Let $\displaystyle E\subset \mathbb{R}^1$ be a non-empty countable set. Suppose $\displaystyle E$ has no isolated points, prove that $\displaystyle \bar E\backslash E$ is dense in $\displaystyle \bar E.$

It is difficult, isn't it?

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- Oct 25th 2012, 05:40 PMxinglongdadaDense-in-itself set
Let $\displaystyle E\subset \mathbb{R}^1$ be a non-empty countable set. Suppose $\displaystyle E$ has no isolated points, prove that $\displaystyle \bar E\backslash E$ is dense in $\displaystyle \bar E.$

It is difficult, isn't it? - Oct 25th 2012, 08:51 PMhollywoodRe: Dense-in-itself set
I think the best approach might be to suppose that $\displaystyle \bar E\backslash E$ is not dense in $\displaystyle \bar E$. Start unraveling the definitions and see what you can prove about the set E.

- Hollywood - Oct 26th 2012, 12:59 AMxinglongdadaRe: Dense-in-itself set
I could not see any property of $\displaystyle E$, if I assume $\displaystyle \bar E'\backslash E$ is not dense in $\displaystyle E$.

I argue as follows. We need only prove the closure of $\displaystyle \bar E'\backslash E$ contains $\displaystyle \bar E.$

Suppose then $\displaystyle x$ not in the closure of $\displaystyle \bar E'\backslash E$, then there exists a $\displaystyle \tilde \delta>0$ , such that for any $\displaystyle \delta\in (0,\tilde \delta)$, the ball $\displaystyle B(x,\delta)\subset (E'\backslash E)^c=E^{'c}\cup E$. Since $\displaystyle E$ is countable, we see $\displaystyle B(x,\delta)\cap E^{'c}\neq \emptyset.$ This, by definition, means $\displaystyle x\in E^{'c-}$ (the closure of $\displaystyle E^{'c}$).

Here, I could only prove $\displaystyle x\in E^{'c-}$, but not $\displaystyle x\in E^{'c}.$

Would you help me out? Thank you. - Oct 26th 2012, 01:25 PMPlatoRe: Dense-in-itself set
Another way to define being

*dense-in-itself*to to say that the set contains no isolated points.

Thus is $\displaystyle E$ dense-in-itself?

Can $\displaystyle \overline{E}\setminus E$ be dense-in-itself? - Oct 26th 2012, 02:20 PMxinglongdadaRe: Dense-in-itself set
Yes, indeed they are equivalent. $\displaystyle E\subset E'$ is equivalent to saying $\displaystyle E$ has no isolated points.

I do not see how to prove the statement. - Oct 26th 2012, 03:06 PMPlatoRe: Dense-in-itself set
- Oct 26th 2012, 03:13 PMhediRe: Dense-in-itself set
E has no isolated points so every point x in cl(E) is limit of a sequence xn of different elements of E.E is countable so in the neigborhood of each xn there is yn in cl(E)-E.yn is converging to x,so we have our claim.

- Oct 26th 2012, 03:18 PMxinglongdadaRe: Dense-in-itself set
Yes, I want to prove this.

- Oct 26th 2012, 03:29 PMxinglongdadaRe: Dense-in-itself set

Why " in the neigborhood of each xn there is yn in cl(E)-E"? I know only

$\displaystyle B(x_n,1/n,x_n+1/n)\subset E^{-c}\cup (\bar E\backslash E)\cup E.$ Thus, by the countability of $\displaystyle E$, I could only see $\displaystyle B(x_n-1/n,x_n+1/n)$ contains an element $\displaystyle y_n$ lies in $\displaystyle \bar E^c$ or $\displaystyle \bar E\backslash E$? - Oct 26th 2012, 06:11 PMPlatoRe: Dense-in-itself set
I HATE THESE TRICKY PROOFS.

Notation: $\displaystyle \mathcal{B}(x;\delta)=\left( {x - \delta ,x + \delta } \right)$ where $\displaystyle \delta>0$.

Note that if $\displaystyle t\in E$ then $\displaystyle \mathcal{B}(t;1)$ contains a point $\displaystyle x_1\in E\setminus\{t\}$ because $\displaystyle E$ has no isolated points.

But $\displaystyle \mathcal{B}(t;1)$ is uncountable so $\displaystyle \left( {\exists y_1 \notin E} \right)\left[ {y_1 \in \mathcal{B}(t;1)} \right]$

So if $\displaystyle \delta_1=1$ then $\displaystyle \exists\delta_2$ such that $\displaystyle \mathcal{B}(t;\delta_2)}\subseteq\mathcal{B}(t; \delta_1)}$ so that $\displaystyle x_1\notin \mathcal{B}(t;\delta_2)}$ AND $\displaystyle y_1\notin \mathcal{B}(t;\delta_2)}$.

Can you see why I HATE this?

There is a sequence of points $\displaystyle \left( {x_n } \right)$ in $\displaystyle E$ such that $\displaystyle \left( {x_n } \right) \to t$.

There is a sequence of points $\displaystyle \left( {y_n } \right)$ in $\displaystyle E^c$ such that $\displaystyle \left( {y_n } \right) \to t$. - Oct 26th 2012, 08:37 PMxinglongdadaRe: Dense-in-itself set
Then you prove that $\displaystyle E\subset \partial E.$ How to prove the statement then?

- Oct 27th 2012, 07:34 AMPlatoRe: Dense-in-itself set
- Oct 27th 2012, 08:39 AMhediRe: Dense-in-itself set
since E has no isolated points, the open sets in cl(E) are uncountable so they must intersect cl(E)-E,because E is countable.(notice that cl(E) is uncountable).

- Oct 27th 2012, 11:35 AMhollywoodRe: Dense-in-itself set
A friend of mine came up with this proof having to do with perfect sets. By definiton, a set E is perfect if E=E', the set of limit points of E. A set is perfect if it is closed and has no isolated points. And if a set is perfect, it is uncountable. This Wikipedia article has some background:

Derived set (mathematics) - Wikipedia, the free encyclopedia

You can also find a proof that a perfect set is uncountable on the web.

Ok. Suppose that $\displaystyle \bar E\backslash E$ is not dense in $\displaystyle \bar E$. Then there exists $\displaystyle x\in\bar{E}$ and $\displaystyle \epsilon>0$ such that $\displaystyle B(x,\epsilon)\cap(\bar E\backslash E)$ is empty. So $\displaystyle \bar{B}(x,\frac{\epsilon}{2})\cap(\bar E\backslash E)$ is empty. This means that in $\displaystyle \bar{B}(x,\frac{\epsilon}{2})$ every point of $\displaystyle \bar{E}$ is also a point of E. So $\displaystyle \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E}$ is a subset of E. But $\displaystyle \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E}$ is closed, and since it is a subset of E, contains no isolated points. So $\displaystyle \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E}$ is perfect, and therefore uncountable. This is a contradiction, so $\displaystyle \bar E\backslash E$ is dense in $\displaystyle \bar E$.

- Hollywood