The Continuum Hypothesis and Number e

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SUMMARY

The discussion centers on the application of the Continuum Hypothesis to the mathematical constant e, specifically through limits involving cardinal numbers. The author questions whether the expression $$\lim_{n \to \infty} (1+\frac{1}{n})^n$$ can be equated to $$\lim_{n \to \infty} (1+\frac{1}{\aleph_0})^{\aleph_0}$$, leading to the conclusion that $$2\aleph_0 > \aleph_0$$. However, the consensus is that this reasoning does not hold, and the expressions cannot be equated as suggested. The discussion ultimately reinforces that $$\mathfrak{c}$$ does not equal $$2\aleph_0$$.

PREREQUISITES
  • Understanding of limits in calculus, particularly involving sequences.
  • Familiarity with cardinal numbers, specifically $$\aleph_0$$ and $$\mathfrak{c}$$.
  • Knowledge of the Continuum Hypothesis and its implications in set theory.
  • Basic understanding of exponential functions and their properties.
NEXT STEPS
  • Study the implications of the Continuum Hypothesis in set theory.
  • Explore the properties of cardinal numbers and their relationships.
  • Learn about limits and their applications in calculus, focusing on sequences approaching infinity.
  • Investigate the mathematical constant e and its derivations from limits.
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Mathematicians, students of advanced mathematics, and anyone interested in set theory and the foundations of mathematics.

Quasimodo
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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: $$ 2\aleph_0 > \aleph_0 $$
 
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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 \ ##.
 
Quasimodo said:
... which is equivalent maybe, I say maybe to writing: $$\ (1+\frac{1}{\aleph_0})^{\aleph_0}=e\\$$

No. Just no.
 
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