Let be a non-empty countable set. Suppose has no isolated points, prove that is dense in

It is difficult, isn't it?

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- Oct 25th 2012, 06:40 PMxinglongdadaDense-in-itself set
Let be a non-empty countable set. Suppose has no isolated points, prove that is dense in

It is difficult, isn't it? - Oct 25th 2012, 09:51 PMhollywoodRe: Dense-in-itself set
I think the best approach might be to suppose that is not dense in . Start unraveling the definitions and see what you can prove about the set E.

- Hollywood - Oct 26th 2012, 01:59 AMxinglongdadaRe: Dense-in-itself set
I could not see any property of , if I assume is not dense in .

I argue as follows. We need only prove the closure of contains

Suppose then not in the closure of , then there exists a , such that for any , the ball . Since is countable, we see This, by definition, means (the closure of ).

Here, I could only prove , but not

Would you help me out? Thank you. - Oct 26th 2012, 02:25 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 dense-in-itself?

Can be dense-in-itself? - Oct 26th 2012, 03:20 PMxinglongdadaRe: Dense-in-itself set
Yes, indeed they are equivalent. is equivalent to saying has no isolated points.

I do not see how to prove the statement. - Oct 26th 2012, 04:06 PMPlatoRe: Dense-in-itself set
- Oct 26th 2012, 04:13 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, 04:18 PMxinglongdadaRe: Dense-in-itself set
Yes, I want to prove this.

- Oct 26th 2012, 04:29 PMxinglongdadaRe: Dense-in-itself set
- Oct 26th 2012, 07:11 PMPlatoRe: Dense-in-itself set
I HATE THESE TRICKY PROOFS.

Notation: where .

Note that if then contains a point because has no isolated points.

But is uncountable so

So if then such that so that AND .

Can you see why I HATE this?

There is a sequence of points in such that .

There is a sequence of points in such that . - Oct 26th 2012, 09:37 PMxinglongdadaRe: Dense-in-itself set
Then you prove that How to prove the statement then?

- Oct 27th 2012, 08:34 AMPlatoRe: Dense-in-itself set
- Oct 27th 2012, 09:39 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, 12:35 PMhollywoodRe: 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 is not dense in . Then there exists and such that is empty. So is empty. This means that in every point of is also a point of E. So is a subset of E. But is closed, and since it is a subset of E, contains no isolated points. So is perfect, and therefore uncountable. This is a contradiction, so is dense in .

- Hollywood