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This post is based upon Exercise 1.58 of Mathematical Proofs by Gary Chartrand et al. #2 and #4 are not questions, but are concerns because I got the right answer therein, but I am irresolute about my reasoning.

Let $I := [0, \infty). $ For each $r \in I$, define: $ A_r := \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 = r^2 \},$
$ B_r := \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \leq r^2 \},$
$ C_r := \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 > r^2 \}.$

Find $ \bigcup_{r \in I} X_r $ and $ \bigcap_{r \in I} X_r $ for $ X = A \, \& \, B \, \& \, C$.

$ \large{1.}$ In my initial attempt, I thought: Since $(0,0) \in A_r$ and $A_r = \text{\{infinitely many circles}\}$,
thus $ \bigcap_{r \in I} A_r = \{(0,0)\}$. However, the given answer with no explanation is $\emptyset$.

While I was rethinking, I realised that there is always another real number between any two numbers. Then I interpreted $A_r$ as infinitely many circles which are infinitesimally spaced apart, so these gaps induce the empty $\bigcap$. Is this reasoning correct?

$ \large{2.}$I understand that $B_r$ contain disks, which must fill the infinitesimal discontinuities between any two circles from $A_r$. My answer parallels the given answer: $\bigcap_{r \in I} B_r = \{(0,0)\}$ .

$ \large{3.}$ Since $ (0,0) \notin C_r$, thus $C_r = \{\text{open disks that extend outside of $r$}\}\backslash (0,0). $
But there are infinitely many $r$ so $ \bigcup_{r \in I} C_r = \mathbb{R}^2 \backslash (0,0) $. This matches the given answer.

$ \large{4.}$However, I am addled by $ \bigcap_{r \in I} C_r $. WLOG, say $r_1 < \color{green}{r_2} $.
Since ${\{x^2 + y^2 > r_1^2\}}\backslash\{x^2 + y^2 = {\color{green}{r_2}}^2\} = \{x^2 + y^2 > {\color{green}{r_2}}^2\}$,
thus $\{x^2 + y^2 > r_1^2\} \supset \{x^2 + y^2 > {\color{green}{r_2}}^2\}$.

So I thought: $\bigcap_{r \in I} C_r \neq \emptyset$. However, the given answer is $\bigcap_{r \in I} C_r = \emptyset$. How and why?

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$\bigcup A_r$ is the union of all circles of any radius $r\ge 0$. Since each point $(x,y)$ is in one such $A_r$ (namely the one with $r=\sqrt{ x^2+y^2}$), we see that $\bigcup A_r=\mathbb R^2$.

Since $A_r\subseteq B_r$, even more so $\bigcup B_r=\mathbb R^2$.

All points exept $(0,0)$ are already in $C_0$, and $(0,0)$ is not contained in any $C_r$. Hence $\bigcup C_r=\mathbb R^2\setminus\{(0,0)\}$


Already $A_1\cap A_2$ is empty, hence even more so $\bigcap A_r$ must be empty.

For any point $(x,y)\ne(0,0)$ we can find $r$ with $(x,y)\notin B_r$, for example let $r=\frac12\sqrt{x^2+y^2}$. On th eother hand, $(0,0)\in B_r$ for all $r$. Hence $\bigcup B_r=\{(0,0)\}$.

For any point $(x,y)$ we can find $r$ with $(x,y)\notin C_r$, for example let $r=\sqrt {x^2+y^2}+1$. Therefore $\bigcap C_r=\emptyset$.

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