1. ## Countable set

hello.

I cant seem to prove the following:

$A,B$ are two different sets such that $A \subset B$
prove that if there exists an injective function $f:A\to B$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap)

can someone give me some hints?
thanks.

2. Originally Posted by parallel
hello.

I cant seem to prove the following:

$A,B$ are two different sets such that $A \subset B$
prove that if there exists an injective function $f:A\to B$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap)

can someone give me some hints?
thanks.
There's something wrong here. First of all there is nothing in your statement to force A to be at least countably infinite, which is a requirement for C to be countably infinite.

For a counter-example let A = {1, 2, 3} and B = {1, 2, 3, 4, 5, 6, 7}. Define an injection $f: A \to B: a \mapsto a$. Then $A \subset B$ and f is an injection, but there exists no set $C \subset A$ that is countably infinite.

-Dan

3. I'm sorry,I typed it incorrectly

should be:

prove that if there exists an injective function $f:B\to A$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap).

thanks

4. Originally Posted by parallel
I'm sorry,I typed it incorrectly

should be:

prove that if there exists an injective function $f:B\to A$.
then there exists a countable infinite set $C \subset A$
(C and A can overlap).

thanks
Which is again not true if both these sets are finite.

5. o.k so I think we can assume they are infinite(I think it's obvious),although I dont even have a clew,how to even start proving this.

6. If we were given that B is infinite then the statement is true.
We can prove that any infinite set has a countablely infinite subset call it $
D \subseteq B$
.
The subset of A you want is $\overrightarrow f (D)$, that is the image of D under f.

7. If $B$ is infinite this implies that $A$ is also infinite and at least as large as $B$ because of the injective map. Now, the property of the integers say there are contained up to cardinality in any infinite set, thus, there exist an injection
$i:\mathbb{Z}\to C$
Then the image of the function,
$i[\mathbb{Z}]\subseteq C$
(I hope that is what you mean by overlap).

8. I dont understand it.

I know that inorder to prove a set is countable,I need to show that there exists an injective function from this set to N(positive integers)

thanks

9. Originally Posted by parallel
I dont understand it.

I know that inorder to prove a set is countable,I need to show that there exists an injective function from this set to N(positive integers)

thanks
I assume that you are told $B$ is an infinite set.

1) $A$ is an infinite set also because the injection from $f:B\to A$ implies that $|B|\leq |A|$ in cardinality. So it must be infinite becuase it is larger than another infinite set (it cannot be finite because any infinite set is larger than any finite set).

2)Therefore, there exists an injection map $i: \mathbb{Z}\to A$. Because $|\mathbb{Z}|$ is the smallest possible infinite set which implies its size is contained in any infinite set. Thus, we state that in terms of an injection (since injection shows that one set is less than another in cardinality).

3)The image of the injection $i [ \mathbb{Z}]$ (image of a function, you should know what that is) is an infinite set that is the size of the integers and is contained in $A$. Thus, $i[\mathbb{Z}]\subset A$ which proves what you were asking.

10. Originally Posted by ThePerfectHacker
1) $A$ is an infinite set also because the injection from.
While that is perfectually true, I think that is the point of this problem.
In other words, I think that is what the student is asked to prove.

11. Thank you very much Plato.

12. Originally Posted by Plato
While that is perfectually true, I think that is the point of this problem.
In other words, I think that is what the student is asked to prove.
I can prove that! If that is what he wants.

I shall show there cannot exist a surjection,
$s: F\to I$
where $F$ is finite and $I$ is infinite.

Assume there is one.
Then since $I$ is infinite $\exists I' \subset I, |I'|=|I|$ (definition of infinity).

Then, the inverse image,
$s^{-1}[I']$ is a proper subset of $F$ (because that set was proper in the infinite set, the inverse image preserve this). But since $|I|=|I'|$ we have $|s^{-1}[I']|=|F|$. Thus, there exists a proper subset of a finite set having the same cardinality which is impossible by the definition of what finiteness is.

13. This is another case where the uniformity of mathematical definitions fails.
In the above, the definition of ‘infinite set’ was used.
Well which definition. This makes hard to help with knowing the text being followed.

There are at least three or maybe four popular definitions:
A set is infinite if it is not finite. A finite set is equipotent to a natural number.
A set is infinite if it is equipotent to a proper subset of itself (called Dedekind infinite).
A set is infinite if it contains a copy of the natural numbers.

I suspect that the purpose of the question was to show that “If an infinite set is mapped injectively of another set then the final set must also be infinite.” From the wording it is difficult to know what definition is being used. In any case my first response works.

14. Originally Posted by Plato
This is another case where the uniformity of mathematical definitions fails.
In the above, the definition of ‘infinite set’ was used.
Well which definition. This makes hard to help with knowing the text being followed.

There are at least three or maybe four popular definitions:
A set is infinite if it is not finite. A finite set is equipotent to a natural number.
A set is infinite if it is equipotent to a proper subset of itself (called Dedekind infinite).
A set is infinite if it contains a copy of the natural numbers.

I suspect that the purpose of the question was to show that “If an infinite set is mapped injectively of another set then the final set must also be infinite.” From the wording it is difficult to know what definition is being used. In any case my first response works.
But all of these definitions are equivalent!
Makes no difference what to use.
(I am only familar with Dedekind infinite).

15. Originally Posted by ThePerfectHacker
But all of these definitions are equivalent!
Makes no difference what to use.
But the definition in use changes the approach to the proof.
That was my point.