MHB Proving that a subset of a countable set is countable

  • Thread starter Thread starter Mathmellow
  • Start date Start date
  • Tags Tags
    Set
Click For Summary
To prove that any subset of a countable set is either finite or countable, one can start by considering a subset V of a countable set S. If V is finite, the proof is straightforward since finite subsets of countably infinite sets exist. In the case where V is infinite, a bijection can be established by defining a mapping from the natural numbers to the elements of V, specifically by selecting the i-th smallest element of V. This demonstrates that any infinite subset of a countable set is also countable. The discussion emphasizes the importance of constructing a bijection to validate the countability of subsets.
Mathmellow
Messages
7
Reaction score
0
I am trying to prove that any subset of a countable set is either finite or countable.

I know that a set $$S$$ is countable if there exists a bijection that takes S to $$\Bbb{N}$$. My first thought was to consider the subset $$V$$ of $$S$$. If $$V$$ is finite we are done, since we can always construct a finite subset of a countably infinite set.
So I guess in the case where $$V$$ is infinite we want to prove that there is a bijection $$\beta: V\to\Bbb{N}$$. However, I am not sure how to do this.

I would really appreciate it if someone could help me with this, so I can feel more comfortable with these types of problems in the future!
 
Physics news on Phys.org
Mathmellow said:
I am trying to prove that any subset of a countable set is either finite or countable.

I know that a set $$S$$ is countable if there exists a bijection that takes S to $$\Bbb{N}$$. My first thought was to consider the subset $$V$$ of $$S$$. If $$V$$ is finite we are done, since we can always construct a finite subset of a countably infinite set.
So I guess in the case where $$V$$ is infinite we want to prove that there is a bijection $$\beta: V\to\Bbb{N}$$. However, I am not sure how to do this.

I would really appreciate it if someone could help me with this, so I can feel more comfortable with these types of problems in the future!

One just has to show that any infinite subset of $\mathbb N$ is in bijection with $\mathbb N$.

Let $A\subseteq \mathbb N$ be infinite. Define a map $f:\mathbb N\to A$ by declaring $f(i)$ to be the $i$-th smallest element of $A$. Then $f$ is a bijection.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
2K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K