基于给定定义定理，求证：对每个$k\in\Bbb{N}$，$\Bbb{N}_k$是有限集

Let's break this down using the definitions and theorems you've given, leaning on proof by contradiction plus the well-ordering principle for natural numbers to make things concrete.

First, recap our tools #

Let's restate the key definitions and results we'll rely on to avoid confusion:

• Definition: $\Bbb{N}_k = {1, 2, ..., k}$
• Definition: A set $S$ is infinite if and only if there exists a function $f:S\to S$ that is injective (one-to-one) but not surjective (onto)
• Lemma: For any $k\in\Bbb{N}$ and any $x\in\Bbb{N}_k$, $\Bbb{N}k - {x} \sim \Bbb{N}{k-1}$ (the two sets are equinumerous, i.e., have the same cardinality)
• Theorems:
  • If $A$ is infinite and $A\sim B$, then $B$ is infinite
  • If $A$ is infinite and there's an injective function $f:A\to B$, then $B$ is infinite
  • Restricting an injective function to a subset of its domain gives another injective function

Step 1: Base case ($k=1$) #

Start with $\Bbb{N}_1 = {1}$. Suppose, for contradiction, that $\Bbb{N}_1$ is infinite. By the infinite set definition, there must be an injective but non-surjective function $f:{1}\to{1}$. But the only possible function here maps $1$ to $1$, which is clearly surjective. That's a contradiction—so $\Bbb{N}_1$ must be finite.

Step 2: Contradiction via minimal counterexample #

Suppose (again, for contradiction) that there exists some natural number $m>1$ such that $\Bbb{N}_m$ is infinite. By the well-ordering principle of natural numbers, there's a smallest such $m$.

Since $\Bbb{N}_m$ is infinite, there exists an injective but non-surjective function $f:\Bbb{N}_m\to\Bbb{N}_m$. Because $f$ isn't surjective, there's some element $y\in\Bbb{N}_m$ that's not in the image of $f$ (i.e., $y\notin f(\Bbb{N}_m)$).

Now, let's look at the image set $f(\Bbb{N}_m)$:

• Since $f$ is injective, the function $f:\Bbb{N}_m\to f(\Bbb{N}_m)$ is a bijection (one-to-one and onto its image). This means $\Bbb{N}_m \sim f(\Bbb{N}_m)$.
• Since $\Bbb{N}_m$ is infinite, our first theorem tells us $f(\Bbb{N}_m)$ is also infinite.

Notice that $f(\Bbb{N}_m)$ is a subset of $\Bbb{N}_m - {y}$ (because $y$ isn't in the image). By our lemma, $\Bbb{N}m - {y} \sim \Bbb{N}{m-1}$.

Since $f(\Bbb{N}_m)$ is infinite and there's an injective function (the identity map) from $f(\Bbb{N}_m)$ to $\Bbb{N}_m - {y}$, our second theorem says $\Bbb{N}m - {y}$ must be infinite. And since $\Bbb{N}m - {y} \sim \Bbb{N}{m-1}$, the first theorem tells us $\Bbb{N}{m-1}$ is infinite too.

But wait—we assumed $m$ was the smallest natural number where $\Bbb{N}m$ is infinite. But $m-1 < m$, and we just showed $\Bbb{N}{m-1}$ is infinite. That's a contradiction!

Step 3: Wrap up our conclusion #

Our assumption that there exists an infinite $\Bbb{N}_k$ leads to a contradiction. Therefore, every $\Bbb{N}_k$ (for $k\in\Bbb{N}$) must be a finite set.

内容的提问来源于stack exchange，提问作者Fernando Nazario