perfect subsets of metric space

Question

Let $(X,d)$ be a metric space. Let $B\subseteq X.$ An accumulation point $y$ of $B$ has the property that for all $\epsilon > 0,$ $(B(y,\epsilon)\backslash \{y\}) \cap A\neq \emptyset.$ A subset $P\subseteq X$ is perfect if $P' = P$ (i.e. $P$ is the set of accumulation points of itself). Determine with proof whether the following are perfect:

1. $[a,b]$ in $(\mathbb{R}, |\cdot|) (a < b \in \mathbb{R})$
2. $\{q\in \mathbb{Q} : 0\leq q \wedge q^2 < 2\}$ in $(\mathbb{Q}, |\cdot |)$
3. The Cantor set $C$ in $(\mathbb{R}, |\cdot|),$ as described on Wikipedia.
4. $\prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$ in $(\ell_1, \lVert \cdot \rVert),$ where $\ell_1$ is the subset of $\mathbb{R}^{\mathbb{N}}$ consisting of elements whose $1$-norms are finite (i.e. elements $x=(x_n)$ so that $\sum_{i} |x_i| < \infty$).

I know that if a subset is closed then it contains all its accumulation points. I think $1$ is perfect. By the above remark it suffices to show that every element is an accumulation point. One can let $c\in [a,b]$ and consider separately the cases where $c=a$ and $a < c$ and then let $\epsilon > 0$. For instance, when $a < c,$ let $N\in \mathbb{N}$ be so that $\frac{1}{N} < \epsilon$ so that $c-\min(c-a, \frac{1}{N}) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified step by step; for example $c-\min(c-a, \frac{1}{N})\neq c$ and is within $\epsilon$ of $c$). If $c=a, c +\min\{b-c, \frac{1}N) \in (B(c, \epsilon)\backslash \{c\})\cap [a,b]$ (as can be verified similarly).

The set in 2 is clearly not closed. Also, it seems that $\sqrt{2}$ is an accumulation point of this set as $\sqrt{2}$ is a real number and $\overline{Q} = \mathbb{R}$ so it is the limit of a sequence in $\mathbb{Q}$ and this sequence can approach $\sqrt{2}$ from below (though explicitly stating what exactly this sequence is would be more convincing). So I think one can show $\sqrt{2}$ is an accumulation point but it's clearly not in the set.

As for $3$ I think it's actually perfect but I'm not really sure how to show every point is a limit point. I think it shouldn't be too hard to show the Cantor set is closed.

For $4$ I'm not sure whether it's perfect. I think it's a countable union of nowhere dense sets (point sets) so that might be useful in determining whether it's closed.

Answer 1

For 1. you could also note that $a=\lim_{y\to \infty} a+(b-a)/y$ and if $a<x\le b$ then $x=\lim_{y\to \infty } x-(x-a)/y.$

For 2. About 100 AD Hero (Heron) of Alexandria wrote of a method of approximating square roots. (Isaac Newton made a major generalization for approximate solutions for many other formulas). If $A>0,$ take $x_1>\sqrt A$ and let $x_{n+1}=\frac {1}{2}(x_n+\frac {A}{x_n}).$ By induction on $n$ & some elementary algebra we have $\sqrt A< x_{n+1}< x_n$ for all $n\in \Bbb N.$ So $L=\lim_{n\to \infty }x_n$ exists and $L\ge\sqrt A >0.$ So $$0<L=\lim_{n\to \infty }x_{n+1}=\lim_{n\to \infty }\frac {1}{2}(x_n+\frac {A}{x_n})=\frac {1}{2}(L+\frac {A}{L})$$ implying $L=\sqrt A.$

Now let $A=2=x_1$ and consider the sequence $(\frac {2}{x_n})_{n\in \Bbb N}.$

For 4. Let $S= \prod_{k=1}^\infty \{0,\frac{1}{2^k}\}$. If $x=(x_n)_n\in S$ and $\epsilon >0,$ take $k_0\in \Bbb N$ with $1/2^{k_0}<\epsilon$ and let

$y=(y_n)_n$ where

$y_n=x_n$ if $n\ne k_0,$ and

$ y_{k_0}=|x_{k_0}-1/2^{k_0}|.$

So $x$ is an accumulation point of $S$.

To show $S$ is closed: Suppose $z=(z_n)_n\in l_1 \setminus S$. Take some $m\in \Bbb N$ with $z_m\not \in \{0,1/2^m\}$ and let $r=\min (|z_m|,|z_m-1/2^m|).$ The open ball $B(z,r),$ centered at $z$ with radius $r,$ is disjoint from $S$ because if $y=(y_n)_n\in B(z,r)$ then $|y_m-z_m|\le \|y-z\|<r$ so $y_m\not \in \{0,1/2^m\}.$ Therefore $z\not\in\overline S.$

Answer 2

You are correct about 1 and 2. For a proof of 3 see here.

For 4, to show that every point of the set is a limit point, for any sequence $(a_n)$ in it, can you find a sequence of sequence (lol) $\{(b_n)_k\}$ in the set such that $$ |(a_n) - (b_n)_k| < \frac{1}{2^{k-1}} ? $$