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  1. Add forgotten +1

    RichardMau5

    diff, v25, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeOct 11th 2021
    • (edited Oct 11th 2021)

    started an Examples-section (here)

    So far we have tries for 0, 1 and 24.

    (We also have an entry that answers to 2, but that’s about a category, not a number – incidentally, I don’t it’s a good idea to call this category by this name, even if many authors do.)

    diff, v27, current

    • CommentRowNumber3.
    • CommentAuthorNikolajK
    • CommentTimeOct 15th 2021
    • (edited Oct 15th 2021)

    In Proposition 3.1., there happens the common faux pas. It currently says

    If any inhabited subset of the natural numbers possesses a minimal element, then the law of excluded middle holds.

    “Any” can be used for \forall, but if that word is put in an antecedent, then it becomes an \exists.

    I quickly tried to search the site and there’s some other, more mildly ambiguous cases. In initial object, it says

    An initial object ∅ is called a strict initial object if any morphism x→∅ must be an isomorphism.

    This is an intended \forall, but I think I can construe a sentence for the same form where this would also turn to an intended \exists:

    “A school class will immediately be put in quarantine if any of the kids must see a doctor.”

    :P

    I think those cases will always be resolved by substituting ’any’ with ’every’.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeOct 15th 2021

    Please fix it where you spot it. Thanks.

    • CommentRowNumber5.
    • CommentAuthorNikolajK
    • CommentTimeOct 15th 2021

    any=>every

    diff, v28, current

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeOct 16th 2021

    Thanks. By the way, since this is inside a proposition (here), one could alternatively use mathematical language, which might be preferable anyways, as in:

    If SS \subset \mathbb{N} with SS inhabited implies that the subset SS contains its minimum, then LEM holds.

  2. if zero is not in the natural numbers then the natural numbers do not form a rig.

    Anonymous

    diff, v31, current

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeFeb 7th 2023

    added pointer to:

    diff, v36, current