Proper superset of a union of the closure of countably many uncountable sets

Question

Consider a countable collection of uncountable subsets $A_1, A_2, A_3, \ldots$ and the subset $B = \cup_{i = 1}^\infty A_i$ of a metric space $X$. $\bar{A}$ denotes the closure of the set $A$. I came up with the following example of a set $B$ as described above such that $\bar{B} \supsetneq \cup_{i=1}^{\infty} \bar{A_i}$, but I am not sure if it is correct, and so I present it for verification:

Let $X = \mathbb{R}^2$, endowed with the standard Euclidean metric. Consider the set $S = \left\{q \in \mathbb{Q} : 0 < q <1\right\}$. Then $S$ is countable. Let $\left\{s_i\right\}$ for $i \in \mathbb{N}$ be an enumeration of $S$.

Define, for each $i \in \mathbb{N}$, $A_i = \left\{w \in \mathbb{R}^2 : \lvert w \rvert = s_i \right\}$. Then $\forall i$:

• $\lvert A_i \rvert = \mathfrak{c}$; consider the bijection $f_i: \lbrack 0,1) \rightarrow A_i$ defined by $f_i(x) = (s_i \cos(\frac{x}{2\pi}), s_i \sin(\frac{x}{2\pi}))$ with the trigonometric argument restrictions of $0 \leq x < 2\pi$.
• $\bar{A_i} = A_i$, clearly, as circles in $\mathbb{R}^2$ are closed sets.
• $\bar{B} = \left\{ x \in \mathbb{R}^2 : \lvert x \rvert \leq 1 \right\} $; embarrassingly, I am having some difficulty formalizing my intuition here. The basic idea in my mind is that for every $x \in \bar{B}$ and $\epsilon > 0$, $\exists k \in \mathbb{Q}: \max (0, \lvert x \rvert - \epsilon) < k < \lvert x \rvert \leq 1$ since $\bar{\mathbb{Q}} = \mathbb{R}$, so $k \in S = s_j$ for some $j \in \mathbb{N}$, and (I am waving hands here) we have that $A_j \cap \left\{a \in \mathbb{R}^2 : \lvert x-a \rvert < \epsilon \right\} \neq \emptyset$. If my example is correct, I would appreciate some simple and clean way to finish the argument here.

Then, in particular, $C = \left\{ 0 \right\} \cup \left\{ x \in \mathbb{R}^2 : \lvert x \rvert = 1 \right\} \subset \bar{B}$, but $C \neq \bar{A_i}$ for any $i$ $\implies C \neq \cup_{i=1}^{\infty} \bar{A_i} \implies \bar{B} \supsetneq \cup_{i=1}^{\infty} \bar{A_i}$.

Although I cannot quite put my finger on it yet, what property of $\mathbb{R}^2$ allowed me to construct $\bar{B}$ as shown given my $A_i$'s (assuming, of course, that my example is correct)? That is, what property characterizes the metric spaces (of uncountable cardinality) in which the $A_i$'s and $S$ constructed in my example would fail to produce $\bar{B}$ as above, and more generally, as a proper superset of the $A_i$'s (where $\bar{B} = \cup_{i=1}^{\infty} \bar{A_i}$)?

Answer 1

I believe your argument works, but it looks like you're doing more work than is necessary. To prove that $\bar{B}\supsetneq \cup_i \bar{A_i}$, you need only show that there exists a limit point of $B$ that is not contained in any $\bar{A_i}$, or equivalently $A_i$ since each $A_i$ is closed. And the origin $O$ is one such limit point because each $A_i$ has a positive radius, so we are done.

If you still want to show that $\bar{B}=\{x\in \mathbb{R}^2:d(O,x)\le 1\}$, here is a basic sketch of the proof:

For any point $x$, consider the line $L(x)$ going through $O$ and $x$ (If $x=O$, consider any line going through $O$). The intersection $C(x)=L(x)\cap B$ will be all points a rational distance $0<q<1$ from $O$. Now there is clearly an isomorphism between $L(x)$ and $\mathbb{R}$, where $C(x)$ is mapped to $((-1,1)\cap \mathbb{Q})-\{0\}$. It can easily be shown that $\bar{C}(x)=[-1,1]$.

Due to this finding, we see that for any point $x$ such that $d(O,x)\le 1$, every open sphere centered at $x$ will contain points in $C(x)$, and hence $B$. And for any point $x$ such that $d(O,x)>1$, there exists an open sphere of radius $d(O,x)-1$ that contains no points of $B$. Therefore, $\bar{B}$ consists precisely of the points $x$ such that $d(O,x)\le 1$.

To see what property of $\mathbb{R}^2$ allowed for the desired condition to be satisfied, let's consider a simpler example. If we are working only in $\mathbb{R}$, then consider the following new definition of $A_i$: $$A_i:=\left[\frac1{i+1},\frac1i\right] (i=1,2,\dots)$$

We see that $B=\cup_iA_i$ has a limit point of $0$ while none of the $A_i$ has a limit point of $0$. In this example, the most important property of $\mathbb{R}$ that allows us to construct this example is simply that there exist uncountable subsets of $\mathbb{R}$. In fact, we can construct this type of example in any metric space where each $\{x:\frac1{n+1}\le d(O,x)\le \frac1n\}$ is an uncountable set.