The proof of the infinity base of $\mathbb{R}^{\infty}$ We know that a finite basis of the finite-dimensional space $\mathbb{R}^n$ is 
$$
\{(1, 0, 0, 0,\ldots,0),\:(0, 1, 0, 0, 0,\ldots,0),\:(0, 0, 1, 0, 0, 0,\ldots, 0),\:\ldots,\:(0, 0, \ldots, 0, 0, 0, 1)\}
$$
What about the infinite-dimensional space $\mathbb{R}^{\infty}$?  I want to know that how is the proof of the infinity basis of $\mathbb{R}^{\infty}$?
 A: I hope there is still someone reading this thread.
By $\mathbb{R}^∞$ I am assuming that you are refering to the direct product of a countable family of $\mathbb{R}$, i.e. it contains elements of the form $(v_1, v_2, ...)$ where there may be infinitely many non-zero $v_i$'s. Unlike the direct sum of a countable family of $\mathbb{R}$, which has $B= \{e_1=(1, 0, 0, ...), e_2=(0, 1, 0, ...), e_3=(0, 0, 1, ...), ... \}$ as a basis, $B$ is not a basis of $\mathbb{R}^∞$, as the vector $V= (1, 1, 1, ...)$ is in $\mathbb{R}^∞$ but it is not a finite linear combination of the $e_i$'s.  Although I do not know how to find a particular basis for $\mathbb{R}^∞$, one can show that a basis of $\mathbb{R}^∞$ must be uncountable by the following statement,
There exists a linear independent subset $M$ of $\mathbb{R}^∞$ with $|M|=|R|$ - (*)
(We use the notation $|S|$ to denote the cardinal number of a set $S$.


*

*$|S|\leq|X|$ means that there is an injection from $S$ to $X$

*$|S|<|X|$ means that $|S|\leq|X|$ and there is no bijection from $S$ to $X$
For more details about cardinal number, please consult Introduction section 8 of T. Hungerford's "Algebra".)
The proof of (*) is contained in lemma 2 of this post by Asaf Karagila. It involves constructing an explicit injection from $\mathbb{R}$ to $\mathbb{R}^∞$ where the image $M$ is linearly independent.

Sketch of proof of (*):
First note that each element in $\mathbb{R}^∞$ can be viewed as a function from the set of non-negative integers $\mathbb{Z}^+$ to $\mathbb{R}$ and vice versa. Now for each real number $c$, let $f_c: \mathbb{Z}^+ \to R$ be $f_c(n)=c^n$ for each $n$. For example, $f_2=(1, 2, 4, 8, ...)$. If a linear combination of $k$ distinct such functions equals the zero element in $\mathbb{R}^∞$, i.e.
$$a_1*f_{c_1} + a_2*f_{c_2} + ... + a_k*f_{c_k} = 0$$
Then evaluate this equation at integers $0, 1, 2, 3, ..., k$ respectively, we get a system of $k$ equations (all equations equal 0) with $k$ variables where the coefficient matrix is invertible. Hence all $a_i$'s must be zero which means that the set of all $f_c$'s is linearly independent. Since this set has cardinality $|\mathbb{R}|$, (*) is proved. QED

There is another proof of (*) in Chap 9 section 5 of Jacobson's Lectures in Abstract Algebra Vol 2 but it is more abstract and does not give us an explicit injection from $\mathbb{R}$ to $\mathbb{R}^∞$ like the one given above.
Finally, note that not only we can have an estimate such as (*) on the dimension of $\mathbb{R}^∞$, one can also prove that
$$\dim_{\mathbb{R}} \mathbb{R}^∞ = |\mathbb{R}^∞| \text{ - (**)}$$
Suppose $W$ is a basis of $\mathbb{R}^∞$. Since $W$ is a subset of $\mathbb{R}^∞$, we must have $|W|<=|\mathbb{R}^∞|$. If we can also show that 
$$|\mathbb{R}^∞|\leq|W| \text{ - (***)}$$
then by Schroeder-Bernstein theorem (Hungerford Introduction Thm 8.6), we have $|W|=|\mathbb{R}^∞|$.
To prove (***), first assign an ordering to each finite subset of $W$.  For example if $\{x, y, z\}$ is a 3-element subset of $W$, then assign an ordered 3-tuple to this set.  You may either assign $(x, y, z)$ or $(x, z, y)$ or $(y, x, z)$, etc. But once you pick a particular order, you must stick with it.  Make similar assignment to each finite subset of $\mathbb{R}^∞$. Now note that for each element $v$ in $\mathbb{R}^∞$, it must be a unique finite linear combination of elements in $W$ with non-zero coefficients. So $v$ is characterized by three sets of information --- 


*

*a positive integer $n$ which is the number of element in $W$ in the linear combination;

*a finite subset of $W$ contains the elements of $W$ involved in the linear combination and hence an ordered n-tuple of elements of $W$;

*an $n$-tuple of real numbers which are the corresponding coefficients.
By this way we have created an injection $F:\mathbb{R}^∞ \to \bigcup_{n\in\mathbb{Z}^+}(W^n\times R^n)$.
Since both $W$ and $\mathbb{R}$ are infinite sets, there are bijections $g_n:W^n \to W$ and $h_n:\mathbb{R}^n \to \mathbb{R}$ for each n by Thm 8.12 (i) in the introduction chapter of Hungerford. Now the map $T:\bigcup_{n\in\mathbb{Z}^+}(W^n\times \mathbb{R}^n) \to \mathbb{Z}^+ \times W \times \mathbb{R}$ given by $u \to (n, g_n(\pi_1(u)), h_n(\pi_2(u)))$, where $\pi_1$ and $\pi_2$ are the obvious projections, is a bijection.  So the composition $G\circ F:\mathbb{R}^∞ \to N^* \times W \times \mathbb{R}$ is an injection.  So we have $|\mathbb{R}^∞|\leq|\mathbb{N}^* \times W \times \mathbb{R}|$.  By definition $|\mathbb{N}^* \times W \times \mathbb{R}|=|\mathbb{N}^*|*|\mathbb{W}|*|\mathbb{R}|$. We also have $|\mathbb{N}^*|*|W|*|\mathbb{R}|=|W|*|\mathbb{R}|$ by Thm 8.11 in the introduction chapter of Hungerford. Combining everything so far, we have
$$|\mathbb{R}^∞|\leq|W||\mathbb{R}|$$
But by (*) we have $|W|\geq|\mathbb{R}|$, so by theorem 8.11 in the introduction chapter of Hungerford, we have $|W||\mathbb{R}|=|W|$ and hence (***) is proved.
Combining (**) and the fact that $|\mathbb{R}|=2^{|\mathbb{N}|}$, $\mathbb{N}$ being the set of natural number, we have 
$\text{dim} \mathbb{R}^∞ = |\mathbb{R}^∞|$
$:= |\mathbb{R}|^{|\mathbb{N}|}$ by definition of cardinality since $\mathbb{R}^∞$ can be view as the set of all functions from $\mathbb{N}$ to $\mathbb{R}$
$= (2^{|\mathbb{N}|})^{|\mathbb{N}|}$
$= 2^{(|\mathbb{N}|*|\mathbb{N}|)}$   by exercise 8.10 (c) introduction chapter, Hugnerford
$= 2^{|\mathbb{N}|}$    by Theorem 8.11 introduction chapter, Hungerford
$ = |\mathbb{R}|$
Hence $\mathbb{R}^∞$ as an $\mathbb{R}$ vector space is a direct sum of an uncountable family (indexed by $\mathbb{R}$ itself) of $\mathbb{R}$'s even though it is a direct product of just a countable family of $\mathbb{R}$'s.
A: One of the most frequently considered infinite-dimensional spaces is $\ell^2(\mathbb R)$, which is the set of all infinite sequences $\mathbf x = (x_1,x_2,x_3,\ldots)$ of real numbers for which $x_1^2+x_2^2+x_3^2+\cdots<\infty$.  One sort of "basis" used for that space is $\{(1,0,0,0,\ldots), (0,1,0,0,\ldots), (0,0,1,0,\ldots),\ldots)$.  This is an example of a Hilbert space.
But that is not a "basis" by one of the most frequently used definitions of "basis": some elements of $\ell^2(\mathbb R)$ cannot be written as a finite linear combination of members of this "basis".  One speaks of certain infinite "linear combinations", but then one must be careful about which kind of convergence is used.  The kind most often used in thinking about this space involves the norm $\|\mathbf x\|=\sqrt{x_1^2+x_2^2+x_3^2+\cdots}$.  One says that $\mathbf x_n$ converges to $\mathbf x$ if $\|\mathbf x_n-\mathbf x\|\to0$.
This is of course not the only such "basis".
If you haven't thought about infinite-dimensional spaces before, it may be rash to assume they're just like finite-dimensional spaces, and your question shows evidence that you're doing that.
