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The wikipedia page on the cardinal numbers says that $\aleph_0$, the cardinality of the set of natural numbers, is the smallest transfinite number. It doesn't provide a proof. Similarly, this page makes the same assertion, again without a proof.

How does one prove there is no smaller transfinite number? Equivalently (I think), why is there no smaller infinite set than the natural numbers?

statusfailed
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4 Answers4

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This is a consequence of the following theorem:

Suppose that $A$ is a set of integers, then either $A$ is finite, or $|A|=|\Bbb N|$.

Since we define $\aleph_0$ to be the cardinality of $\Bbb N$, this means that every infinite subset of a set of size $\aleph_0$ is itself of size $\aleph_0$, and so there cannot be a smaller infinite cardinal.

Note that the above proves that $\aleph_0$ is a minimal element of the infinite cardinals. There is no smaller. To prove that it is in fact the smallest of the infinite cardinals we need to use some other set theoretical assumptions (e.g. every two cardinals are comparable) which are commonly assumed throughout mathematics nowadays.

The proof of the aforementioned theorem is simple, by the way. Suppose that $A$ is infinite, then the map $a\mapsto |\{a'\in A\mid a'<a\}|$ is a bijection between $A$ and $\Bbb N$. The proof of that is by induction.

Asaf Karagila
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    Curious that you did not mention the key use of the axiom of choice... – Andrés E. Caicedo Oct 06 '13 at 21:16
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    @Andres: I thought about it, but I feel that it might cause more damage than help to someone inexperienced with infinite sets. I'm still tempted to make that edit. – Asaf Karagila Oct 06 '13 at 21:17
  • As a parenthetical remark, at least. – Andrés E. Caicedo Oct 06 '13 at 21:21
  • When you're right, you're right. :-) – Asaf Karagila Oct 06 '13 at 21:24
  • I found a related question here. Is the "every two cardinals are comparable" assumption the same as $2^{\aleph_\alpha} = \aleph_{\alpha+1}$? How does it relate to the axiom of choice? Further reading would be as welcome as a concrete answer :-) – statusfailed Oct 06 '13 at 22:43
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    @statusfailed: The statement $2^{\aleph_\alpha}=\aleph_{\alpha+1}$ is known as the Generalized Continuum Hypothesis, or GCH. It implies the axiom of choice, which is equivalent to the assertion "every two cardinals are comparable". But GCH is in fact stronger than the axiom of choice, and it is possible that GCH fails and the axiom of choice holds. In order to have that $\aleph_0$ is indeed the smallest infinite cardinal we need a very small fragment of the axiom of choice to hold, but it is all very technical. My recommendation is to forget it for now. Just remember it requires some choice. – Asaf Karagila Oct 06 '13 at 22:46
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    It would be good to explicitly mention AC, even though the equivalent statement that all cardinals are comparable has already been mentioned as "commonly assumed". The reason is that, when people make this assumption, they are almost always thinking of AC, not comparability of cardinals. – Andreas Blass Aug 02 '14 at 16:19
  • What I totally object is using interchargably "cardinality" and size regarding infinite sets. Cantor's cardinality IS NOT size. – Anixx Feb 17 '21 at 20:50
  • @Anixx: This opinion is contrary to the standard convention in set theory. And with that, there is nothing more to discuss. – Asaf Karagila Feb 17 '21 at 21:28
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One property of cardinals says that if there is an injection $f:A \to B$ then the cardinality of $A$ is less than or equal to the cardinality of $B$. If $B$ is infinite, then you can choose any $b_1 \in B$ and define $f(1) = b_1$, and then choose different $b_2 \in B$ so that $f(2) = b_2$, and so forth and you will get an injection $f:{\mathbb N} \to B$ if $B$ is infinite (because for every $n$, you have infinitely many choices left in $B$ for $f(n)$). So $|B| \geq |{\mathbb N}|$ if $B$ is infinite, thus $| \mathbb{N}|$ is the smallest infinite cardinal.

user2566092
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$\Bbb N$ is the union of all finite ordinals $[n]:=\{0,1,2,\dots,n-1\}$.

Hence, if a set $A$ is not finite, then each $[n]$ embeds into it. In particular, we can define embeddings $f_n:[n]\hookrightarrow A$ on top of each other, i.e. satisfying $f_n(k)=f_{n-1}(k)$ for all $k<n-1$. But that altogether (considering $\bigcup_nf_n$) gives an embedding of $\Bbb N$ into $A$, so that $\aleph_0\le |A|$.

Berci
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Well a set $M$ is infinite if for every finite subset $U \subsetneq M$ there exists a $x \in M, x\notin U$. Consequently, you will be able to construct a sequence $(x_n)_{n \in \mathbb{N}} $ of distinct elements in $M$. Therefore $|M| \ge \aleph_0 $

Hyperplane
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    This is not true as written. Replace $\subsetneq$ with $\subseteq$, and it will be fine now. – Andrés E. Caicedo Oct 06 '13 at 21:30
  • @AndrésE.Caicedo I can not get why we should replace $\subsetneq$ with $\subseteq$. Could you please explain your statement? From my point of view, $U$ is finite subset of $M$ and $M$ is infinite, thus $U \neq M$ and consequently $U \subsetneq M$ is justified. – Akira Sep 09 '18 at 10:36
  • @LeAnh You are trying to prove $M$ is infinite. You cannot assume it to begin with. The statement, as written, is true of some finite sets $M$. – Andrés E. Caicedo Sep 09 '18 at 12:30
  • @AndrésE.Caicedo I got your point. Is Hyperplane's answer correct? If so, why did you write This is not true as written.... I think Hyperplane wrote $U \subsetneq M$ since: if $U=M$ then there does NOT exist $x \in M, x \notin U$. – Akira Sep 09 '18 at 12:52
  • @LeAnh No, the answer is not correct. That is why I said it is not true and why I said there are finite sets satisfying the statement the answer claims characterizes infinity. – Andrés E. Caicedo Sep 09 '18 at 13:17
  • @AndrésE.Caicedo Oh I see now: For every subset $U \subsetneq M$, we always have $\exists x \in M, x \notin U$ by the definition of $\subsetneq$. So Hyperplane just restated the property of $\subsetneq$. But I don't understand why you wrote Replace $\subsetneq $ with $\subseteq $, and it will be fine now. Please explain! – Akira Sep 09 '18 at 13:33
  • @LeAnh You should figure it out on your own, it is more productive. – Andrés E. Caicedo Sep 09 '18 at 13:34
  • @AndrésE.Caicedo please check if my understanding is correct! I figured it out, but I very need your confirm. – Akira Sep 09 '18 at 14:02