$n^{\mathrm{th}}$ Derived sets

Asked Nov 05 '20 at 02:19

Active Nov 05 '20 at 12:40

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I'm trying to construct a countable compact metric space such that for all $n$, its nth derived set is non-empty.

My attempt: Let $A = \{0\} \cup \left\{\frac{1}{n}: n \geq 2\right\}$. Consider $$A_1 = A \cup \left\{\frac{1}{n} + \frac{1}{k}\left(\frac{1}{n-1}-\frac{1}{n}\right): k \geq 2\right\}.$$ I'm basically constructing a sequence that approaches each point in $A$ from the right. Repeat this process; for $i \in \mathbb{N}$, $A_{i+1}$ is the union of $A_i$ along with all the sequences that approach each point in $A_i$ from the right. However, $\bigcup_{ i \geq 1}A_i$ is not closed in this case.

Is there any hint or suggestion?

edited Nov 05 '20 at 12:40

asked Nov 05 '20 at 02:19

ddth

If you just need the $n$-the derived set to be non-empty for a fixed $n$, look at the set $A_n$ constructed in this answer and this one. – Brian M. Scott Nov 05 '20 at 02:33
I've looked at them before posting. However, I'm trying to find a countable compact metric space $A $ such that for any $n \in \mathbb{N}$, $A^{(n)} \neq \emptyset$. Is there any suggestion? Thank you! – ddth Nov 05 '20 at 02:42
Are you at all familiar with infinite ordinal numbers? – Brian M. Scott Nov 05 '20 at 02:51
Unfortunately, I'm not. – ddth Nov 05 '20 at 02:53
1

See the set ${E_2}$ in this answer. Something similar to $E_2 \cup {2}$ will work. (You'll need to include right limiting endpoints at each construction stage leading up to this, I think, to ensure the set is closed.) – Dave L. Renfro Nov 05 '20 at 14:16

$n^{\mathrm{th}}$ Derived sets

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