Dense-in-itself set

**xinglongdada** · October 25th 2012, 05:40 PM

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

It is difficult, isn't it?

**hollywood** · October 25th 2012, 08:51 PM

I think the best approach might be to suppose that $\bar E\backslash E$ is not dense in $\bar E$ . Start unraveling the definitions and see what you can prove about the set E.

- Hollywood

**xinglongdada** · October 26th 2012, 12:59 AM

I could not see any property of $E$ , if I assume $\bar E'\backslash E$ is not dense in $E$ .

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

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

Would you help me out? Thank you.

**Plato** · October 26th 2012, 01:25 PM

Originally Posted by xinglongdada

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

Another way to define being dense-in-itself to to say that the set contains no isolated points.

Thus is $E$ dense-in-itself?

Can $\overline{E}\setminus E$ be dense-in-itself?

**xinglongdada** · October 26th 2012, 02:20 PM

Yes, indeed they are equivalent. $E\subset E'$ is equivalent to saying $E$ has no isolated points.

I do not see how to prove the statement.

**Plato** · October 26th 2012, 03:06 PM

Originally Posted by xinglongdada

Prove that $\overline{E}\backslash E$ is dense in $\overline{E}.$

Well frankly I was confused by the title of the thread and the question above.

If you want to prove the above question, can you show that

$\overline{\overline{E}\setminus E}=\overline{E}~?$

**hedi** · October 26th 2012, 03:13 PM

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.

**xinglongdada** · October 26th 2012, 03:18 PM

Yes, I want to prove this.

**xinglongdada** · October 26th 2012, 03:29 PM

Originally Posted by hedi

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.

Why " in the neigborhood of each xn there is yn in cl(E)-E"? I know only
$B(x_n,1/n,x_n+1/n)\subset E^{-c}\cup (\bar E\backslash E)\cup E.$ Thus, by the countability of $E$ , I could only see $B(x_n-1/n,x_n+1/n)$ contains an element $y_n$ lies in $\bar E^c$ or $\bar E\backslash E$ ?

**Plato** · October 26th 2012, 06:11 PM

Originally Posted by xinglongdada

Why " in the neigborhood of each xn there is yn in cl(E)-E"? I know only
$B(x_n,1/n,x_n+1/n)\subset E^{-c}\cup (\bar E\backslash E)\cup E.$ Thus, by the countability of $E$ , I could only see $B(x_n-1/n,x_n+1/n)$ contains an element $y_n$ lies in $\bar E^c$ or $\bar E\backslash E$ ?

I HATE THESE TRICKY PROOFS.
Notation: $\mathcal{B}(x;\delta)=\left( {x - \delta ,x + \delta } \right)$ where $\delta>0$ .
Note that if $t\in E$ then $\mathcal{B}(t;1)$ contains a point $x_1\in E\setminus\{t\}$ because $E$ has no isolated points.
But $\mathcal{B}(t;1)$ is uncountable so $\left( {\exists y_1 \notin E} \right)\left[ {y_1 \in \mathcal{B}(t;1)} \right]$

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

Can you see why I HATE this?

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

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

**xinglongdada** · October 26th 2012, 08:37 PM

Then you prove that $E\subset \partial E.$ How to prove the statement then?

**Plato** · October 27th 2012, 07:34 AM

Originally Posted by xinglongdada

Then you prove that $E\subset \partial E.$ How to prove the statement then?

Read reply #7.

**hedi** · October 27th 2012, 08:39 AM

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).

**hollywood** · October 27th 2012, 11:35 AM

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

- Hollywood

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