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.
To complete the preceding answerspicture (the obvious remaining part). If ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete the preceding answers (the obvious remaining part). If ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete the picture (the obvious remaining part). If ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete Robin Chapman's answerthe preceding answers (the obvious remaining part): if. If${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete Robin Chapman's answer (the obvious remaining part): if${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete the preceding answers (the obvious remaining part). If${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
To complete Robin Chapman's answer (the obvious remaining part): if ${S\subset\mathbb R}$ is well ordered, then it is countable: indeed it has countable cofinality. Thus well-ordered subsets of R are exactly countable ordinals.
