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I've seen many ways to define $e$ using Dedekind cuts. Would the following method work, too?

$\forall n \in \mathbb{N}, \, A_n = \{x \in \mathbb{Q} \mid x \lt \left(1 + \frac{1}{n}\right)^n \} $

$e = \bigcup_{n = 1}^\infty A_n$

Matthew Towers
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talopl
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  • Related: https://math.stackexchange.com/q/488800/42969 – Martin R Jul 22 '22 at 18:00
  • Since $(1+1/n)^n$ is increasing, rational, and converges to $e,$ this works. You can write it as: $$e=\left{x\in\mathbb Q\mid \exists n:x<\left(1+\frac1n\right)^n\right}$$ You could also write it as: $$e=\left{x\in\mathbb Q\mid \exists n:x<\sum_{k=0}^n\frac1{k!}\right}$$ – Thomas Andrews Jul 22 '22 at 18:04
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    It's a little easier to show that $A_n=\sum_{k=0}^{n}\frac1{k!}$ is increasing and bounded above than it is to show $B_n=\left(1+\frac1n\right)^n$ is increasing and bounded above. – Thomas Andrews Jul 22 '22 at 18:06

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