# Verification and help to simplify an argument about closure of some sets.

Hi everyone I'd like to know if what I have so far is correct, I think is much work for something which is too simple I would appreciate any advice or whatever. Moreover, I have doubt in (3) and (4), hopefully are correct. Thanks in advance.

Definition: (Closure) Let $X$ be a subset of $\mathbb{R}$. The closure of $X$, denote it $\overline{X}$, is define to be the set of all the adherent points of $X$. A point $x$ is said to be adherent to $X$, iff for each given $\varepsilon>0$ there is some $y\in X$ for which $|x-y|\le \varepsilon$.

Lemma: The closure of $\mathbb{N}$ is $\mathbb{N}$. The closure of $\mathbb{Z}$ is $\mathbb{Z}$. The closure of of $\mathbb{Q}$ is $\mathbb{R}$. The closure of $\mathbb{R}$ is $\mathbb{R}$. The closure of $\varnothing$ is $\varnothing$.

Proof:

(1) First we have to show that $\overline{\mathbb{N}}=\mathbb{N}$. Clearly for each $n\in \mathbb{N}$, $n$ is $\varepsilon$-adherent to $\mathbb{N}$ for any $\varepsilon>0$. So, it lies in $\overline{\mathbb{N}}$, i.e., $\overline{\mathbb{N}}\supset \mathbb{N}$.

Now we have to show that $\overline{\mathbb{N}}$ only contains element of $\mathbb{N}$. Suppose for the sake of contradiction that there exists an element $n'\in \overline{\mathbb{N}}$ such that $n'\notin \mathbb{N}$. Then $|n-n'|>0$ for all $n\in \mathbb{N}$. Let define the set of all $|n-n'|$, the set is non-empty and bounded below by zero. Then it has a greatest lower bound, let call it $b$. Clearly $b\ge 0$. We have to show that all the cases lead a contradiction.

If $b>0$, setting $\varepsilon= b/2$, implies that $n'$ is not $\varepsilon$-adherent to any $n\in \mathbb{N}$, in particular $n'$ is not adherent to $\mathbb{N}$. If $b=0$, then either $b$ lies in the set or does not, if $b$ lies in the set, it is a least element and would imply $|n-n'|=0$ for some $n\in \mathbb{N}$ and so $n=n'$ which contradicts our hypothesis. If $0$ is not in the set, we shall show that we can reached a contradiction. We define recursively the sequence $n_0=0$ and $n_j=\text{min}(\{n>n_{j-1}: |n-n'|\le 1/j \})$, the sequence is well-define because the set $\{n>n_{j-1}: |n-n'|\le 1/j \}\not= \varnothing$ (if were empty it has a minimum element which is not zero, contradicting that $0$ is its greatest lower bound). Then for each $n_j$ we have $|n_j-n'|\le 1/j$, thus $(n_j)\rightarrow n'$, i.e., $(n_j)$ converges to $n$. But $(n_j)$ is a subsequence $\mathbb{N}$, so it has to be unbounded and therefore not convergent. In either case we have a contradiction. Thus $\overline{\mathbb{N}}\subset \mathbb{N}$ and hence $\overline{\mathbb{N}}= \mathbb{N}$ as desired.

(2) For the second assertion notice that $\mathbb{Z}= \mathbb{N}\cup \mathbb{-N}$. The set $\mathbb{-N}$ define by $\mathbb{-N}=\{-n:n\in \mathbb{N}\}$. Thus we must have $\overline{\mathbb{Z}}=\overline{ \mathbb{N}\cup \mathbb{-N}}=\mathbb{N}\cup \overline{ \mathbb{-N}}$. But by the same argument as above we can show that $\overline{ \mathbb{-N}}=\mathbb{-N}$. Thus $\overline{\mathbb{Z}}=\mathbb{Z}$ as desired. The proof is based in the fact that $\overline{X} \cup \overline{Y}= \overline{X \cup Y}$, which will show below.

(3) We have to show that $\overline{\mathbb{Q}}=\mathbb{R}$. Let $x\in \mathbb{R}$, and $\varepsilon>0$ be arbitrary. Then $\exists q\in \mathbb{Q}$ such that $x<q<x+\varepsilon$, i.e, $d(x,q)\le \varepsilon$ since $\varepsilon$ was arbitrary this shows that $x$ is an adherent point of $\mathbb{Q}$. For the other inclusion we have to show that $\overline{\mathbb{Q}}\subset \mathbb{R}$. Let $q'\in \overline{\mathbb{Q}}$ and let $f: \mathbb{N}\rightarrow \mathbb{Q}$ be a bijective map. Define a sequence of natural numbers recursively by setting $n_0=0$ and

$$n_j= \text{min}\{n\in \mathbb{N}: d(q',q_f(n))\le 1/j \text{ and }\, n\not= n_i \text{ for all }\, i< j\}$$

Since $q'$ is an adherent point and $f$ is a bijective map the set is non-empty and each $n_j$ is well-defined. Thus the sequence $(q_{f(n_j)})$ converges to $q'$ and hence $q'$ is a real number as desired.

(4) We shall show that $\overline{\mathbb{R}}=\mathbb{R}$. Let $x'\in \overline{\mathbb{R}}$ our task is prove that $x\in \mathbb{R}$. Let define $$X_n:=\{x\in \mathbb{R}: x'\le x\le x'+1/n \}$$

Clearly for each $n\ge 1$, the set $X_n$ is non-empty. Then using the AC$_\omega$ (Axiom of Countable Chocie) we can find a sequence $(x_n)_{n=1}^\infty$ such that $x_n \in X_n$. Thus $(x_n)\rightarrow x$ and by the completeness the result follows.

(5) $\varnothing= \overline{\varnothing}$. Suppose for the sake of contradiction that $\overline{\varnothing} \not= \varnothing$, then it has an element $x'$. So $x'$ is adherent to the empty set, but this implies that there exist an element in $\varnothing$ for which $x'$ is $\varepsilon$-adherent for every $\varepsilon$, which is a contradiction. $\Box$

Now we shall show the two auxiliary claims which were used in the proof of the above lemma.

Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$.

Proof claim 1: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon>0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$

Claim 2: If $X,Y\subset \mathbb{R}$. Then $\overline{X} \cup \overline{Y}= \overline{X \cup Y}$.

Proof claim 2: Let $x' \in \overline{X \cup Y}$, i.e., $x'$ is and adherent point of $X \cup Y$. Then $\exists x\in {X \cup Y}$ s.t. $d(x,x')\le \varepsilon, \forall\varepsilon>0$. Without loss of generality suppose that $x\in X$, so $x'\in \overline{X} \subset \overline{X} \cup \overline{Y}$. Thus $\overline{X \cup Y} \subset \overline{X} \cup \overline{Y}$. Now since $X \cup Y\subset \overline{X \cup Y} \subset \overline{X} \cup \overline{Y}$, we can conclude that $\overline{\overline{X \cup Y}} =\overline{X} \cup \overline{Y}$. But since $\overline{\overline{X \cup Y}}=\overline{X \cup Y}$ the result follows. $\Box$

Any suggestion? Do you think is correct? or Am I totally off track? Do exist for (4) a way to keep away the Axiom of Countable Choice I cannot find one? Please, any help would be great.

• Is there anything wrong with my proof? – Jose Antonio Jan 20 '14 at 6:57

Let $x \in \overline{\mathbb{N}}$. Assume that $x \notin \mathbb{N}$, and pick maximal $m$, minimal $n$, both in $\mathbb{N}$, such that $m < x < n$. Then $|x-m| = x-m$ and $|x-n| = n-x$. Let $\epsilon = \min \{ (n-x)/2, (x-m)/2 \}$. Then there exists a natural number $k$ such that $|x-k| < \epsilon$. Contradiction.
• OK, I understand good one :). For $\mathbb{Z}$ I think is better use a lemma to show $\overline{X\cup Y} = \overline{X} \cup \overline{Y}$. – Jose Antonio Jan 20 '14 at 1:08
• @Svinepels: I know this is an old topic, but I'm working on the same exercise and reading your answer I don't understand why after writing $|x-k|<\varepsilon$ you claim that this is a contradiction; could you explain this point in more detail? – lorenzo Dec 13 '16 at 23:44
• If $|x-k|<\epsilon$ then $k$ is closer to $x$ than both $m$ and $n$. This contradicts the fact that $m$ was chosen to be the biggest natural number less than $x$and $n$ the least natural number bigger than $x$. – Ulrik Dec 14 '16 at 21:02