Is open set in Banach space countable? Is open set in Banach space countable? If there exists those kind of open set, could you give me an example? If it is not, how to prove that all open sets in Banach space are uncountable? Thank you.
 A: In response to the comment stream:  it appears to me this question is neither too simple nor too difficult.
Caveat:  Based on the comments of Eric Wofsey (see below), I believe I need to qualify my remarks here by saying that I am assuming the Banach spaces under consideration are non-trivial, by which I mean they contain more vectors than the single element $0$.  I make the following
Observation:  Let $B \ne \{0\}$ be a Banach space.  Then for every $r \in \Bbb R_{\ge 0}$, $\exists v \in B, \Vert v \Vert = r$; for if $r = 0$, we may take $v = 0$; if $r > 0$, choose any $0 \ne v \in B$; then $\Vert v \Vert \ne 0$ and we may set $w = (r/\Vert v \Vert)v$; then $\Vert w \Vert = \Vert (r/\Vert v \Vert)v \Vert = r/\Vert v \Vert \Vert v \Vert = r$.  Furthermore, if $B \ne \{0\}$ the set $S(v, \epsilon) = \{w \in B \mid \Vert w - v \Vert = \epsilon$ is non-empty, since there is a $y \in B$ with $\Vert y \Vert = \epsilon$; then $\Vert (y + x) - x \Vert = \Vert y \Vert = \epsilon$.
Hopefully this Observation clarifies the discussion somewhat.
So I read this question as, "Is there a countable open set in any non-trvial Banach space?"
And with this reading, I answer it it the negative, based on the following:
Let $U$ be open in the Banach space $B$, and let $x_0 \in U$.  Since $U$ is open, there is a non-empty open ball (i.e., one with $\epsilon \ne 0$)
$B(x_0, \epsilon) \subset \bar B(x_0, \epsilon) \subset U, \tag 1$
i.e., the closure $\bar B(x_0, \epsilon)$ of $B(x_0, \epsilon)$ is contained in $U$.  For $t \in (0, 1)$, the sphere
$S(x_0, t\epsilon) = \{x \in B \mid \Vert x - x_0 \Vert = t\epsilon \} \subsetneq \bar B(x_0, \epsilon). \tag 2$
For any fixed element $y \in S(x_0, \epsilon) = \partial{\bar B(x_0, \epsilon)}$,
$x_0 + t(y - x_0) \in S(x_0, t\epsilon). \tag 3$
Now it is easy to see that if $t_1 \ne t_2$, we have
$x_0 + t_1(y - x_0) \ne x_0 + t_2(y - x_0); \tag 4$
otherwise,
$x_0 + t_1(y - x_0) = x_0 + t_2(y - x_0) \Longrightarrow t_1(y - x_0) = t_2(y - x_0)$
$\Longrightarrow (t_1 - t_2)(y - x_0) = 0 \Longrightarrow y - x_0 = 0 \Longrightarrow y = x_0, \tag 5$
a contradiction.  Thus(4) binds, and this implies the map sending $t \in (0, 1) \mapsto x_0 + t(y - x_0)$ is injective.  Since $(0, 1)$ is uncountable, the set of vectors
$\{ x_0 + t(y - x_0) \mid t \in (0, 1) \} \subset U \tag 6$
is uncountable; thus $U$ contains an uncountable subset, and hence cannot itself be countable.
A: Let $S$ be a non-empty open subset of the Banach space $B,$ where $B\ne \{0\}.$
If $0\not \in S$ we may take some $v\in S$ with $v\ne 0 .$ 
If $0\in S$ we can still take some $v\in S$ with $v\ne 0,$ because for some $r>0$ we have $\{y\in B: \|y-0\|<r\}\subset S,$ and there exists $u\in B$ with $u\ne 0,$ so we may take  $v=ru/(2\|u\|).$ 
So with $0\ne v\in S,$ take $r>0$ such that $\{u\in B:\|u-v\|<r\}\subset S.$ For $t\in [0,r/\|v\|)$ let $f(t)=v(1+t).$ Then $f(t)\in S.$ 
Now $f$ is $1$-to-$1$ because $f(t_1)=f(t_2)\implies (t_1-t_2)v=0\implies t_1=t_2$ (because $v\ne 0$).
So $f$ is an injection from the uncountable set $[0, r/\|v\|)$ into S, so $S$ is uncountable.  
