A while ago I got confused about the cardinality of open balls in metric spaces and erroneously thought all open balls were uncountable. It turns out this is an obvious mistake – i.e. if $(X,d)$ is a metric space and $|X|\leq\aleph_{0}$ then clearly no subset of $X$ can have cardinality of the continuum. Because all open balls in $(\mathbb{R}^{k},d_{k})$ are uncountable I think I got lazy and just assumed this to be true for all spaces. Since all open sets are unions of open balls then all open sets in $(\mathbb{R}^{k},d_{k})$ are also uncountable. See the question here;
Showing all open sets in $\mathbb{R}^{n}$ are uncountable – extend to general metric space?
Since I was also learning about the space $(\mathcal{C},[0,1])$ the previous question got me wondering whether all open balls of $\mathcal{C}$, and hence all open sets of $\mathcal{C}$, are uncountable? I think they are, and so this is my first question.
I adapted the proof that worked for $(\mathbb{R}^{k},d_{k})$, but of course encountered some new issues and the following question
Is the cardinality of $\mathcal{C}[0,1]$ the same as the cardinality of $\mathbb{R}$?
helped me out, in particular the above question claimed all functions in $\mathcal{C}$ are determined by values in the set $\mathbb{Q}\cap[0,1]$ (think I am happy with this part of the proof), and hence
$$|\mathcal{C}|\le|\Bbb R|^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}
=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}=|\Bbb R|=\aleph$$
I used this upper bound for $\mathcal{C}$ in my proof, however I used a different argument since I could not easily understand where $|\mathcal{C}|\le|\Bbb R|^{\aleph_0}$ comes from (the rest I understand).
Given the above difficulties I tried the following: noting for any two sets $A$ and $B$ that $|A\cap B|\leq |A\times B|=|A|\times|B|$ then $|\mathbb{Q}\times[0,1]|\leq|\mathbb{Q}|\times|[0,1]|=\aleph_{0}\aleph$ must be an upper bound for the number of functions in $\mathcal{C}$. Since $\aleph_{0}\aleph=\aleph$ then $\mathcal{C}\leq|\mathbb{Q}\times[0,1]|\leq\aleph$ must hold.
So my second question is, how is it that $\mathcal{C}$ being determined by values in the set $\mathbb{Q}\cap[0,1]$ implies $|\mathcal{C}|\le|\Bbb R|^{\aleph_0}$ (and hence that $\mathcal{C}\leq\aleph$), and is the above alternate method OK as a proof of this?
My third question is whether the following proof is correct that all open balls of $\mathcal{C}$ are uncountable?
Proof;
Choose a $f\in\mathcal{C}$ and any $\epsilon>0$ and let $B_{d_{\infty},\epsilon}(f)$ and $B'_{d_{\infty},\epsilon}(f)$ be open $d_{\infty}$-balls about $f$ of, respectively, $\mathcal{C}$-functions and those $\mathcal{C}$-functions that are constant functions. The following relationships hold;
\begin{align*}
B'_{d_{\infty},\epsilon}(f)\subseteq B_{d_{\infty},\epsilon}(f)\subseteq\mathcal{U}(\mathcal{C})\subseteq\mathcal{C}
\end{align*}
Let $B_{d,\epsilon}(x^{*})\subset\mathbb{R}$ be the open $d$-ball about $x^{*}:=\text{sup}\{f(t):t\in[0,1]\}$. Note that any $x\in B_{d,\epsilon}(x^{*})$ can be mapped to a constant function $g(t):\equiv x$ for all $t\in[0,1]$ and hence $|g(t)-x^{*}|=|x-x^{*}|=d(x,x^{*})<\epsilon$ for all $t\in[0,1]$. Thus we have
\begin{align*}
d_{\infty}(g,f)&=\text{sup}\{|g(t)-f(t)|:t\in[0,1]\}\\
&=\text{sup}\{|x-f(t)|:t\in[0,1]\}\\
&\leq|x-x^{*}|\\
&=d(x,x^{*})\\
&<\epsilon,
\end{align*}
which implies $g\in B'_{d_{\infty},\epsilon}(f)$ since $g\in\mathcal{C}$. Thus $B_{d,\epsilon}(x^{*})\subseteq B'_{d_{\infty},\epsilon}(f)$. With the above in mind, for all $x\in B_{d,\epsilon}(x^{*})$ define the bijective mapping $\phi(x)\mapsto h_{x}(t)\equiv x$ for all $t\in[0,1]$
so that $\phi:B_{d,\epsilon}(x^{*})\longrightarrow B_{d,\epsilon}(x^{*})\subseteq B'_{d_{\infty},\epsilon}(f)$ holds. We then have $\aleph=|B_{d,\epsilon}(x^{*})|\leq |B'_{d_{\infty},\epsilon}(f)|$. Furthermore given $\delta: B'_{d_{\infty},\epsilon}(f)\longrightarrow B'_{d_{\infty},\epsilon}(f)$, $\delta(f)\mapsto f$ is a bijective mapping and $B'_{d_{\infty},\epsilon}(f)\subseteq \mathbb{R}$ where $\mathbb{R}=\phi(\mathbb{R})$ is viewed as the set of all constant functions, then $|B'_{d_{\infty},\epsilon}(f)|\leq|\mathbb{R}|=\aleph$. Thus we have $\aleph\leq|B'_{d_{\infty},\epsilon}(f)|\leq\aleph$ which implies $|B'_{d_{\infty},\epsilon}(f)|=\aleph$. Now using the fact that $B'_{d_{\infty},\epsilon}(f)\subseteq B_{d_{\infty},\epsilon}(f)$ and the identity mapping $\delta$ previously defined, we have $|B'_{d_{\infty},\epsilon}(f)|\leq |B_{d_{\infty},\epsilon}(f)|$. From the previous result that $\aleph=|B'_{d_{\infty},\epsilon}(f)|$ this then implies $\aleph\leq |B_{d_{\infty},\epsilon}(f)|$.
Due to the density of $\mathbb{Q}$ in $\mathbb{R}$, for any $x\in\mathbb{R}\backslash\mathbb{Q}$ we can construct a sequence of rationals $\{q_{n}\}_{n}$ such that $q_{n}\longrightarrow x$. By the continuity of any $f\in\mathcal{C}$ we then have $\text{lim}_{n}f(q_{n})=f(\text{lim}_{n}q_{n})=f(x)$ which shows each continuous function on $[0,1]$ is determined by its values on the countable
set $\Bbb Q\cap[0,1]$. Noting for any two sets $A$ and $B$ that $|A\cap B|\leq |A\times B|=|A|\times|B|$ then $|\mathbb{Q}\times[0,1]|\leq|\mathbb{Q}|\times|[0,1]|=\aleph_{0}\aleph$ must be an upper bound for the number of functions in $\mathcal{C}$. Since we have $\aleph_{0}\aleph=\aleph$ then $\mathcal{C}\leq|\mathbb{Q}\times[0,1]|\leq\aleph$ must hold. Since $\delta:B_{d_{\infty},\epsilon}(f)\longrightarrow B_{d_{\infty},\epsilon}(f)\subseteq\mathcal{C}$ is a bijection then $|B_{d_{\infty},\epsilon}(f)|\leq|\mathcal{C}|=\aleph$ must be true. Thus we have $\aleph\leq|B_{d_{\infty},\epsilon}(f)|\leq\aleph$ and so we conclude $|B_{d_{\infty},\epsilon}(f)|=\aleph$
EDIT: Alternate proof as suggested in comments
Choose any $f\in\mathcal{C}$ and $\epsilon>0$. Then $B_{\epsilon}(f)$ is an open ball in $\mathcal{C}$. Given that open balls in $\mathcal{C}$ are convex, for any $t\in[0,1]$ and any $g,h\in\mathcal{C}$ we have $tg+(1-t)h:=m_{t}\in B_{\epsilon}(f)$. Since there is one $m_{t}$ for every $t\in[0,1]$ then $\aleph=|[0,1]|\leq |B_{\epsilon}(f)|\leq|\mathcal{C}|$. Since $|\mathcal{C}|=\aleph$ this means $B_{\epsilon}(f)=\aleph$ must hold.
Best Answer
Easier: balls in $C[0,1]$ are convex; pick two distinct points $f$ and $g$ in your ball and look at the convex combination $h_t = t f + (1-t) g$. How many different $h_t$ do you have?