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We define $A$ to be a finite set if there is a bijection between $A$ and a set of the form $\{0,\ldots,n-1\}$ for some $n\in\mathbb N$.

How can we prove that a subset of a finite set is finite? It is of course sufficient to show that for a subset of $\{0,\ldots,n-1\}$. But how do I do that?

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    $\begingroup$ I wrote this answer to a now-deleted question. I realized that I don't recall many proofs of the pigeonhole principle (anywhere) which are longer than "obviously true.", so I decided to post the question with an answer. $\endgroup$ – Asaf Karagila Nov 18 '12 at 1:40
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    $\begingroup$ I welcome other people to write answers as well. $\endgroup$ – Asaf Karagila Nov 18 '12 at 1:58
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The proof is essentially the pigeonhole principle, and it is proved by induction.

Let us denote $[n]=\{0,\ldots,n-1\}$. What we want to prove is in two parts:

  1. If $A\subseteq[n]$ then either $A=[n]$ or there is a bijection between $A$ and $[m]$ for some $m<n$.

  2. If $m<n$ then there is no bijection from $[n]$ into $[m]$. Equivalently we want to prove that if $f\colon[n]\to[n]$ is injective then it is a surjective. Also we may want to prove that if $f\colon[n]\to[m]$ then $f$ is not injective, or $f\colon[m]\to[n]$ then $f$ is not surjective.

The first part is not very difficult, we define by induction $f(0)=\min A$; and $f(k+1)=\min A\setminus\{f(0),\ldots,f(k)\}$. Either $f$ is a bijection between $A$ and $[n]$ or the induction had to stop before and $f$ is a bijection between $[m]$ and $A$ for some $m<n$. Note that this is not begging the question because $A$ is a set of natural numbers, and it is a subset of $[n]$ so it cannot contain more elements than $[n]$ (more in the actual sense, not in the sense of cardinality).

The second part is tricky because it seems so obvious. This is where the pigeonhole principle is actually proved.

We prove this part by induction. For $n=0$ this is obvious because $[0]=\varnothing$.

Assume that for $[n]$ it is true that if $f\colon[n]\to[n]$ is injective then it is surjective. We want to show this is also true for $[n+1]$.

Let $f\colon[n+1]\to[n+1]$ be an injective function. There are two cases:

  1. If $f(k)\neq n$ for all $k<n$ then restricting $f$ to $[n]$ is an injective function from $[n]$ into $[n]$. By the induction hypothesis we have to have that the restriction of $f$ to $[n]$ is surjective, therefore $f(n)\notin[n]$ (otherwise $f$ is not injective) and therefore $f(n)=n$ and so $f$ is surjective as well.

  2. If $f(k)=n$ for some $k<n$, by injectivity this $k$ is unique. We define $g$ as follows: $$g(x)\begin{cases} f(n) & x=k\\ n & x=n\\ f(x) & x\neq k,n\end{cases}$$ It follows that $g$ is also injective, and for all $k< n$ we have to have $g(k)\neq n$ (because $g(n)=n$ and $g$ is injective). Therefore by the previous case we know that $g$ is surjective. It follows that $f$ is also surjective, as wanted.


A word on the axiom of choice and finiteness. The above proof shows that finite sets are Dedekind-finite. There are other ways of defining finiteness, all which are true for finite sets, but may also be true for infinite sets. For example "every surjection is a bijection" might fail for infinite Dedekind-finite sets; or "every linearly ordered partition is finite"; etc.

Assuming the axiom of choice, or actually the axiom of countable choice, is enough to conclude, however, that all Dedekind-finite sets are finite. Therefore the above forms of finiteness are equivalent once more. (The reverse implication is false, by the way, it is consistent that the axiom of countable choice fails, but every Dedekind-finite set is finite.)

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    $\begingroup$ There is another proof in my mind. Am I allowed to write it here Sir? $\endgroup$ – mrs Nov 18 '12 at 2:07
  • $\begingroup$ @Babak: See my second comment on the question. :-) $\endgroup$ – Asaf Karagila Nov 18 '12 at 2:09
  • $\begingroup$ Techically, the question seems to ask only for 1. $\endgroup$ – Michael Greinecker Nov 18 '12 at 2:17
  • $\begingroup$ @Michael: True, but it simpler to prove this using th second point. At least if you want a full proof. :-) $\endgroup$ – Asaf Karagila Nov 18 '12 at 2:19
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    $\begingroup$ @FreeMind: It's a recursive definition. $f(0)=\min A$, etc. This is what is written exactly after that line you quote. The recursion is over $\{0,\ldots,n-1\}$, since $A$ is a subset of that. Either at some point we are finished, or $A$ equals $\{0,\ldots,n-1\}$. $\endgroup$ – Asaf Karagila Sep 22 '18 at 7:14
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Here is a slightly different proof of the pigeonhole principle:

Suppose $f:[n]\to[n]$ is an injection that is not surjective for $n>0$. We can assume without loss of generality that $n$ is not in the range of $f$. For if it is, we map the element that is mapped to $n$ to some element not in the range instead and get a new injection that is not surjective with $n$ not in the range. We now show that $f|[n-1]$ is injective, but not surjective. Trivially, the restriction of an injection is an injection. But $f(n)\in[n-1]$ is not in the range of $f|[n-1]$.

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  • $\begingroup$ Note that this argument also uses induction. $\endgroup$ – Asaf Karagila Nov 18 '12 at 2:20
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    $\begingroup$ @Asaf Of course, the natural numbers are essentially defined by an indctive property. $\endgroup$ – Michael Greinecker Nov 18 '12 at 2:22
  • $\begingroup$ Yes, I just made this remark because the induction is not very apparent in your proof, and one could think that the proof is not inductive. :-) $\endgroup$ – Asaf Karagila Nov 19 '12 at 0:11
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Let $I_n = \{0, 1, ... n-1\}$, and $B\subset A$. Then there exists an injection from $B\to I_n$ since there's an bijection from $A\to I_n$. Since there is such a $n$ there exists(*) a minimal $j$ such that there is an injection from $B\to I_j$ and if that were not an surjection there we could construct a injection from $B\to I_{j-1}$ (unless $j=0$, but then $I_j=\emptyset$ and the function is trivially a surjection).


If you instead use Dedekind-finite as a definition for finite. That is that every injective function on $A$ is surjective, then it becomes rather obvious.

If $B\subseteq A$ were not finite there would be a injective mapping from $B$ to $C\subsetneq B$ and we could expand this to $A$ by taking the identity map outside $B$ then we would have an injective mapping from $A$ to a proper subset of $A$ which means that $A$ is infinite which contradicts the assumption.


(*) This relies on that every non-empty set $X$ (of $j$ such that there is an injection $B\to I_j$) of natural numbers has a minimal element which can be proven by induction.

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  • $\begingroup$ It appears that the cut-to-the-chase approach is to just get the words injection, surjection, bijiection out as quickly as possible. $\endgroup$ – CopyPasteIt Aug 21 '17 at 12:39
  • $\begingroup$ Why is there a minimal $j$ like that? This assumes that $B$ is already finite. $\endgroup$ – Asaf Karagila Aug 21 '17 at 12:45
  • $\begingroup$ @AsafKaragila That can be proven by induction... $\endgroup$ – skyking Aug 21 '17 at 12:47
  • $\begingroup$ Obviously. I proved that by induction in my answer. And I agree it's fine to leave this gap. But one should point out this essential gap. What you have done is write "assume a subset of a finite set is finite, then a subset of a finite set is finite." $\endgroup$ – Asaf Karagila Aug 21 '17 at 12:49
  • $\begingroup$ @as I wonder now if I need to point out any gaps in my answer. I guess no response means I am OK. $\endgroup$ – CopyPasteIt Aug 21 '17 at 12:51
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Note: This is a rework of an effort I made over a year ago. It wasn't my intention but I wound up developing the fundamental background to understanding finite cardinals.

I do not think this proof uses induction.


We use the notation $[n] = \{0,1,\ldots,n-1\}$

Recall the following fact:

Proposition 1: Let $f: A \to B$ be a function and $\beta: B \to B$ be a bijective transformation of the target. Then $f$ is $\text{1:1}$ if and only if $\beta \circ f$ is $\text{1:1}$.

Proposition 2: If $\tau: [j] \to [k]$ is an injection, then $j \le k$.
Proof
For any injection $\sigma: [j] \to [k]$ let $S(\sigma) = max(\{s \, | \, \sigma(t) = t \text{ for } t \le s\}$. For the injection $\tau$, let

$\tag 1 T = max(\{S(\beta \circ \tau \, | \, \beta \text{ is a bijective transformation of } [k]\}$

The integer $T$ must be equal to $j-1$. Assume to get a contradiction that $T \lt j-1$ and select a transformation so that $S(\beta \circ \tau) = T$. Let $\beta \circ \tau (T+1) = u$, so that $u \gt T$. Let $\rho$ be the transformation that interchanges $u$ with $T + 1$. Then $S(\rho \circ \beta \circ \tau) \ge T+1$, which is absurd.

We have shown the existence of an insertion $\iota: [j] \to [k]$, so $j \le k$. $\quad \blacksquare$

Proposition 3: If there is a bijective correspondence $f$ between $[n]$ and a set $B$, then there is no injective mapping $g$ from $[n+1]$ into $B$.
Proof
Assume $g$ is injective and consider the function $f^{-1} \circ g$. By proposition 2, $n+1\le n$, but that doesn't make sense. $\quad \blacksquare$

Using the above theory it is easy to show the following theorem:

Theorem 4: If a set $B$ is finite then there exists a $C \in \mathbb N$ such that for $m \lt C$ there exists proper injections from $[m]$ into $B$, for $m = C$ there exist a bijection between $[C]$ and $B$, and for $m \gt C$ there exists no injections from $[m]$ into $B$.

The following is now also easy to prove:

Proposition 5: If $B$ is not a finite set, then for every $m \ge 0$ the set $[m]$ can be non-surjectively injected into $B$.

Proposition 6: If $A$ is a finite set and $B$ is a subset, then $B$ is finite.
Proof
Assume, to get a contradiction, that $B$ in not finite. Let $f: A \to [n]$ be a bijective mapping. By proposition 5, there is an injection $g:[n+1]\to B$. Since we can naturally insert $B$ into $A$, we have an injection from $[n+1]$ to $[n]$. By proposition 2, this is impossible. $\quad \blacksquare$

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