Proof: If S is denumerable, then S is equinumerous w/ a proper subset

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Homework Help Overview

The discussion revolves around the concept of denumerability in set theory, specifically addressing the properties of denumerable sets and their relationship with natural numbers. Participants are exploring definitions and implications of being denumerable, countable, and infinite.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the definitions of denumerable and countable sets, particularly whether a denumerable set must be infinite. There is discussion about the implications of bijections with the natural numbers and the classification of sets as finite or infinite.

Discussion Status

Some participants have provided clarifications on definitions and relationships between different types of sets. There is ongoing exploration of the terminology and its implications, with no explicit consensus reached on all points raised.

Contextual Notes

Participants are navigating potentially confusing terminology regarding countable and uncountable sets, and there are differing interpretations of definitions from the textbook. The discussion includes the need for clarity on the nature of rational numbers and their countability.

playboy
The question is:

" Prove that if S is denumerable, then S is equinumerous with a proper subset of itself"

To begin, I am confused with the term denumerable because the textbook gives some diagram which is throwing me off. So can somebody clarify this for me:

A set S is denumerable if S and N (natural numbers) are equinumerous. That is, their is a BIJECTIVE function f: N---->S

The textbook says N is INFINITE. So if N is infite, that means that S HAS TO BE INFINTE if S is denumerable?

Now I'm reading the books diagram like this:

Countable sets are FINITE OR DENUMERABLE
Infinite sets are UNCOUNTABLE OR DENUMERABLE

So is this true: "if a set S is denumerable, then it HAS TO BE INFINITE?"
 
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If a set is in bijection with N is it infinite? Of course.
 
matt grime said:
If a set is in bijection with N is it infinite? Of course.

If a set is Denumerable, then it is in bijection with N. That means it has to be infite also. That is what i thought. So every denumerable set is infinite.

But the textbook says this:

Countable sets are FINITE OR DENUMERABLE
Infinite sets are UNCOUNTABLE OR DENUMERABLE

Is the above true? If an infine set is Denumerable, then how can a countable set be Denumerable?
 
The *definition* of countable is that it is either finite or denumerable. It is just a definition.

Definition: a set is countable if it can be placed in bijection with some subset (including N itself) of the natural numbers, otherwise it is uncountable.
 
Maybe it would help to look at the terms for the three kinds of sets this way.
Code:
1)   [b]finite[/b]        not denumerable        countable 

2) not finite       [b]denumerable[/b]           countable

3) not finite      not denumerable     [b]not countable[/b]
Your book let's not finite = infinite and not countable = uncountable.

But just as a warning, not everyone uses these terms the same way -- and it's somewhat confusing. If you learn yours, you can figure out what other people are talking about. One other use is
Code:
1)  [b]finite[/b]       not countable    not uncountable

2) not finite      [b]countable[/b]      not uncountable

3) not finite    not countable     [b]uncountable[/b]
This way, not countableuncountable and not uncountable = at most countable. I prefer the way your book does it because it fits better with the ways negation normally works in English (with not, un-, in-) and doesn't unnecessarily complicate things, e.g., by making something both not countable and at most countable.
 
Last edited:
Thank you honestrosewater for that post, it helps.

My TA gave examples of countable numbers
Q = {m/n, m,n in Z}

I was trying to tell him that "m,n in Z" can't make it countable?

I say this because a set is countable if f:N--->s is surjective or if f:S---->N is injective.

so shouldn't it be Q = {m/n, m,n in N and n cannot be 0}
 
Well, modulo the issue that you can't have zero as a denominator, I don't get you at all.

Firstly Q means the rational numbers, and you have to have negative rationals.

Secondly, why does having Z bother you? Z obviously has a bijection with the naturals, or an injection into the naturals, so there is no problem at all. Q is simply (a subset of) the product of two countable sets, and is hence countable.
 

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