Is natural numbers set $\mathbb N$ infinite set?

A set with uncountable number of elements is called an infinite set.

Is that the set of all natural numbers, $\Bbb N=\text{{$1,2,3,\ldots$}}$ infinite set?

As far i know $\Bbb N$ is "countably" infinite or denumerable.

That is, i can't say $\Bbb N$ is "uncountable".

Then how is $\Bbb N$ an infinite set?

• The word "countable" doesn't mean that you can count it in your lifetime. It means that it is the "same size" (or smaller than) as $\mathbb{N}$. In particular $\mathbb{N}$ itself is countable. – vadim123 Dec 21 '14 at 4:10
• A set with an uncountable number of elements is an infinite set, but infinite sets need not be uncountable. – user_of_math Dec 21 '14 at 4:11

Sets can be grouped into three sizes: finite, countably infinite, and uncountably infinite. The natural numbers are countably infinite, and thus infinite. In fact, the very definition of being countably infinite is based on the natural numbers: a set is countably infinite if and only if there is a bijection between that set and the natural numbers.

• If $S=\text{{$\ldots,-3,-2,-1$}}$,then is it countably infinite?as negative numbers are not counting number?and if the function is $f(s)=s^3$.So that there is no bijection between $S$ and $\Bbb N$ – user 31466 Dec 26 '14 at 9:27
• @Leaf: The negative integers are countable, yes. Here's one function giving a bijection between them and the natural numbers: $f(n)=-n.$ Of course not all functions are bijections, you only need one. – Charles Dec 26 '14 at 18:14

Finite set is a set that for some natural number $k$, there is a bijection between the set and $\{1,\ldots,k\}$.

An infinite set is a set which is not finite. $\Bbb N$ is not finite, so it is infinite.

Countably infinite set is a set that has a bijection between itself and $\Bbb N$. An uncountable set is a[n infinite] set which is not countable. So an infinite set can be countably infinite or uncountable.

(If it bothers you that the definition of finiteness appeals to the natural numbers there are set theoretical definitions which makes no such appeal. For example Tarski's definition $A$ is finite if and only if every non-empty $U\subseteq\mathcal P(A)$ has a $\subseteq$-minimal element; or Dedekind's definition $A$ is finite if and only if every injection from $A$ to itself is a bijection.

Interestingly the equivalence between the definition in my first line, and Tarski's requires no use of the axiom of choice, whereas the equivalence with Dedekind's definition does require us to use the axiom of choice.)

• why do i need the set to be bijective to be finite? Can't only the surjectve property fulfill the criteria of being finite?That is , do i need the set to be injective to be finite? – user 31466 Dec 21 '14 at 5:42
• Every non-empty set has a surjection onto $\{1\}$. If you mean a surjection from $\{1,\ldots, k\}$, it turns out to be equivalent. But using bijection is the definition used commonly. – Asaf Karagila Dec 21 '14 at 7:22

Finite/infinite and countable/uncountable are in different level when we try to describe "how large" a set is. $$\begin{equation} "How\ large"\ a\ Set\ is \left\{ \begin{array}{ll} finite&\\ infinite& \left\{ \begin{array}{ll} countable&\\ uncountable& \end{array} \right. \end{array} \right. \end{equation}$$ Which means, if we try to describe how large a set is, firstly, we should see whether it is finite or infinite.

For finite sets, we can easily say which one is larger by comparing how many elements they contained. However, for the sets with infinite elements, one may not compare them by the usual way.

But somehow, we still want to know which one is larger. Then, we use countable and uncountable to describe them.

For example, $\mathbb{N}$ and $\mathbb{R}$ are both infinite sets, but it is obvious that $\mathbb{R}$ contains more than $\mathbb{N}$. Since there are much real numbers between any two integers, but this doesn't hold for integers, like between 2.1 and 2.2.

In fact, we use "Cardinality" to say how large a set is, and the Cardinality of $\mathbb{N}$ is denoted by $\aleph_0$, and $\mathbb{R}$ is equal to $\aleph_1$, if you believe continuum hypothesis.

• The cardinality of $\Bbb R$ is not defined as $\aleph_1$. – Asaf Karagila Dec 21 '14 at 7:25
• It is still not true that $\aleph_1$ denotes the cardinality of the reals. All that CH tells you is that it is equal, but the definitions for either cardinal are still the correct definitions, and not as you are saying that here. In either case, the fact that you believe in something not entirely widely accepted doesn't mean that you have to present your belief as true. This is called dishonesty. – Asaf Karagila Dec 21 '14 at 7:40
• Open a set theory book. Is that enough proof? – Asaf Karagila Dec 21 '14 at 8:08
• I do have my own opinion but (1) I don't forcefeed them to other people online, instead I prefer to point out possible options; (2) Did it occur to you that my opinion might be that Platonism and definitive truth values are overrated? In any case, I am talking about the fact that your answer presents $\aleph_1$ AS THE DEFINITION of the cardinality of the continuum. This is not true, and never was true. Not even in the context of believing any type of continuum hypothesis. – Asaf Karagila Dec 21 '14 at 8:39
• I have told you, you just insist otherwise. The cardinality of $\Bbb R$ is not denoted by $\aleph_1$. It is equal to $\aleph_1$ UNDER ADDITIONAL HYPOTHESIS THAT YOU BELIEVE BUT OTHER PEOPLE HAVE NOT ACCEPTED EXCLUSIVELY. – Asaf Karagila Dec 21 '14 at 9:04