A, B are countable. P(s), power set, and f: A-> not always countable?

In summary, the power set P(s) and function f:A->B are not necessarily countable when sets A and B are countable. P(s) is only countable if A and B are finite, while f:A->B is countable as long as A and B are countable. The set of all functions f:A->B may not be countable due to the diagonal argument, and may be what the professor was referring to in the homework question.
  • #1
Tony11235
255
0
Suppose A and B are countable. Explain why P(s), power set, and f: A ->B are not necessarily countable.

P(s) is only countable if A and B are finite, am I am correct? Otherwise, the power set of an infinite set is not countable. As for f: A -> B, doesn't f have to be a bijection?
 
Physics news on Phys.org
  • #2
I assume you're regarding the function f as a set of ordered pairs. I would think that |f| = |A|, so if A is countable then so is f. I don't know what s is, but I believe that if s is finite, then P(s) is finite, and if s is infinite, then P(s) is uncountable.
 
  • #3
He may have meant the set of all functions f:A-->B is not necessarily countable. In that case it is true because, for example, you can represent each function from the positive integers to the positive integers as an infinite sequence of integers, and you can use a diagonal argument to say that the set of all sequences of integers is not countable. For the power set of A the integers, you can represent each subset as a bit string.
 
  • #4
There were initially two quetions but one of them is sort of obvious, but the question is, assumming that A and B are countable sets, under what conditions would f:A->B NOT be countable ?
 
Last edited:
  • #5
Do you really mean the function f:A-->B and not the set of all functions f:A-->B? As AKG said if you are only talking about the function, then it is always countable.
 
  • #6
0rthodontist said:
Do you really mean the function f:A-->B and not the set of all functions f:A-->B? As AKG said if you are only talking about the function, then it is always countable.

I was talking about the function from sets A to B. I would think that as long as A and B are countable sets, that f:A-->B would always be countable. So why would my professor ask the question, on a homework, "Suppose sets A and B are countable. Explain why P(s) and f:A-->B are not necessarily countable". Any idea?
 
  • #7
My only guess is that he was talking about the set of all functions f:A-->B. You should ask him.
 
  • #8
Tony11235 said:
Suppose A and B are countable. Explain why P(s), power set, and f: A ->B are not necessarily countable.

P(s) is only countable if A and B are finite, am I am correct? Otherwise, the power set of an infinite set is not countable. As for f: A -> B, doesn't f have to be a bijection?

BTW The set of integers is countable, but not finite. I think that's what's going on here...

-Dan
 

Related to A, B are countable. P(s), power set, and f: A-> not always countable?

1. What does it mean for A and B to be countable?

Countable sets are sets that can be put into a one-to-one correspondence with the set of natural numbers. This means that each element in the set can be counted and assigned a unique number.

2. What is P(s) or the power set?

The power set of a set s is the set of all subsets of s, including the empty set and the set itself. It is denoted by P(s) or 2^s.

3. What is the significance of f: A-> not always countable?

A function f from set A to set B is not always countable if there exists an element in set B that is not mapped to by any element in set A. This means that the cardinality of set B is greater than the cardinality of set A.

4. Can a power set be countable?

No, a power set cannot be countable. This is because the cardinality of the power set of a set with n elements is 2^n, which is always greater than the cardinality of the original set.

5. How do A, B and P(s) relate to each other?

A and B are countable sets, meaning they can be put into a one-to-one correspondence with the natural numbers. P(s) is the power set of set s, which contains all possible subsets of s. Depending on the cardinality of set s, P(s) may or may not be countable.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
23
Views
893
  • Calculus and Beyond Homework Help
Replies
1
Views
585
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
267
  • Topology and Analysis
Replies
2
Views
1K
  • Topology and Analysis
Replies
2
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
3K
Replies
3
Views
435
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
Back
Top