Help with Countability Questions for Real Numbers

In summary: Cardinality is a property of sets, not of maps. So even though the empty set and the empty map don't have the same cardinality, they are still both countable.In summary, the empty set is not countable, and the empty map is the only possible way to map the empty set onto a subset of the real numbers.
  • #1
jackbauer
10
0
Hi people, I need some help with these questions please:

1.Is the set of all x in the real numbers such that (x+pi) is
rational, countable?

I don't think this is countable, isn't the only possible value for x = -pi, all other irrationals will not make x+pi rational i thought?

2.Is the set of all x in the real numbers, such that for all k, (x+the square root of k) is not a natural number, countable?

Again I don't think this is countable because if the square root of k is irrational, like it is for k=2, then you can add infinitely many values of x to root k for which the sum of x and root k is not natural.

Lastly, is every infinite subset of the power set of the naturals uncountable?
I don't think so because P(N), the power set of N, has cardinality 2^(aleph nought) and is countable, hence it's subsets will be countable.

Could anyone offer some advice,

Cheers,
JB
 
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  • #2
jackbauer said:
1.Is the set of all x in the real numbers such that (x+pi) is
rational, countable?

I don't think this is countable, isn't the only possible value for x = -pi, all other irrationals will not make x+pi rational i thought?

What about 1/2 - pi? 1/3 - pi? 6/5 - pi?
 
  • #3
Countability of infinite sets is possible. The natural numbers are infinite and are certainly countable. However, 2^N is not countable
 
  • #4
jackbauer said:
Hi people, I need some help with these questions please:

1.Is the set of all x in the real numbers such that (x+pi) is
rational, countable?

I don't think this is countable, isn't the only possible value for x = -pi, all other irrationals will not make x+pi rational i thought?

think again. and are you asserting that finite sets are not countable? It is a matter of convention but one to get straight at the start.

2.Is the set of all x in the real numbers, such that for all k, (x+the square root of k) is not a natural number, countable?

Again I don't think this is countable because if the square root of k is irrational, like it is for k=2, then you can add infinitely many values of x to root k for which the sum of x and root k is not natural.

why are you adding 'x' to the square root of 'k'? reread the question.

Lastly, is every infinite subset of the power set of the naturals uncountable?
I don't think so because P(N), the power set of N, has cardinality 2^(aleph nought) and is countable,

that is false, P(N) is uncountable, and is I'd wager the only set you've been shown is uncountable.


Could anyone offer some advice,

Cheers,
JB

you need to reread your notes and questions far more carefully.
 
  • #5
The empty set is finite, so it must be countable!
But which bijective mapping do we actually have here??
Does it depend on whether we take 0 as a natural number or not?
I've never thought of it before...
 
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  • #6
symplectic_manifold said:
The empty set is finite, so it must be countable!
But which bijective mapping do we actually have here??
Does it depend on whether we take 0 as a natural number or not?
I've never thought of it before...

Well, for any set [itex]X[/itex], the empty map is an injection of [itex]\emptyset[/itex] into that set, so [itex]|\emptyset |\leq |X|[/itex]
 
  • #7
Hm, why injection?? You meant bijection, didn't you? You need it to show "not larger", so we biject a set X onto a subset Z of Y to show that cardX<=cardY (since cardX=cardZ)
Nevertheless I think I got it. I forgot about the empty map.
So we can map the empty set onto a subset of any set X using the empty map. Thus we get the inequality.
Although we don't need the inequality, do we? We simply biject the empty set onto the set of natural numbers by the empty map. It does the whole job, doesn't it?! Hence, countability!
 
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  • #8
symplectic_manifold said:
Although we don't need the inequality, do we? We simply biject the empty set onto the set of natural numbers by the empty map.

You seem to be confused. There is no bijection from the empty set to any set that isn't the empty set, including the natural numbers.
 
  • #9
CrankFan said:
You seem to be confused. There is no bijection from the empty set to any set that isn't the empty set, including the natural numbers.

So is the empty map useless, obscure?
But how can we speak about cardinality and countability then??
OK, Guys I seem to need more info on this matter. Give me some links or references, please! :rolleyes:
 
  • #10
A bijection between two sets implies they have exactly the same cardinality. The empty set has cardinality zero, while any non-empty set has positive cardinality. That doesn't make the empty set or the empty map useless, obscure, etc. For some references, check out these wiki sites:
http://en.wikipedia.org/wiki/Cardinality
http://en.wikipedia.org/wiki/Empty_set

- edit -
Notably,
We say that two sets A and B have the same cardinality if there exists a bijection, i.e. an injective and surjective function, from A to B. [...] We say that a set A has cardinality greater than or equal to the cardinality of B (and B has cardinality less than or equal to the cardinality of A) if there exists an injective function from B into A. We say that A has cardinality strictly greater than the cardinality of B if A has cardinality greater than or equal to the cardinality of B, but A and B do not have the same cardinality, i.e. if there is an injective function from B to A but no bijective function from A to B.
 
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  • #11
Well, for any set LaTeX graphic is being generated. Reload this page in a moment., the empty map is an injection of LaTeX graphic is being generated. Reload this page in a moment. into that set, so LaTeX graphic is being generated. Reload this page in a moment.

OK, NateTG thought in terms of an injection into a set, while I thought about a bijection onto a subset of a given set.

There is no bijection from the empty set to any set that isn't the empty set, including the natural numbers.

So the empty map is always an injection, right.

Nimz, if I'm not mistaken from the quotation follows that the strict inequality in NateTG's post is justified too!

What about a map from the empty set that goes to the empty set. Is it defined?

And how is countability of the empty set explained in the end...by "every finite set is countable"?
 
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  • #12
symplectic_manifold said:
Nimz, if I'm not mistaken from the quotation follows that the strict inequality in NateTG's post is justified too!

In the context of my previous post, [itex]X[/itex] is an arbitrary set. Since [itex]X = \emptyset[/tex] is a possibility, the inequality cannot be strict.

What about a map from the empty set that goes to the empty set. Is it defined?

Yes. The only possible function from the empty set to any other set is the empty function, and in the usual formalism, it's defined.

(A brief digression.)
Formally a function [itex]F:A \rightarrow B[/itex] is a set of ordered pairs [itex]F=\{(a,b)\}[/itex] where [itex]a \in A[/itex] and [itex]b \in B[/itex] and [itex]\forall a \in A[/itex] there is exactly one [itex]b \in B[/itex] such that [itex](a,b) \in F[/itex].

If [itex]A = \emptyset[/itex] then the only possible function is [itex]F=\emptyset[/itex].

And how is countability of the empty set explained in the end...by "every finite set is countable"?

Here's a proof that the cardinality of any finite set is less than or equal to the cardinality of the natural numbers:
Proof by induction on the cardinality of the finite set:
Case 0:
The cardinality of the finite set is [itex]0[/itex].
The finite set is [itex]\emptyset[/itex] and the emtpy function is suffficient.

Case N:
The cardinality of the finite set is [itex]|N| > 0[/itex].
Clearly there is some element [itex]x \in N[/itex]. Since [itex]N[/itex] is finite we have [itex]|N|-1 = |N - \{x\}|[/itex].
By the inductive hypothesis we know that there is some injective function [itex]F_{N - \{x\}}:(N-\{x\}) \rightarrow \mathbb{N}[/itex], so we can construct
[tex]F_N:N \rightarrow \mathbb{N}[/tex]
as follows:
[tex]F_N(n)=\left \begin{array}{cc}1&\mbox{ if }$n=x$\\F_{N - \{x\}}(n)+1&\mbox{otherwise}\end{array}\right[/tex]
 
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  • #13
symplectic_manifold said:
OK, Guys I seem to need more info on this matter. Give me some links or references, please! :rolleyes:

http://www.math.uchicago.edu/~mileti/teaching/math278/settheory.pdf [Broken]
 
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1. What are countability questions for real numbers?

Countability questions for real numbers refer to the concept of determining whether a set of real numbers is countable or uncountable. A set is countable if it is possible to assign a unique natural number to each element in the set, while a set is uncountable if there is no way to assign a unique natural number to each element.

2. How do I determine if a set of real numbers is countable?

To determine if a set of real numbers is countable, you can use the following methods:

  • Listing method: List out all the elements in the set and see if you can assign a unique natural number to each element. If you can, then the set is countable.
  • Bijection method: Use a one-to-one correspondence between the set and the natural numbers. If such a correspondence exists, then the set is countable.
  • Diagonalization method: Use Cantor's diagonal argument to show that the set is uncountable. If you cannot find a one-to-one correspondence between the set and the natural numbers, then the set is uncountable.

3. Are all subsets of real numbers countable?

No, not all subsets of real numbers are countable. Some subsets, such as the set of all real numbers between 0 and 1, are uncountable. This is because there are an infinite number of real numbers between 0 and 1, making it impossible to assign a unique natural number to each one.

4. Why is it important to determine if a set of real numbers is countable?

Countability of a set of real numbers has important implications in mathematics. Countable sets have a well-defined order and can be easily manipulated and studied, while uncountable sets can be more complex and difficult to work with. Additionally, the concept of countability is closely tied to the concept of infinity, which has many applications in various branches of mathematics.

5. Is there a connection between countability and the size of a set of real numbers?

Yes, there is a connection between countability and the size of a set of real numbers. Countable sets are considered to be smaller than uncountable sets, as there are more elements in an uncountable set than in a countable set. This relationship is further explored through the concept of cardinality, which is a measure of the size of a set.

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