Why did Munkres treat the case $A = \emptyset$ separately? (“Topology 2nd Edition” by James R. Munkres.)

elementary-set-theorygeneral-topologylogic

I am reading "Topology 2nd Edition" by James R. Munkres.

Munkres wrote:

"$A$ is finite if it is empty or if there is a bijection $$f : A \to \{1,\dots,n\}$$ for some positive integer $n$.".

Munkres didn't write:

"$A$ is finite if there is a bijection $$f : A \to S_n := \{x \in \mathbb{Z}_{+} \mid x < n\}$$ for some positive integer $n$.".

Note that $S_1 = \emptyset$ and there is a bijection $$f : A \to S_1 = \emptyset$$ if and only if $A = \emptyset$.

Why?

enter image description here

Best Answer

Only Munkres can answer your question "why"? Anyway, his definition is correct and your definition is correct, so it is a matter of taste which you prefer.

But do you really believe that your definition is more transparent? You first define the set $S_n$ and then observe that it is empty if for $n = 1$. This step is nothing else than thinking about the two cases $n >1$ and $n = 1$. And, by the way, it would be easier to define $S_n = \{ x \in \mathbb Z \mid 1 \le x \le n\}$. Then $S_0 = \emptyset$ and $S_n = \{1,\ldots,n \}$ for $n \ge 1$. Using $S_n$ is close to what Munkres does: You have the empty set and the nonempty sets $\{1,\ldots,n \}$ for positive integers $n$. These are the prototypes of finite sets.

Related Question