# Dense in R.

WatchPage 1 of 1

Go to first unread

Skip to page:

There exists a pair of disjoint subsets of Q such that both are dense in R.

True or false?

The statement is true but how can I find an example of such disjoint subsets of Q?

True or false?

The statement is true but how can I find an example of such disjoint subsets of Q?

0

reply

Report

#2

(Original post by

There exists a pair of disjoint subsets of Q such that both are dense in R.

True or false?

The statement is true but how can I find an example of such disjoint subsets of Q?

**Infinity...**)There exists a pair of disjoint subsets of Q such that both are dense in R.

True or false?

The statement is true but how can I find an example of such disjoint subsets of Q?

1

reply

Report

#3

**Infinity...**)

There exists a pair of disjoint subsets of Q such that both are dense in R.

True or false?

The statement is true but how can I find an example of such disjoint subsets of Q?

0

reply

(Original post by

Write down some rationals p/q,cancel the gcd(p,q) so its in lowest terms. Is there some distinction between the denominators in each one? The distinction will partition Q and will also be dense

**physicsmaths**)Write down some rationals p/q,cancel the gcd(p,q) so its in lowest terms. Is there some distinction between the denominators in each one? The distinction will partition Q and will also be dense

0

reply

Report

#5

(Original post by

I am getting infinite rationals this way. How will I define this set?

**Infinity...**)I am getting infinite rationals this way. How will I define this set?

2/3, 5/6 ,1/3, 4/9 ,7/18, 1/10 we have simplified the fractions. Now each denominator is either even or odd. What would the partition be? Why would each of them be dense...

0

reply

(Original post by

if i write some rationals down just so we get an idea

2/3, 5/6 ,1/3, 4/9 ,7/18, 1/10 we have simplified the fractions. Now each denominator is either even or odd. What would the partition be? Why would each of them be dense...

**physicsmaths**)if i write some rationals down just so we get an idea

2/3, 5/6 ,1/3, 4/9 ,7/18, 1/10 we have simplified the fractions. Now each denominator is either even or odd. What would the partition be? Why would each of them be dense...

0

reply

Report

#7

(Original post by

I partitioned Q in two sets. A = Rationals Which have even denominator. B = which have odd denominator.. I can take integers as numerator for both sets. Both are dense in R. But they are not disjoint.

**Infinity...**)I partitioned Q in two sets. A = Rationals Which have even denominator. B = which have odd denominator.. I can take integers as numerator for both sets. Both are dense in R. But they are not disjoint.

0

reply

(Original post by

They are dense if u take gcd(p,q)=1

**physicsmaths**)They are dense if u take gcd(p,q)=1

1

reply

Report

#9

**Infinity...**)

I partitioned Q in two sets. A = Rationals Which have even denominator. B = which have odd denominator.. I can take integers as numerator for both sets. Both are dense in R. But they are not disjoint.

Spoiler:

Show

For example, A = numbers with finite decimal representations that end in 1, B = numbers with finite decimal representations that end in 2 works.

1

reply

(Original post by

For a slightly different solution, you could look at numbers with finite decimal representations (e.g. 123, 3.14159, etc):

**DFranklin**)For a slightly different solution, you could look at numbers with finite decimal representations (e.g. 123, 3.14159, etc):

Spoiler:

Show

For example, A = numbers with finite decimal representations that end in 1, B = numbers with finite decimal representations that end in 2 works.

Can we represent such sets in roster form?

0

reply

Report

#11

I don’t know the answer to this yet, maybe we would have to do some topology to try form a counterexample(if one exists)

0

reply

Report

#12

Nope, this is not true in general. Take the discrete metric on any set and take the set itself.

1

reply

Report

#13

It’s true in this form

Take (X,d) to be any metric space.

If

X has a countable dense subset

X has no isolated points

Then it’s true

Take (X,d) to be any metric space.

If

X has a countable dense subset

X has no isolated points

Then it’s true

0

reply

(Original post by

Nope, this is not true in general. Take the discrete metric on any set and take the set itself.

**physicsmaths**)Nope, this is not true in general. Take the discrete metric on any set and take the set itself.

0

reply

Report

#15

(Original post by

yep, I took N as set and also as its subset. So we can't split N in such disjoint subsets.

**Infinity...**)yep, I took N as set and also as its subset. So we can't split N in such disjoint subsets.

0

reply

(Original post by

This example doesn’t worn since N isn’t dense in R

**physicsmaths**)This example doesn’t worn since N isn’t dense in R

0

reply

N is dense in N itself since it contains all its points and N has no limit point. so closure of N = N.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top