Define the closure of a set $A\subset \Bbb R$ as $clA.$ Show that $x\in clA,$ iff $d(x,A)=0$ where $d(x,A)=inf\{|x-a|:a\in A\}.$

Question

Define the closure of a set $A\subset \Bbb R$ as $clA.$ Show that $x\in clA,$ iff $d(x,A)=0$ where $d(x,A)=\text{inf}\{|x-a|:a\in A\}.$

I tried solving the probllrm like this:

We know that $clA=A\cup L,$ where $L$ is the set of limit points (or derived set) of $A.$

Let $x\in clA.$ This implies $x\in A$ or $x\in L.$ If $x\in A,$ then $d(x,A)=\text{inf}\{|x-a|:a\in A\}=0,$ trivially and we are done. If $x\in L,$ then it means that $x$ is a limit point of the set $A,$ and which implies that $\exists (a_n)\in A$ such that $\lim a_n=x.$ So, for all $\epsilon \gt 0$ there exists a $K\in \Bbb N$ such that if $n\geq K,$ $|a_n-x|\lt\epsilon.$ If we place $|a_n-x|=p$ in a set $d(x,A),$ then it follows, that $inf(d(x,A))=0,$ which proves the fact that:" $x\in clA,$ if $d(x,A)=0$ where $d(x,A)=\text{inf}\{|x-a|:a\in A\}.$"

Now, we wish to prove the converse part i.e " If $d(x,A)=0$ where $d(x,A)=\text{inf}\{|x-a|:a\in A\},$ then, $x\in clA.$"

So, let $x\in \Bbb R$ such that $d(x,A)=0.$ Let $S=\{|x-a|:a\in A\}.$ If, $0\in S,$ then it simply implies that $\exists a\in A,$ such that $|x-a|=0\implies x=a\in A.$ Thus, $x\in A\cup L=clA,$ and we are done.

Now if, $0\notin S,$ then we claim $x\in clA.$ We try to show that this claim is true.

If, $d(x,A)=0$ and $0\notin S$ then, for a particular $\epsilon \gt 0$ there must exist an $a\in A$ such that $|x-a|\lt \epsilon.$ For if, such an $a\in A$ does not exist, then it means $\forall a\in A,$ $|x-a|\geq \epsilon,$ implies that $\text{inf}(S)=\epsilon,$ as opposed to $0$ as, $\epsilon\gt 0,$ a contradiction. So, this is never the case. Hence, if in particular, we choose, $\epsilon =\frac 1n,n\in \Bbb N$ then it means $\exists a_n\in A$ such that $|a_n-x|\lt \frac 1n.$ We thus, get a convergent sequence $(a_n)\in A$ such that $\lim a_n=x.$ This implies, that $x$ is a limit point of $A$ i.e $x\in L\implies x\in clA.$ This completes the proof.

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I hope my solution is correct. However, I want to know if I can improve the proof to make it more understandable or is it good to go, as this. Any suggestions to improve the proof or to make it simpler will be highly appreciated. Lastly, if something looks/is wrong with my proof, then please do let me know.

Answer 1

It is correct but there are better ways to put it, yes, especially this part here.

If we place $|a_n-x|=p$ in a set $d(x,A)$, then it follows that $\inf(d(x,A))=0$ which proves the fact that: $x\in\operatorname{cl}A$ if $d(x,A)=0$ where $d(x,A)=\inf\{|x-a|:a\in A\}$.

The idea is correct, but the above could have been written better. Also on a technical note, $d(x,A)$ is the infimum so you should have just written $d(x,A)$ instead of $\inf(d(x,A))$, and the set as $\{|x-a|:a\in A\}$. Regardless, here is a better way to put it.

We want to show $\inf\{|x-a|:a\in A\}=0$. For any $\varepsilon>0$, note that there is $K$ such that $|x-a_K|<\varepsilon$ so any $\varepsilon>0$ is not a lower bound for $\{|x-a|:a\in A\}$. But since $|x-a|>0$ for all $a$, $0$ is clearly a lower bound. Therefore, $0$ is the infimum of the set $\{|x-a|:a\in A\}$, as desired.

Everything else is fine. Hope it helps. :)

Answer 2

Some improvements that I'd suggest:

In the forward direction:

1. You don't need to consider $x\in A$ and $x\in L$ separately. What you do for $x\in L$ case works for $x\in A$ as well. In this case, $a_n=x$ for all $n$.
2. I know that you have the right idea but I don't understand what you did for $x\in L$ case. You said: There exists $(a_n)$ in $A$ such that $\lim a_n=x$. Okay. This implies that $\lim (a_n-x)=0$ and which in turn implies $\lim |a_n-x|=0$. Remember that $0\leq d(x,A)\leq |a_n-x|$ for all $n$ so $0\leq d(x,A)\leq \lim|a_n-x|=0$.

In the Backward direction:

1. Again you don't need to consider $A,L$ separately so you don't need $S$. You just need what you wrote in the end. You are using the definition of infimum in the $d(x,A)$. For every $n\in \mathbb N$, $\frac1n>0=d(x,A)$ so it is not a lower bound so there exists $a_n\in A$ such that $|a_n-x|<\frac1n$ which proves that $a_n\to x$. So $x\in \text{cl}(A)$.

I also used to do separate cases when I studied it. So I understand you.

Answer 3

Let $(X,d)$ be a metric space and $A\subset X$. Then,

$\begin{align} x\in\bar A&\iff \forall n\in\Bbb Z^+,\;B(x,\frac 1n)\cap A\neq\emptyset\\ &\iff d(x,A)=0. \end{align}$

Notation:

$B(x,\frac 1n)$: The open ball with center $x$ and radius $\frac 1n$.

$\bar A$: The closure of $A$.

$d(x,A)=\inf\{d(x,a)|a\in A\}$.