英文字典中文字典


英文字典中文字典51ZiDian.com



中文字典辞典   英文字典 a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p   q   r   s   t   u   v   w   x   y   z       







请输入英文单字，中文词皆可：

uncountable    

安装中文字典英文字典查询工具!


中文字典英文字典工具:
选择颜色:
输入中英文单字

































































英文字典中文字典相关资料:


  • Uncountable vs Countable Infinity - Mathematics Stack Exchange
    My friend and I were discussing infinity and stuff about it and ran into some disagreements regarding countable and uncountable infinity As far as I understand, the list of all natural numbers is countably infinite and the list of reals between 0 and 1 is uncountably infinite
  • 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 )
  • 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
  • linear algebra - Uncountable Basis? - Mathematics Stack Exchange
    In particular some naturally occurring vector spaces can be shown to have an uncountable basis using the axiom of choice And we can show that this use of the axiom of choice is in fact necessary If we're mentioning the axiom of choice, perhaps it should also be pointed out that the axiom of choice is in fact equivalent to the statement "Every
  • Does an uncountable discrete subspace of the reals exist?
    Several of the answers given rely on completeness or compactness, but these properties are not needed For example there is no uncountable discrete subspace of the irrationals The reals are a second-countable space, so any subspace is also second-countable, which prevents the subspace from having an uncountable discrete subspace
  • Why do we not need measures to be uncountably additive?
    As long as uncountable many sets have positive measure the union will have infinite measure (unlike for a countable sum) Thanks for any help and suggestions! Edit: I know that in the end uncountable additivity would imply that every subset of $\mathbb{R}$ has Lebesgue measure $0$, but I feel like the author of the notes has something else in
  • 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





中文字典-英文字典  2005-2009