A good example of set A such that $(A')\cap (A')'=\emptyset$ in $\mathbb{R}^2$

Question

I tried to show that $A=\{1/n:n\in\mathbb{N}\}$ is such that $A'=\{0\}$ but $(A')'=\emptyset$ but I can't imagine a set in $\mathbb{R}^2$ with the same property and obviusly non trivial, like $A=\{(0,1/n):n\in\mathbb{N}\}$ or similary, Can you please provide some example and hints for prove it if you consider that necesary?

Answer 1

The only way this is possible is if $A^{\prime\prime}$ is empty, since $A^{\prime\prime} \subseteq A^\prime$ for any $A \subseteq \mathbb R^2$. Below let $d$ denote the usual Euclidean metric on $\mathbb R^2$.

• Suppose that $x \in A^{\prime\prime}$, and let $\varepsilon > 0$. Then there is a $y \in A^\prime$ with $0 < d(x,y) < \varepsilon$. Set $\delta = \min \{ d(x,y) , \varepsilon - d(x,y) \} > 0$. Then there is a $z \in A$ with $0 < d(y,z) < \delta$. Note now that $$d(x,z) \leq d(x,y)+d(y,z) < d(x,y) + \delta \leq d(x,y) + (\varepsilon - d(x,y)) = \varepsilon.$$ Also since $d(y,z) < \delta \leq d(x,y)$ it must be that $x \neq z$. Therefore $x \in A^\prime$.

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The above proof works for any metric space. This set inclusions holds more generally for T₁-spaces.

• Let $X$ be T₁, and let $A \subseteq X$. Suppose $x \in A^{\prime\prime}$. Let $U$ be any open neighborhood of $x$. Then there is a $y \in U \cap A^\prime$ with $y \neq x$. Since $X$ is T₁, it follows that $U \setminus \{ x \}$ is an open neighborhood of $y$. As $y \in A^\prime$ there is a $z \in ( U \setminus \{ x \} ) \cap A$ with $z \neq y$. Clearly $z \in U \cap A$ and $z \neq x$. Therefore $x \in A^\prime$.