What things are you given? Are you given that R is uncountable?
You use bijections to prove sets are COUNTABLE, not uncountable. In this case you won't be able to find a bijection, i think.
Hi, can friends here please help me with this question? I'm really struggling with it.
Prove the set of irrational number R\Q is uncountable.
I know the idea is try to find the bijection function, right? But I just cannot do it. Please help me with it. Thanks a lot.
Yes, sorry that I was a bit drunk, haha, of course, bijection is for COUNTABLE, I knew that, thanks.
And yes, I was given R is uncountable. I guess I have the idea is that prove union or intersection of uncountable is uncountable as well. But I just really don't know how to do in details. Please help me. Thanks a lot.
I was given R is uncountable. I guess I have the idea is that prove union or intersection of uncountable is uncountable as well. But I just really don't know how to do in details. Please help me. Thanks a lot.
Not necessarily so. A bijection from a set A to the set of real numbers between 0 and 1 would prove that A is uncountable. A bijection always shows that two sets have the same cardinality.
Sorry: this was in response to Vsalev's post: "You use bijections to prove sets are COUNTABLE, not uncountable. In this case you won't be able to find a bijection, i think."
I see what you're doing here. The set = , which we know is uncountable. We also know that is countable. But a number must either be rational or irrational, so it follows logically that must be uncountable.
Now, proving this formally... Could one more or less just follow the logic here, or is there a more rigorous approach? For instance, is it acceptable just to state = (which is obvious), or should we go about proving this first?
that's a perfectly valid proof as it stands. The union of two countable sets is countable. (Have you proved that yet?)
If the set of all irrational numbers were countable then the set of all real numbers, the union of the set of irrational numbers and the set of rational numbers, would be the union of two countable set and so countable. That contradicts the fact that the set or all real numbers is uncountable.
You can do this with a variant of Cantor's diagonal slash (I will be leaving out most of the hard work to leave something for the reader to do, that and I'm too lazy to do it myself)
We will be working with reals in the interval and with decimal expansions, so the decimal expansion of a real in is .
I will use the expressions "the decimal expansion of is periodic" or " is periodic" to mean the eventually the decimal expansion of becomes periodic.
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Lemma
A real in is rational if and only if it is periodic
Proof
Excersise for the reader (it can be looked up)
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Suppose the set or irrational numbers is enumerable and that is an enumeration. Let denote the th term in the decimal expansion of
Now we combine Cantor's slash argument with an idea of Liouville's, define the real with decimal expansion:
Then and is not periodic, so as differes from each irrational in our enumeration at at least one place in its' decimal expansion it is not in our enumeration and is irrational. hence we have the required contradiction ...
CB
We can also use . Note that this is a surjection.
Noting that , if then . Thus, letting be the restriction of to we see that is a surjection onto something with cardinality that of . This is sufficient.
(This is sufficient because if there exists a surjection from to then . Similarly, if there exists an injection from to then . Clearly the latter holds, and the former holds by what I just wrote about. Thus, , as required.)