors over $\mathbb{Q}$. And $\pi$ does not satisfy any polynomial equation $p(\pi)=0$ with $p(t) \in \mathbb{Q}[t]$.

Question: Is there an "elementary" proof that $\pi$ is not algebraic over $\mathbb{Q}$?

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Hint : Using Zorn's Lemma one can prove that ,
Every vector space has a basis .
So , $\mathbb{R}$ over $\mathbb{Q}$ has some dimension .
Now, question is what could possibly a basis ?
The elements $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},.......... ,$ can be shown to be the linearly independent , but they certainly do not span $\mathbb{R}$ as we also need elements like $\pi , {\pi}^2,{\pi}^3,.......$, etc . which also form a linearly independent set.
In fact, one can show that , the subspace of $\mathbb{R}$ generated by any countable subset of $\mathbb{R}$ must be countable .
Because $\mathbb{R}$ itself is uncountable , so no countable set can be a basis for $\mathbb{R}$ over $\mathbb{Q}$.
Hence , the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is uncountable i,e, infinite and there is no particular way to find such type of basis in general as they contain uncountable number of element . such type of basis are called Hamel basis.

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  • $\begingroup$ To the down-voter: please leave a comment explaining the down-vote. $\endgroup$ Sep 5, 2021 at 22:09
  • $\begingroup$ Please provide additional details in your answer. As it's currently written, it's hard to understand your solution. $\endgroup$
    – Community Bot
    Sep 5, 2021 at 22:18

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