Showing all open sets in $\mathbb{R}^{n}$ are uncountable - extend to general metric space?

Question

I would like to ask for a clarification to the comments to the question here ;

Proof that every open subset of $\mathbb{R}^n$ is uncountable

As far as I can tell the comment implies for $\textbf{x}\in\mathbb{R}^{n}$ the mapping $f(\textbf{t})\mapsto \textbf{x}+\textbf{t}(1,0,0,...,0)$ for all $|\textbf{t}|<\epsilon$ is a bijective and hence any open set $U\in\mathbb{R}^{n}$ satisfies $|U|=\aleph$. However I could not make this work for me. Instead, letting $e_{1}=(1,0,0,...,0)$,...,$e_{n}=(0,0,0,...,1)$ be the basis vectors for $\mathbb{R}^{n}$ then I believe the mapping $f(\textbf{t})\mapsto \textbf{x}+\sum_{i=1}^{n}\textbf{t}\textbf{e}_{i}$ is in fact required. Where am I going wrong? Below is my "proof" of my claim;

Finally, can the basis vector in the referenced question, or (if I am somehow correct) can the basis vectors in my question, be replaced by basis vectors in a general metric space to come up with a very similar proof - i.e. are all open sets in a metric space uncountable?

************ proof - I switched here to $k$ rather than $n$ *******

For any $\textbf{y}\in\mathbb{R}^{k}$ and $\epsilon>0$ let $B_{\epsilon}(\textbf{y})$ be an open ball centered at $\textbf{y}$. For $i=1,2...,k$ let $\textbf{e}_{i}\in\mathbb{R}^{k}$ be a vector with a $1$ in the $i^{th}$ component and zeroes elsewhere - i.e. $\textbf{e}_{i}$ is the $i^{th}$ basis vector for $\mathbb{R}^{k}$. Now for all $t_{i}$ satisfying $|t_{i}|<k^{-1/2}\epsilon$ we have

\begin{align*} d_{k}\left[\textbf{y},\textbf{y}+\sum_{i=1}^{k}t_{j}\textbf{e}_{j}\right]&= d_{k}\left[\sum_{i=1}^{k}y_{i}\textbf{e}_{i},\sum_{i=1}^{k}y_{i}\textbf{e}_{i}+\sum_{i=1}^{k}t_{i}\textbf{e}_{i}\right]\\ &=\sqrt{\sum_{i=1}^{k}\left(y_{i}(\textbf{e}_{i})_{i}-y_{i}(\textbf{e}_{i})_{i}-t_{i}(\textbf{e}_{i})_{i}\right)^{2}}\\ &=\sqrt{\sum_{i=1}^{k}\left(-t_{i}\right)^{2}}\\ &=\sqrt{\sum_{i=1}^{k}t_{i}^{2}}\\ &<\sqrt{k\cdot k^{-1}\epsilon^{2}}\\ &=\epsilon. \end{align*}

Accordingly for $\textbf{t}:=(t_{1},...,t_{k})$ satisfying $|t_{i}|<k^{-1/2}\epsilon$ for all $i=1,...,k$ (which implies $||\textbf{t}||_{k}=\sqrt{\sum_{i=1}^{k}t_{i}^{2}}<\sqrt{k\cdot k^{-1}\epsilon^{2}}=\epsilon$), if we define the vector $\textbf{y}_{\textbf{t}}=\textbf{y}+\sum_{j=1}^{k}(\textbf{t})_{j}\textbf{e}_{j}$ then the above equation means $d_{k}[\textbf{y},\textbf{y}_{\textbf{t}}]<\epsilon$ which implies $\textbf{y}_{\textbf{t}}\in B_{\epsilon}(\textbf{y})$. Thus defining the mapping $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$, $f_{\textbf{y}}(\textbf{t})\mapsto \textbf{y}_{\textbf{t}}$, we have $f_{\textbf{y}}((-\epsilon,\epsilon))\subseteq B_{\epsilon}(\textbf{y})$. Conversely choose a $\textbf{x}\in B_{\epsilon}(\textbf{y})$ which implies $d_{k}[\textbf{x},\textbf{y}]:=d<\epsilon$, and write $\textbf{x}$ and $\textbf{y}$ in terms of the basis vectors $\textbf{x}=\sum_{i=1}^{k}\textbf{x}_{i}\textbf{e}_{i}$ and $\textbf{y}=\sum_{i=1}^{k}\textbf{y}_{i}\textbf{e}_{i}$ so that $d_{k}^{2}[\textbf{x},\textbf{y}]$ can be written as

\begin{align*} d_{k}^{2}[\textbf{x},\textbf{y}]&=\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)^{2}(\textbf{e}_{i})_{i}^{2}\\ &=\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)^{2}\\ &=d^{2}\\ &<\epsilon^{2}. \end{align*}

Defining $t_{i}(\textbf{x},\textbf{y}):=(\textbf{x}_{i}-\textbf{y}_{i})$, and in turn $t(\textbf{x},\textbf{y})=\left(t_{i}(\textbf{x},\textbf{y})\right):=((\textbf{x}_{1}-\textbf{y}_{1}),...,(\textbf{x}_{k}-\textbf{y}_{k}))$, using the above equation implies $||t(\textbf{x},\textbf{y})||_{k}=d_{k}[\textbf{x},\textbf{y}]=d<\epsilon$. Thus we have

\begin{align*} \textbf{x}&=\textbf{y}+\sum_{i=1}^{k}\left(\textbf{x}_{i}-\textbf{y}_{i}\right)\textbf{e}_{i}\\ &=\textbf{y}+\sum_{i=1}^{k}t_{i}(\textbf{x},\textbf{y})\textbf{e}_{i}\\ &=\textbf{y}+\textbf{t}(\textbf{x},\textbf{y})\textbf{e}_{i}\\ &=\textbf{y}_{\textbf{t}(\textbf{x},\textbf{y})}, \end{align*}

for some $||t(\textbf{x},\textbf{y})||_{k}<\epsilon$, which leads to the conclusion $\textbf{x}\in f_{\textbf{y}}((\epsilon,\epsilon))$. Accordingly $B_{\epsilon}(\textbf{y})\subseteq f_{\textbf{y}}((-\epsilon,\epsilon))$ and so the conclusion $f_{\textbf{y}}((-\epsilon,\epsilon))= B_{\epsilon}(\textbf{y})$ follows. Thus $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$ is an onto function. Furthermore for all $\textbf{s},\textbf{t}\in\mathbb{R}^{k}$ since $(\textbf{y})_{i}+(\textbf{s})_{i}=(\textbf{y})_{i}+(\textbf{t})_{i}$ if and only if $(\textbf{s})_{i}=(\textbf{t})_{i}$ for all $i=1,...,k$ implies $\textbf{y}+\sum_{j=1}^{k}(\textbf{s})_{j}\textbf{e}_{j}=\textbf{y}+\sum_{j=1}^{k}(\textbf{t})_{j}\textbf{e}_{j}$ if and only if $\textbf{s}=\textbf{t}$ then $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$ is also an injective function, and hence is a bijection. Now $|(-\epsilon,\epsilon)|=\aleph$ is well known [and I do not prove it here] and so $|-\epsilon,\epsilon|=|B_{\epsilon}(\textbf{y})|$ (i.e. $(-\epsilon,\epsilon)$ is equipotent to $B_{\epsilon}(\textbf{y})$). By Definition of equality of set cardinalities we conclude $|(-\epsilon,\epsilon)|=| B_{\epsilon}(\textbf{y})|=\aleph$.

Answer 1

Taking onboard all the comments above, it seems every open ball in $\mathbb{R}^{k}$ is uncountable [proof below as per the original proposition in the linked question], and since all non-empty open sets in $\mathbb{R}^{k}$ are a union of open balls, and unions of uncountable sets produce another uncountable set, then we conclude all non-empty open sets in $\mathbb{R}^{k}$ are uncountable.

Finally it has been shown by examples that in general open sets in metric spaces need not be uncountable.

Proof that all open balls in $\mathbb{R}^{k}$ are uncountable;

For any $\textbf{y}\in\mathbb{R}^{k}$ and $\epsilon>0$ let $B_{\epsilon}(\textbf{y})$ be an open ball centered at $\textbf{y}$. Let $\textbf{e}_{1}\in\mathbb{R}^{k}$ be the vector with a $1$ in the $1^{st}$ component and zeroes elsewhere - i.e. $\textbf{e}_{1}$ is the $1^{st}$ basis vector for $\mathbb{R}^{k}$. Now for all $t$ satisfying $|t|<\epsilon$ we have

\begin{align*} d_{k}\left[\textbf{y},\textbf{y}+t\textbf{e}_{1}\right]&= d_{k}\left[\sum_{i=1}^{k}y_{i}\textbf{e}_{i},\sum_{i=1}^{k}y_{i}\textbf{e}_{i}+t\textbf{e}_{1}\right]\\ &=\sqrt{\sum_{i=1}^{k}\left(y_{i}(\textbf{e}_{i})_{i}-y_{i}(\textbf{e}_{i})_{i}-t(\textbf{e}_{1})_{i}\right)^{2}}\\ &=\sqrt{\left(-t(\textbf{e}_{1})_{1}\right)^{2}}\\ &=\sqrt{t^{2}}\\ &=|t|\\ &<\epsilon. \end{align*}

Accordingly if we define the vector $\textbf{y}_{t}=\textbf{y}+t\textbf{e}_{1}$ then the above equation means $d_{k}[\textbf{y},\textbf{y}_{t}]<\epsilon$ which implies $\textbf{y}_{t}\in B_{\epsilon}(\textbf{y})$. Thus defining the mapping $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow B_{\epsilon}(\textbf{y})$, $f_{\textbf{y}}(t)\mapsto \textbf{y}_{t}$, we have $f_{\textbf{y}}((-\epsilon,\epsilon)):=S\subseteq B_{\epsilon}(\textbf{y})$.

Furthermore for all $s,t\in\mathbb{R}$ since $\textbf{y}+s\textbf{e}_{1}=\textbf{y}+t \textbf{e}_{1}$ if and only if $s=t$ then $f_{\textbf{y}}:(-\epsilon,\epsilon)\longrightarrow S$ is an injective and obviously onto function, and hence is a bijection. Now $|(-\epsilon,\epsilon)|=\aleph$ is well known [and I do not prove it here] and so by the definition of comparing cardinalities of sets we have $\aleph=|-\epsilon,\epsilon|\leq |B_{\epsilon}(\textbf{y})|$. But since $|B_{\epsilon}(\textbf{y})|<|\mathbb{R}^{k}|=\aleph$ [proof omitted but using bijective functions again] we conclude $\aleph=|B_{\epsilon}(\textbf{y})|$.

Proof that all open sets in $\mathbb{R}^{k}$ are uncountable;

From the above proof and the definition of comparing cardinalities of sets using bijective functions, the definition of the open sets $\mathcal{U}(\mathbb{R}^{k})$ implies for any $U\in\mathcal{U}(\mathbb{R}^{k})$ that $\aleph\leq |U|$, for choosing any $u\in U$ there exists a $\epsilon_{u}$ such that $B_{\epsilon_{u}}(u)\subseteq U$ where $|B_{\epsilon_{u}}(u)|=\aleph$ which means the bijection $\phi:B_{\epsilon_{u}}(u)\longrightarrow B_{\epsilon_{u}}(u)\subseteq U$, $\phi(u)\mapsto u$, injects a set with cardinality of the continuum into $U$. Furthermore since $|\mathbb{R}^{k}|=\aleph$, the fact that $U\subset\mathbb{R}^{k}$ and the existence of the bijection $\phi:U\longrightarrow U\subseteq \mathbb{R}^{k}$, $\phi(u)\mapsto u$ combined the definition of comparing set cardinalities implies $|U|\leq\aleph$. Thus we have $|U|=\aleph$

Answer 2

If $U$ is a non-empty open subset of $\Bbb R^n$, let $\vec{x}=(x_1,\ldots, x_n)$ be a point of $U$. There is some $r>0$ so that $B(\vec{x}, r) \subseteq U$ by the definition of openness.

For $t \in (-r,r)$ define $f(t) = (x_1+t, x_2, \ldots, x_n) = \vec{x}+t\vec{e}_1$. Then $$d(f(t), \vec{x}) = \sqrt{t^2 + 0^2 \ldots 0^2} = |t| < r$$

so that $f(t) \in B(\vec{x},r)$ and so $f$ injects the uncountable (size continuum) set $(-r,r)$ into $U$ and so $|U| \ge \mathfrak{c}$ and in fact $|U|=\mathfrak{c}$ (I don't use the old school notation $\aleph$ for continuum but rather the modern $\mathfrak{c}$).

Of course a metric space can be countable too (e.g. $\Bbb Q$ in the Euclidean distance), so the idea that all open balls in all metric spaces will be uncountable is nonsense. Metric spaces can have finite open balls even (when they have isolated points; consider $\Bbb Z$ in its usual metric).