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I had the following idea but it doesn't look like a proper proof:

Let's say $\mathbb N$ is finite. Now consider an arbitrary number $k$ in $\mathbb N$ such that it is the biggest number in the set $\mathbb N$. Now let's consider the number $k+1$. $k+1$ is in $\mathbb N$ and is greater than $k$. Hence we can conclude that $\mathbb N$ has no greatest element because I can always add $1$, and thus is infinite.

EDIT 1: I just want to prove it using basic set theory , without using cardinality or other complicated stuff.

EDIT 2: I want to know if this is a formal proof that proves that the natural number set is infinite, supposed we do not know anything about the natural number. In other words i want to know if this can work as stand alone proof.

EDIT 3: I can use the following : If the size of the natural numbers set $\mathbb{N}$ is less than or equal to the size of a set $F$ then $F$ is infinite. $\mathbb{N}$ is less than or equal to $F$ iff there is an injective function from $\mathbb{N}$ to $F$

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    It is correct. If it would be finite, you would be able to find a largest element, which is not possible as you proved. – 2017-02-06
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    What is the definition of infinite that you are using? Typically in set theory, a set is called infinite if it is in bijection with a proper subset of itself. – 2017-02-06
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    @Dylan it's also to define infinite as "non-finite", where finite means "in bijection with $\{1,\dots,n\}$ for some $n$" – 2017-02-06
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    @Dylan: No, that is not the typical definition in set theory. What you define is called _Dedekind-finiteness_ -- it is equivalent to ordinary "finite" if the Axiom of Choice is assumed, but not without it. – 2017-02-06
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    Your approach would be correct *if you have previously proved* that a finite set must have a greatest element. – 2017-02-06
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    Can you say what exactly doesn't seem "proper" to you about this proof? – 2017-02-06
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    @EricWofsey I want to know if this is a formal proof that proves that the natural number set is infinite, supposed we do not know anything about the natural number. In other words i want to know if this can work as stand alone proof. – 2017-02-06
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    @Alcakram: from some of the comments, the answer is clearly no. – 2017-02-06
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    You can't prove anything at all about the natural numbers if you don't know _something_ about the natural numbers... – 2017-02-06
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    @YvesDaoust why? – 2017-02-06
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    You say you want to avoid using "cardinality or other complicated stuff", but in that case, what is your definition of "finite"? For that matter, what is your definition of "the set of natural numbers"? – 2017-02-06
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    @Alcakram: because of $k$. – 2017-02-06
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    "... without using cardinality ... ." How do we say how many elements are in a set (in particular, that there are infintely many) without considering its cardinality? We could avoid using the _word_ "cardinality," just as we can avoid using the word "number" when counting from 1 to 100, but you can't say anything about the size of a set without using cardinality any more than you can count to 100 without using numbers. – 2017-02-06
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    @EricWofsey We can use the following : If the size of the natural numbers set N is less than or equal to the size of a set F then F is infinite. N is less than or equal to F iff there is an injective function from N to F – 2017-02-06
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    @DavidK We can use the following : If the size of the natural numbers set N is less than or equal to the size of a set F then F is infinite. N is less than or equal to F iff there is an injective function from N to F – 2017-02-06
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    Well, then, there is an injective function ($x\mapsto x$) from the natural numbers N to the natural numbers N, so the size of N is less than or equal to the size of N, so N is infinite. Done. (That's a use of cardinality, by the way.) – 2017-02-06
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    @DavidK Oh yes I think that is the answer that I was looking for. Thanks – 2017-02-06

3 Answers 3

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This argument works, though depending on what facts you are allowed to assume there may be some details that need to be filled in. In particular, you should probably say more about how you know that any finite subset of $\mathbb{N}$ has a greatest element (so if $\mathbb{N}$ were finite, then it would have a greatest element). For instance, if your definition of "finite" is "in bijection with $\{m\in\mathbb{N}:m

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As others have pointed out, you're assuming the lemma that every totally ordered finite set has a maximal element. From your comments I would guess that that's not an okay lemma to assume, so you'll either need to prove it or come up with a different proof.

There's a very simple proof that follows from your third edit

EDIT 3: I can use the following : If the size of the natural numbers set $\mathbb{N}$ is less than or equal to the size of a set $F$ then $F$ is infinite. $\mathbb{N}$ is less than or equal to $F$ iff there is an injective function from $\mathbb{N}$ to $F$

This gives rise to the easiest proof based on your current knowledge base. Observe that $f:\mathbb{N}\to\mathbb{N}$ given by $x\to x$ is an injective function from $\mathbb{N}$ to $\mathbb{N}$. Therefore by letting $F=\mathbb{N}$ in your lemma, you find out that $\mathbb{N}$ is infinite.

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Yes, this is a working proof -- assuming you already know (that is, have proved) that a finite totally ordered set has a maximal element.