# Linear ordered dense subset.

1. Jan 10, 2013

### cragar

1. The problem statement, all variables and given/known data
If a linear ordered set P has a countable dense subset, then
$|P| \leq 2^{\aleph_0}$
3. The attempt at a solution
because it has a linear order then all elements of P can be compared x<y .
And because it is dense that mean that I can find an element f
such that x<f<y for any x or y.
so we can partition this set into countable many pieces.
now either their are countably many elements between
any two elements. And the union of countably many things
with a countable number of objects is countable.
Or their are an uncountable number of things in between
any of the 2 dense elements, but this union would
be $2^{\aleph_0}$ if anything larger was in between
then the cardinality would not work.
This is a little informal, just want to know if this is on the right track.

2. Jan 10, 2013

### micromass

How will you define this partition??

3. Jan 10, 2013

### cragar

Can i just take an arbitrary partition will some element w.
And because the set is linear ordered, their will be either 0 elements to the left or
at most $2^{\aleph_0}$ to the left or right. I guess im assuming that.
does this look ok? Im thinking of using something like if I have a countable dense
subset on an interval I can only have $2^{\aleph_0}$ limit points.
all thought I will probably need to prove that.

4. Jan 10, 2013

### micromass

I don't really get it. Can you write it out more formally?

5. Jan 10, 2013

### cragar

I guess I am going to pick a random element from the set call it x.
now I have partitioned the set into 2 sub sets.
I call them E and U. E is everything less than or equal to x and U is everything greater than x. Now either there is 0 elements in E or countable or 2^N,
and U must either have finite, countable or 2^N.
I was going to assume one of the partions have more than 2^N elements
and then say this was a contradiction. But I dont think that really proves it and I havn't
used the fact that the set has a countable dense subset.
Im not really sure what to do next.
Maybe I could assume there were more than 2^N elements
in between and 2 of my countable elements and then someohow get a contradiction by showing that
I couldn't embed a countable subset into a set with more than 2^N elements.

6. Jan 10, 2013

### gopher_p

This is the direction you want to head in. Also, make sure to look up (or recall) the definition of the cardinal $2^{\aleph_0}$. You wouldn't want to try to prove something that's true by definition.

7. Jan 11, 2013

### cragar

so $2^{\aleph_0}$ is the set of all possible combinations of the natural numbers. I still dojnt really know how to prove what I want.
I guess beacuse we have a countable dense subsets between any 2 of our
countable elements that their exists at most $2^{\aleph_0}$
limits points and there are only a countable many partitions in out set.

8. Jan 11, 2013

### gopher_p

Yes. I would characterize $2^{\aleph_0}$ as the cardinality of the power set of the natural numbers (or, for that matter, any set of cardinality $\aleph_0$). In general, $2^{|X|}$ is the cardinality of the power set of $X$.

You have some good ideas (mixed in with some not-as-good ones), so let's try to get them organized. First let's give names to the pertinent sets to make things easier to read and write; let $P$ be the linearly ordered set and $A$ be the countable subset of $P$ which is dense in $P$, and let $\bar{A}$ be the set of limit points of $A$.

1) Forget about dense part for the moment. Can you show that $|\bar{A}|\leq2^{\aleph_0}$ using only the definition of limit point, the fact that $|A|=\aleph_0$, and our definition for $2^{\aleph_0}$? So your proof here should really depend only on the cardinality of $A$ and not at all on its relationship to the ambient space $P$.

2) Now for the dense part ... Can you show that $P=\bar{A}$ using the definition of dense?

Out of curiosity, what definitions for limit point and dense are you using? Part 2 could potentially end up being very easy.

Last edited: Jan 11, 2013