Please show me a basis for $\{x | x = (x_1, x_2, \cdots, ), x_i \in \mathbb{R}\}$. Does a basis really exists? Let $S = \{x | x = (x_1, x_2, \cdots, ), x_i \in \mathbb{R}\}$.
Then, $S$ is a vector space under component-wise addition and scalar multiplication.  
I heard that $S$ has a basis.  
I tried to find a basis for $S$, but I was not able to find a basis.
I wondered there is no basis for $S$.
To tell the truth, I was not able to find even a spanning set of $S$.  
For example, $\{(1, 0, \cdots,), (0, 1, \cdots, ), \cdots\}$ is not a basis.  
Please show me a basis for $S$.
 A: A spanning set is the whole space. Every vector space has a basis. This can be proved using Zorn's Lemma. But an explicit construction of a basis may not be possible. In this case also we cannot explicitly write down a basis. 
A: An example of a spanning set for $S$ is all of $S$ itself - that always works. But of course it is not a basis. 
If you pose no restrictions on your sequences, then $S$ is not even the closure of the span of countable many sequences. This is since $l^{\infty}$ is not separable. 
As you have noted, the space $S$ is a vector space and thus it has a basis. However, the theorem that any vector space has a basis uses (and is equivalent to) the axiom of choice. Hence, despite knowing that a basis exists, this basis may be very inaccessible and non-constructive.
Another example for a space which does not seem to have a well behaved basis is $C[0,1]$ the space of continuous functions on the unit interval.
A: Here's a basis (whose construction uses the axiom of choice).
Define an equivalence relation on $S$ as follows: say that $x \sim y$ iff there exists an $N \in \Bbb N$ and $a,b \in \Bbb R$ not both zero such that $a x_n = b y_n$ for all $n > N$.
Let $\mathcal E$ denote the set of all equivalence classes over $S$ except for the equivalence class of zero.  For every $E \in \mathcal E$, we can apply the axiom of choice to select a sequence $x(E) \in E$.  Let $\mathcal B_1 = \{x(E) | E \in \mathcal E\}$.  Let $\mathcal B_2$ be the "non-basis" that you define, i.e.
$$
\mathcal B_2 = \{(1,0,0,\dots,),(0,1,0,\dots,),\dots\}.
$$
The set $\mathcal B = \mathcal B_1 \cup \mathcal B_2$ is a basis.
