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  1. I really liked the reasons already posted on the page why it is wrong to consider the empty set as connected. I just added one more which I also find compelling and which is related to enumerative combinatorics.

    Abdelmalek Abdesselam

    diff, v10, current

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeOct 16th 2018

    NIce, thanks!

    • CommentRowNumber3.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 17th 2018

    Never thought about that, but it makes sense – thanks!

    • CommentRowNumber4.
    • CommentAuthorTim_Porter
    • CommentTimeOct 17th 2018

    Strangely enough there is an empty space missing in the first line of empty space as it says Theempty space. Fun!

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeOct 17th 2018

    Tim, I saw that too, but didn’t see how to fix it.

    • CommentRowNumber6.
    • CommentAuthorAli Caglayan
    • CommentTimeOct 17th 2018

    I have fixed it for now by removing the link but this is another bug to tell Richard.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeOct 20th 2018
    • (edited Oct 20th 2018)

    after the lines

    that the logarithm of exponential generating functions of some type of objects should be the exponential generating function for the connected objects of that type. Since this logarithm has no constant term, this suggests the empty object should not count as connected. This result is also known in the physics literature as the linked-cluster theorem

    I added the pointer:

    (see this Prop. at geometry of physics – perturbative quantum field theory).

    diff, v13, current

    • CommentRowNumber8.
    • CommentAuthorGuest
    • CommentTimeMar 2nd 2021

    I do not understand the counterpoint about the uniqueness of connected component decomposition failing when the empty set is connected. The decomposition is defined to be into maximal (path-)connected components in order to be unique. Since the empty set is never maximal, it would never appear in such a decomposition. It would be like saying the decomposition is non-unique since [0,2]=[0,1)∪[1,2][0,2] = [0,1) \cup [1,2].

    • Mike Earnest
    • CommentRowNumber9.
    • CommentAuthorHurkyl
    • CommentTimeMar 2nd 2021

    Re #8, the decomposition you refer to is meant to be a decomposition into a coproduct; i.e. the ∪\cup is meant to be a disjoint union. So [0,1)∪[1,2]=[0,2][0,1) \cup [1,2] = [0,2] is not the sort of decomposition it alludes to.

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeMar 3rd 2021

    I have changed the “∪\cup“s to “⊔\sqcup“s (here).

    Maybe it was technically correct as it was, given that the ambient text suggests that the spaces in the union are assumed to be disjoint. Or maybe it’s not so clear, since the text is discussing a counter-factual property of the empty set.

    diff, v15, current

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