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The following is the question I would like some guidance concerning:

Let $i,j\in N$ such that $i>j$. Prove that there is no injective function $f$ from {$1,...,i$} to {$1,...,j$}.

There are numerous proofs for this online which utilize the notion of cardinality, which is what I understand concerns the size of finite sets. I am not allowed to use such a notion. I have attempted 'adapting' proofs that are inspired by posts such as:

There exists no injective function from the power set of A to A

But haven't been successful because in this question we cannot assume one set is the power set of another, and so the strategy of constructing a particular set to prove a given proposition does not work.

I welcome any and all comments.

Community
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Jhon Deo
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2 Answers2

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We can prove this by simultaneous induction on $i$ and $j$.

Suppose the claim fails. Let $j_0$ be the least natural number such that for some $i>j_0$, there is an injection from $\{1, . . . , i\}$ into $\{1, . . . , j_0\}$. Let $i_0$ be the least such natural number, and $f$ such an injection. Clearly we have $i_0>1$, so in particular $\{1, . . . , i_0\}$ is nonempty.

So what? Well, without loss of generality, $f(i_0)=j_0$. Now think about $\{1, . . . , i_0-1\}$. Restricting $f$ to this set gives an injection from $\{1, . . . , i_0-1\}$ to $\{1, . . . , j_0-1\}$. But $j_0-1<i_0-1$, and this contradicts the assumed minimality of $j_0$, so we're done.

Noah Schweber
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  • Just seeing this here and finding it very useful for my own purposes, so thank you so much. Just to be sure I'm following: we argue that if the claim is false then the set of $j$ for which there exists an $i > j$ s.t. ${1,...,i}$ injects into ${1,...,j}$ is nonempty and thus, by WOP, there is a least such $j$: $j_0$. By definition of $j_0$ the set of $i$ s.t. ${1,...,i}$ injects into ${1,...,j_0}$ is nonempty and thus, by WOP, there is a least such $i$: $i_0$. Now $j_0 \geq 1$ by virtue of the claim, whence $i_0 > j_0 \implies i_0 > 1$. This fact establishes... – EE18 Mar 04 '24 at 23:36
  • ... that $i_0 - 1 \in \Bbb N \setminus {0}$ so that ${1,...,i_0 - 1}$ is well-defined. Further, if $f$ is the injection which exists by hypothesis between ${1,...,i_0}$ and ${1,...,j_0}$ then the restriction of $f$ to ${1,...,i_0 - 1}$ is likewise an injection into ${1,...,j_0}$, contradicting the minimality of $i_0$. Is this what you had in mind? – EE18 Mar 04 '24 at 23:38
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Let $I = \{1,\ldots,i\}$ and $J = \{1,\ldots,j\}$ be sets with $i > j$ and suppose $f : I \to J$ is an injective function. Then for every $n,m \in I$, with $n \ne m$, we know $f(n), f(m) \in J$ and $f(n) \ne f(m)$.

But since we have $i$ elements in $I$ which $f$ maps to at most $j$ elements in $J$, by the pigeonhole principle, there exist distinct $m,n \in I$ such that $f(m) = f(n)$. This contradicts injectivity of $f$.

Alexis Olson
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  • This is circular - the pigeonhole principle is exactly what you're trying to prove! – Noah Schweber Oct 12 '16 at 05:58
  • @NoahSchweber OP said without cardinality, not without pigeonhole principle. If you want to prove the pigeonhole principle, that is easily found. – Alexis Olson Oct 12 '16 at 06:03
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    You're missing my point: I'm saying that relying on pigeonhole here is circular. The pigeonhole principle is proved . . . using the result in the OP! The whole driving force behind the contradiction in the proof of the pigeonhole principle is that cardinality makes sense: we say $S$ has size $n$ if there is a bijection between $S$ and ${1, . . . , n}$. In order to derive a contradiction from assuming that pigeonhole fails, we need that size is well-defined: if $S$ has size $n$, it does not have size $m$ for any $m<n$. But this is exactly what the OP is asking about! – Noah Schweber Oct 12 '16 at 06:10
  • @NoahSchweber I think I understand your point, but explain exactly which part of the pigeonhole proof uses the OP's result? – Alexis Olson Oct 12 '16 at 06:14
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    The proof of pigeonhole goes roughly: "Suppose $S$ is an $n$-element set, and the pigeonhole principle fails. . . . Then $S$ is an $m$-element set, for $m<n$; contradiction." My point is that getting a contradiction from "$S$ is an $n$-element set and $S$ is an $m$-element set for some $m<n$" requires us to already know some facts about cardinality: basically, that we can't count the number of elements in a finite set in two different ways and get two different answers; in particular, that you can't have an injection from a large set into a small set, which is the claim the OP is asking about. – Noah Schweber Oct 12 '16 at 06:16
  • Essentially: the pigeonhole principle's proof relies on cardinality of finite sets being well defined. How do you define the cardinality of a finite set? How do you prove that every finite set has a unique cardinality? – Noah Schweber Oct 12 '16 at 06:17
  • @NoahSchweber Cardinality of finite sets follows from bijective set equivalence forms an equivalence relation and defining $n = |{1,\ldots,n}|$ as the canonical equivalence class representative. Comparing cardinalities is related by definition to the existence of injective functions between sets. Got it. – Alexis Olson Oct 12 '16 at 06:35