# I The Continuum Hypothesis and Number e

#### Quasimodo

Summary
The Continuum Hypothesis and Number e
Summary: The Continuum Hypothesis and Number e

Now, I must ask a very stupid question:

When taking: $$\lim_{_{n \to \infty} } (1+\frac{1}{n})^n=e\\$$ the $n$ we use take its values from the set: $\left\{ 1,2,3 ... \right\}$ which has cardinality $\aleph_0$, which is equivalent maybe, I say maybe to writing: $$\ (1+\frac{1}{\aleph_0})^{\aleph_0}=e\\$$
Upon: $$\lim_{_{n \to \infty} } (1+\frac{1}{n})^{2n}=e^2\\$$ we take, $$\ (1+\frac{1}{\aleph_0})^{2\aleph_0}=e^2\\$$
So, since two equal power bases give two different results, we have to assume that their exponents are different hence: $$2\aleph_0 > \aleph_0$$

#### BvU

Homework Helper
The question being "since the continuum hypothesis says that the smallest cardinal number $>\aleph_0$ is $\mathfrak c$, have I now proven that ${\mathfrak c} = 2 \aleph_0$ " ?

With a possible successor "so with ${\mathfrak c} = 2^ {\aleph_0} = 2 \aleph_0$ " ?

As you guessed: no and no. And you may not write $\ (1+\frac{1}{\aleph_0})^{\aleph_0}=e \$.

Homework Helper
... which is equivalent maybe, I say maybe to writing: $$\ (1+\frac{1}{\aleph_0})^{\aleph_0}=e\\$$
No. Just no.

• weirdoguy

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