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  • real analysis - Proving that the interval $(0,1)$ is uncountable . . .
    I'm trying to show that the interval $(0,1)$ is uncountable and I want to verify that my proof is correct My solution: Suppose by way of contradiction that $(0, 1)$ is countable Then we can crea
  • set theory - Union of Uncountably Many Uncountable Sets - Mathematics . . .
    By definition, uncountable means the set is not countable There are no other choices So, if you take 1 or more uncountable sets, it will stay in the biggest class, uncountable Even if you take uncountably many sets that are uncountable, there's no where above uncountable to go Uncountable isn't a cardinal
  • set theory - What makes an uncountable set uncountable? - Mathematics . . .
    The question in its current form is a bit unclear, but I'll try to address your concern below Assuming we are working with the axioms of $\mathsf {ZFC}$, we can construct an uncountable set by taking the power set of any set with infinitely many elements (e g $\mathbb{N}, \mathbb{Z}, \mathbb{R}$, etc )
  • analysis - Proving the open interval $(0,1)$ is uncountable . . .
    However, it is perhaps more common that we first establish the fact that $(0, 1)$ is uncountable (by Cantor's diagonalization argument), and then use the above method (finding a bijection from $(0, 1)$ to $\mathbb R)$ to conclude that $\mathbb R$ itself is uncountable
  • Proving a set is uncountable - Mathematics Stack Exchange
    Stack Exchange Network Stack Exchange network consists of 183 Q A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers
  • Uncountable $\sigma$-algebra - Mathematics Stack Exchange
    As a secondary question: I think this result is supposed to be used to show that if $\mathcal A$ is a $\sigma$-algebra with infinitely many elements, then it's uncountable, but I wasn't able to show that the property mentioned above (the one I'm trying to show) was satisfied in this case
  • The sum of an uncountable number of positive numbers
    The question is not well-posed because the notion of an infinite sum $\sum_{\alpha\in A}x_\alpha$ over an uncountable collection has not been defined The "infinite sums" familiar from analysis arise in the context of analyzing series defined by sequences indexed over $\mathbb{N}$, and the series is defined to be the limit of the partial sums
  • Proving that R is uncountable - Mathematics Stack Exchange
    By construction, this sequence is different from all the others This contradicts the assumption that this set of sequences is countable Hence, $[0,1]$ must be uncountable, and therefore $\mathbb{R}$ must be uncountable
  • elementary set theory - Uncountable minus countable set is uncountable . . .
    So, yes, we need the negation of uncountable The definition of uncountable is not countable So not uncountable is countable Now, the continuum hypothesis is about the "different types of uncountable", but we are not concern about that Uncountable means cardinality strictly greater than $\aleph_0$ $\endgroup$ –





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