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Suppose $A$ is a finite set of cardinality $n$. And Let $B$ be a subset of $A$ and the cardinality of $B$ equals $n$. Then $B=A$.

Many texts use this fact very frequently but it seems that they just take it for granted. How can I prove this rigorously? Any help will be appreciated.

Asaf Karagila
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Mathcho
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3 Answers3

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Since $B \subseteq A$, we can partition $A = B \cup (A \setminus B)$. These sets are disjoint. Taking cardinalities, we see $n = n + \left|A \setminus B\right|$, which implies $|A \setminus B| = 0$, hence $A \setminus B = \emptyset$, so $A = B$.

Mark
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    In my opinion, it is obvious that B\A is empty since B is a subset of A. Maybe we gotta show A\B is empty... – Mathcho May 29 '15 at 17:42
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    Woops, $A \setminus B$ is what I meant. – Mark May 29 '15 at 17:43
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    God this is beautiful...thanks mark you're truly my savior. I really appreciate it. – Mathcho May 29 '15 at 17:47
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    You're welcome :) Glad it helped. – Mark May 29 '15 at 17:51
  • "These sets are disjoint." You mean $B$ and $A \setminus B$, right? (I'm sure that seems basic, but on my first reading, I had to stop and think about it quite a bit. Forgive me. I'm a lowly programmer who needed this result to make some code more efficient. ;) I don't do math with sets every day.) – jpmc26 Jul 12 '17 at 00:34
  • @jpmc26 Yes. Stopping and thinking about things is pretty central to most reading of math :) – Mark Jul 12 '17 at 07:34
  • @Mark I'm a programmer. The holy grail for us is things so obvious and straightforward that you can read through without having to resolve ambiguities or fill in missing details. =p Even in my physics years, I did problems that way because I found I made mistakes too easily without flushing stuff out. My point was that while it might seem obvious to you, it isn't to others who aren't as focused in the field. – jpmc26 Jul 12 '17 at 08:26
  • @Mark Well done. I have seen some needlessly complicated proofs of this result that relied on induction or properties of bijective functions, but this is pure elegance. – Display name Oct 21 '23 at 19:45
  • As the OP asks for a rigorous proof, I want to mention, that this proof strongly relies on two assumptions: the cardinality is unique and for two disjoint sets $A,B$ we have that $|A\cup B|=|A|+|B|$. Both are statements you actually have to prove. I would rather suggest Asaf Karagila's answer. – Philipp Oct 25 '23 at 14:29
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You have hit on a very subtle point, which many people can miss when they first run into this issue. Yes, we need to prove these things.

This is a consequence of the pigeonhole principle. If $B\subseteq A$, and $A$ is finite, then either $A=B$ or else there are no injective functions from $A$ into $B$; but if $|A|=|B|$ then by definition there is a bijection from $A$ into $B$.

So this, in turn, becomes a question as to how to prove the pigeonhole principle. The answer is by induction on the number of elements, and you can find a complete proof here.

Community
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Asaf Karagila
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Let's prove the contrapositive statement. Suppose that $A$ has a proper subset with the same cardinality as $A$ (in other words, $A$ is Dedekind infinite.) We will show that $A$ is infinite.

Take a proper subset $B$ of $A$ with the same cardinality as $A$. Then there is a bijection $f : A \to B$ and an element $a \in A \setminus B$. Recursively define a function $h : \mathbb{N} \to A$ by

  • $h(0) = a$
  • $h(i+1) = f(h(i))$.

Using the fact that the function $f$ is injective and its range does not contain $a$, one can prove by induction on $j \in \mathbb{N}$ that the elements $h(0),\ldots,h(j)$ are distinct, so $h$ is an injection from $\mathbb{N}$ to $A$. The existence of such an injection means that $\left|A\right| \ge \aleph_0$, so $A$ is infinite.

(There is a detail hiding here, namely the fact that there cannot be an injection from $\mathbb{N}$ to a finite set. This can be proved by induction on the cardinality of the finite set by an argument similar to the one that Asaf mentions.)

Trevor Wilson
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