Show that there is some linear function on any infinite dimensional Banach space that is unbounded. Let $(V,||\cdot||)$ be an infinite dimensional Banach space. Show that there is some linear function $\phi:V \rightarrow \mathbb{R}$ that is unbounded.
I have seen a proof somewhere. It goes like this: Let $B$ be an uncountable basis for $V$, take any countable subset $\{e_n: n \in \mathbb{N}\}$ from $B$ and define $f(e_n)=n||e_n||$ and $f(x)=0$ for other basis vectors.
I have some questions about this proof.
(1) Can we always extract a countable subset from a uncountable set?
(2) I don't think $f$ is linear. $f(e_i + e_j)=n||e_i+e_j||$ which may not be equal to $n||e_i||+n||e_j||$.
 A: 1) well, as soon as you accept a basis on your infinite dimensional vectorspace, you have to accept the axiom of choice, which gives you such a subset.
2) it is defined on the basis, this means that you extend by linearity, i.e. $$v=\sum_{i=1}^n \alpha_i v_i \mapsto \sum_{i=1}^n \alpha_i f(v_i)$$
and so this is clearly linear. It is something that is used quite often and should be usual for you if you are already at banachspaces. If not, I recommend you revise that.
A: (1) It can be constructed as the limit of finite subsets
(2) It defines the evaluation only of the basis, I guess "extend by linearity" is implied.
A: *

*$B$ has one element, call it $b_1$. $B_1 = B\setminus{\{b_1\}}$ is not empty, since otherwise $B$ is finite. $B_1$ has one element, call it $b_2$. $B_2 = B_1\setminus{\{b_2\}}$ is similarly not empty. Inductively we construct an infinite sequence $b_1,b_2,\dots$ in $B$. 

*Let $f_1,f_2$ be any numbers, Let $e_1,e_2 $ be linearly independent in a dimension 2 vector space $V$,   and $f:\{e_1,e_2\}\to \{f_1,f_2 \} $ be defined by
$$ f(e_i) = f_i.$$ You should verify that if $x$ has the basis expansion $x=x_1 e_1 + x_2e_2$,
$$ F: V\to \mathbb R,\quad F(x) :=  x_1 f_1 + x_2 f_2$$
is a well-defined linear function. Moreover, you should check that $F=f$ on the common domain of definition of two points $\{e_1,e_2\}$. $F$ is called the linear extension of $f$.

*Exercise: extend the above definition to any dimension...


Remark - note that when you said

$$ f(e_i + e_j) = n \|e_i + e_j\|$$ 

$n$ makes no sense at all. The $n$ is 
the index of some basis vector. You seem to be treating the definition as if you know what $n$ is, and then define $f(x) = n\|x\|$. This function is clearly not linear, yes, but this is not related to the linear extension of the function $$f:\{e_1,e_2,\dots\}\to\mathbb R, \quad f(e_) := \|e_\|.$$
Second remark - when you define $f$ to be zero for 'other basis vectors', this refers to the basis vectors in $B\setminus \{e_1,e_2,\dots\}$.
