Cardinality?

Question

Can anyone prove that the set of real numbers between 0 and 1 is uncountable?

Answer 1

By definition, a set is countable if there exists a one-to-one correspondence between its elements and the set of natural numbers. Suppose that the real numbers between 0 and 1 are countable, then we could make a list in which each of them has a sequential number, e.g.

1 --> 0.5987029560904534....
2 --> 0.1985709165402452....
3 --> 0.9759376012498702....
and so forth.

Now construct a decimal number K as follows: take the n-th decimal of the n-th number, and pick a *different* digit for the n-th decimal in K.

In our example, the 1st decimal of the 1st number is 5, so we pick a different digit (say, 6).
The 2nd decimal of the 2nd number is 9, so we pick anything except 9 (say, 0).
The 3rd decimal of the 3rd number is 5, so we pick a digit not five (say, 7).
And so forth.

Claim: our number K = 0.607...... does not occur on the list. Proof: for every n, it is unequal to number n on the list because its n-th decimal is different.

Therefore, our one-to-one correspondence does not include the number K -- contradiction. Therefore the assumption that the real numbers between 0 and 1 are countable is false.

(Technically, we should take care of the fact that 0.199999999999.... and 0.2000000000.... are the same number. However, this does not really affect the reasoning.)

Answer 2

What she said is true.

The decimal numbers between 0 and 1 are infinite.

because we would have to start with the smallest possible closest to 0. and as u can see

0.0000000000000000000000000000000000000000000
0000000000000000000000000001

Where do we find the .000001???? we cant. Therefore just right there u can see that there are an infinite amount of numbers between 0 and 1.

Since there are infinitely numbers between the 0 and 1 its uncountable.

Now how to Prove it formally, I'm not extactly sure. But basically take what I said and make it formal by putting it in the format of a proof.

For Example The smallest # after 0 is 0.1^inifinite power.
Therefore being uncountable.

Answer 3

The entire set of real numbers is uncountable. It is possible to devise a mapping of (0, 1) onto (0, infinity) which is one-to-one and onto.

Answer 4

The continuum hypothesis states that there is no cardinal variety which lies strictly between the cardinality of the integers and the cardinality of the genuine numbers. The continuum hypothesis is per the similar old axioms of set idea. The NEGATION of the continuum hypothesis is likewise per the similar old axioms of set idea. those 2 info at the same time (that are deep theorems) mean that the continuum hypothesis can not be proved, nor disproved in the similar old axioms of set idea. Your problem 2 asks you to instruct the continuum hypothesis (as others talked about). I only had to remark further with regard to the independence of the continuum hypothesis from the different axioms of set idea. I have not heard the continuum hypothesis said as an axiom earlier -- i don't think of it somewhat is favourite. yet there are a lot of theorems of the kind: assuming the continuum hypothesis, then such and such is real ..

Answer 5

Well, [0,1] is certainly infinite. So suppose it is countably infinite and listed as a_1, a_2, a_3, ....

Pick a closed interval I_1 that is a subset of [0,1] and does not contain a_1. Then find a closed interval I_2 that is a subset of I_1 and does not contain a_2. Keep going like this. Then the intersection of all of these closed intervals is non-empty (take the least upper bound of the left-hand endpoints) but contains none of the a_n. This is a contradiction to the assumption that [0,1] is listed by all the a_n. Thus [0,1] is uncountable.

The Cantor diagonalization argument works also, but this one is a bit closer to the axioms for the real numbers. Just knowing that [0,1] is infinite is just not enough. ou have to show it is not countably infinite.

Answer 6

Yep, that is what Cantor did. Any such number can be written in the form 0.(a1)(a2)(a3)....... where (a1), (a2) denotesingle digit numbers which can be 0,1,....,9. Now assume this set of real numbers is countable, then we can list them as
X1=0.(a11)(a12)(a13).....
X2=0.(a21)(a22)(a23).....
X3=0.(a31)(a32)(a33)...
....
Now we will construct a number X=0.(b1)(b2)(b3)... which can not be on this list. For this we will set bN to be a single digit number which is not equal to aNN. Say one more than aNN if aNN is not 9 and 0 if aNN is 9. Then X is not on the given list. Why? Because X is different from any XN at the N-th decimal digit. So it is different from all the numbers in the list yielding a contradiction.

Answer 7

you just need a bijection between that interval and the real numbers themselves.

Answer 8

Use "Cantor diagonal argument". Read the link below for a nice proof.

Answer 9

What is there to prove? The numbers between any two other numbers are infinite.

Forza JUVE!!!!!