Subset of a finite set is finite

Question

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?

Answer 1

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.

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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.)

Answer 2

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 $f|[n-1]$.

Answer 3

For any $n\in N$ let $[n]$ represent the set $\{0,\ldots,n-1\}$.

Proposition: For all natural numbers $n$, all subsets of $[n]$ are finite.

We prove it by induction on $n$. For $n=0$ the only subset of $[0]$ is $\emptyset$ which is finite.
Now assume that all subsets of $[n]$ are finite. Let $A$ be a subset of $[n+1]$.
If $n \notin A$ then $A \subseteq [n]$ and $A$ is finite by the induction hypothesis.
If $n \in A$ then $A-\{n\}\subseteq [n]$ and is finite by the induction hypothesis, which means there is a bijection $f$ from $[k]$ to $A-\{n\}$ for some natural number $k$. Define a bijection $g \colon [k+1] \to A$ as follows:

$$ g(x)= \begin{cases} f(x), & \text{if $x<k$} \\ n, & \text{ if $x=k$} \end{cases} $$

Since there is a bijection from $[k+1]$ to $A$, $A$ is finite. Thus all subsets of $[n+1]$ are finite.

Answer 4

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).

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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.

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(*) 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.

Answer 5

For any $k \in \Bbb N$ let $[k]$ represent the set $\{0,\ldots,k-1\}$.

Lemma 1: If $A$ is a finite set then so is $A \cup \{b\}$.
Proof
Let $\sigma: A \xrightarrow{\sim} [n]$ be a bijective correspondence.
If $b \in A$ there is nothing to do.
Otherwise, define $\tau: A \cup \{b\} \to [n+1]$ as follows:

$\tau(x) = \left\{\begin{array}{lr} \;\;\;\sigma(x)\, \;\;\text{ |} & \text{for } x \in A\\ \;\;\;n \;\;\;\;\;\;\;\text{ |} & \text{for } x = b \end{array}\right\}$

It is elementary to check that $\tau$ is a bijection. $\quad \blacksquare$

Proposition 2: If $A \subset [n]$ then $A$ is a finite set.
Proof
We can assume that $A \ne \emptyset$.
For $0 \le m \le n$ define $A_m = A \cap [m]$ so that $A = A_n$.
Each $A_m$ with $m \lt n$ is either equal to $A_{m+1}$ or is equal to $A_{m+1} \setminus \{m\}$.
To get a contradiction, assume that $A$ is infinite.
By lemma 1 and step-by-step finite decremental induction, if $A_n$ is infinite then all the sets $A_m$ with $0 \le m \le n$ must also be infinite. But

$\quad A_0 = A \cap [0] = A \cap \emptyset = \emptyset$

and we've arrived at a contradiction. $\quad \blacksquare$

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Remark: We can also use lemma 1 to supply a direct proof that uses straightforward induction; for $n \in \Bbb N$, define the statement

$P(n): $ If a set $A$ can be put into bijective correspondence with $[n]$ and $B \subset A$ then there exist $m \in \Bbb N$ such that $B$ can be put into bijective correspondence with $[m]$.

Answer 6

I have run into this old question and was surprised that noone seemed to have said the following.

The actual definition of finite set is the following: A set $A$ is finite if every injection $A\rightarrow A$ is a bijection.

(Note: this definition does not require the set $\mathbb{N}$)

Now let $B$ a finite set and $A\subset B$.

Suppose that $A$ is not finite. Then, by definition there exists a function $f:A\rightarrow A$ that is injective but not surjective. Now define $F:B\rightarrow B$ as follows. $$ F(x)=\begin{cases} f(x) & \text{if $x\in A$}\\ x & \text{if $x\in B\setminus A$} \end{cases} $$ Clearly $F$ is injective but not surjective contradicting the finiteness of $B$.

Answer 7

Isn't this just a simple Induction?

We wan't to prove for every $n \in \mathbb{N}$: Every subset $A \subset \{k \in \mathbb{N}|0 \leq k \leq n-1\}$ is finite.

$n=0: A \subset \{k \in \mathbb{N}| 0 \leq k \leq -1\}= \emptyset$ implies $A =\emptyset$ which is finite (take the empty function as bijection).

$n \mapsto n+1:$ For $A \subset \{k \in \mathbb{N}| 0 \leq k \leq n\}$ we get that $A\setminus \{n\} \subset \{k \in \mathbb{N}| 0 \leq k \leq n-1\}$ is finite by induction hypothesis. So there exists a $m \in \mathbb{N}$ and a bijection $f: \{k \in \mathbb{N}| 0 \leq k \leq m-1\} \to A \setminus \{n\}$. We extend this function on $\{k \in \mathbb{N}| 0 \leq k \leq m\}$ by setting $f(m):= n$. This function remains bijective and hence $A$ is finite.

Answer 8

For any $k \in \Bbb N$ let $[k]$ represent the set $\{0,\ldots,k-1\}$.

Proposition 1: If $A$ is finite and $B \subset A$ then $B$ is finite.
Proof
To get a contradiction, take the minimal $n$ for which there exist sets $A$ and $B$ and a function $\sigma: [n] \to A$ such that $B \subset A$ and $B$ is not finite.
It must be true that $B \subsetneq A$ and so let $a \in A$ with $a \notin B$.
Let $j_a = \sigma^{-1}(a)$ and define

$\tau(x) = \left\{\begin{array}{lr} \;\sigma(x)\, \;\;\;\;\;\;\text{ |} & \text{for } x \lt j_a \\ \;\sigma(x+1) \; \text{ |} & \text{for } x \ge j_a \end{array}\right\}$

The function $\tau$ is a bijective correspondence between $[n-1]$ and $A \setminus \{a\}$; also, $B \subset A \setminus \{a\}$.
But then $n$ is not the minimal integer and we've reached a contradiciton. $\quad \blacksquare$

Answer 9

The set-up follows the definitions in

Hrbacek, Karel; Jech, Thomas, Introduction to set theory, Pure and Applied Mathematics, Marcel Dekker 220. New York, NY: Marcel Dekker (ISBN 0-8247-7915-0/hbk). ix, 291 p. (1999). ZBL1045.03521..

On p. 41: The set of natural numbers is the set $$\{x:x\in I\text{ for every inductive set I}\}.$$

Claim:

Every natural number $n$ is the set of all smaller natural numbers, i.e., $$n=\{m\in\mathbb{N}:m<n\}=\{0,1,\ldots,n-1\}.$$

Indeed, the claim is true for $n=0$ because $0$ is defined to be $\emptyset$. Assume the claim is true for $n$. Then, $$n+1=n\cup\{n\}=\{m\in\mathbb{N}:m<n\}\cup\{n\}=\{m\in\mathbb{N}:m<n+1\}.$$ The Induction Principle now asserts the truth of the claim for all $n\in\mathbb{N}$.

On p.65: Sets $A$ and $B$ are equipotent (have the same cardinality) if there is a bijective function $f:A\to B$. We denote this by $|A|=|B|$.

On p.69: A set $A$ is finite if it is equipotent to some natural number $n\in\mathbb{N}$. We then define $|A|=n$ and say that $A$ has $n$ elements.

On p.70 the authors use a version of the Recursion Theorem (Exercise 3.5, p.52 in the book) to prove what is asked by the OP, I offer a different proof:

If $A$ is a finite set and $B\subseteq A$, then $B$ is finite.

Answer to the question:

Since $A$ is finite, there is a bijection $f:A\to n$ for some $n\in\mathbb{N}$. Then $B$ and $f[B]$ are equipotent, and $f[B]\subseteq n$. If $f[B]=n$, we are done. Otherwise $f[B]$ is a proper subset of $n$, and using the claim above, it suffices to show that

every proper subset $X$ of a natural number $n$ is equipotent to some member of $\{0,1\ldots,n-1\}$.

Indeed, consider the property $\mathbf{P}(x)$: "any proper subset of $x$ is equipotent to a member of $\{0,1,\ldots,x-1\}$.

Clearly, $\mathbf{P}(0)$ holds trivially.

Assume that every proper subset of $n$ is equipotent to a member of $\{0,1,\ldots,n-1\}$. We have to show that any proper subset of $n+1$ is equipotent to a member of $\{0,1,\ldots,n\}$. Let $X$ be a proper subset of $n+1$; we have the following possibilities:

1. If $n\not\in X$, then $X$ is a proper subset of $n$ and the induction hypothesis tells us that $X$ is equipotent to some $m\in n$. Therefore, $m\in n+1$.
2. if $n\in X$ we have two possibilities: If $X=n$ then $X$ is equipotent to $n\in n+1$. Otherwise, $X=Y\cup\{n\}$ where $Y$ is a nonempty proper subset of $n$. The induction hypothesis tells us that $Y$ is equipotent to some $m\in n$. Let $g:Y\to m$ be a bijection. Then $\{g,\{(n,m)\}\}$ is a compatible system of functions, so that $g\cup\{(n,m)\}$ is an injective function with domain $X$ and range $m+1$. Therefore, $X$ is equipotent to $m+1$, and since $m\in n$, we have $m+1\in n+1$.

We have concluded that $\mathbf{P}(n+1)$ holds for either possibility, so the Induction Principle now completes the proof.