# Dense in R.

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Thread starter 2 years ago
#1
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
2 years ago
#2
(Original post by 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?
You might like to start by thinking of how you can get a single proper subset of Q that is dense in R, and try and build on that.
1
2 years ago
#3
(Original post by 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?
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
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Thread starter 2 years ago
#4
(Original post by 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
I am getting infinite rationals this way. How will I define this set?
0
2 years ago
#5
(Original post by Infinity...)
I am getting infinite rationals this way. How will I define this set?
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...
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Thread starter 2 years ago
#6
(Original post by 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...
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
2 years ago
#7
(Original post by 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.
They are disjoint if u take gcd(p,q)=1
0
Thread starter 2 years ago
#8
(Original post by physicsmaths)
They are dense if u take gcd(p,q)=1
They will be disjoint if gcd(p,q) = 1. Okay. I got it. Thanks.
1
2 years ago
#9
(Original post by 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.
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.
1
Thread starter 2 years ago
#10
(Original post by 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.
Oh yes.
Can we represent such sets in roster form?
0
2 years ago
#11
(Original post by Infinity...)
Oh yes.
Can we represent such sets in roster form?
This raises a different question, can any set A dense in B be split into two disjoint sets which are also dense in B?
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
2 years ago
#12
Nope, this is not true in general. Take the discrete metric on any set and take the set itself.
1
2 years ago
#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
0
Thread starter 2 years ago
#14
(Original post by physicsmaths)
Nope, this is not true in general. Take the discrete metric on any set and take the set itself.
yep, I took N as set and also as its subset. So we can't split N in such disjoint subsets.
0
2 years ago
#15
(Original post by Infinity...)
yep, I took N as set and also as its subset. So we can't split N in such disjoint subsets.
This example doesn’t worn since N isn’t dense in R
0
Thread starter 2 years ago
#16
(Original post by physicsmaths)
This example doesn’t worn since N isn’t dense in R
I took N as space.
0
Thread starter 2 years ago
#17
N is dense in N itself since it contains all its points and N has no limit point. so closure of N = N.
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