From my “watched a YouTube video” understanding of Gödel’s Incompleteness Theorem, a consistent mathematical system cannot prove its own consistency, and any seemingly consistent system could always have a fatal contradiction that invalidates the whole system, and the only way to know would be to find the contradiction.

So if at some point our current system of math gets proven inconsistent, what happens next? Can we tweak just the inconsistent part and have everything else still be valid or would we be forced to rebuild all of math from basic logic?

  • CanadaPlus@lemmy.sdf.org
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    7 minutes ago

    Our current system? So ZFC?

    We move to a different, probably weaker system. Certain mathematicians studying hyper-infinite monstrosities like Banach spaces will be sad. Life will go on, though.

    Existing applied math should all work purely in the Von Neumann universe of finite sets, if you’re forced to go that far. Saying there’s actually an infinite number of points in a line is a kind of luxury; it’s just one that feels right to most mathematicians. A number that’s simply much larger than is worth bothering with can work the same way. In the same spirit, you can probably get rid of most of the Von Neumann universe without breaking practical things.

    If you mean actual practical math breaks, I dunno. In any reasonably complex inconsistent formal system, all statements can be proven true. Like 10 = 20. So, can I just grow more fingers somehow?

    Edit: Since this is Lemmy, it’s worth pointing out that Godel’s incompleteness theorem has a kind of interchangeability with the fact some questions are undecidable by Turing machine.

    Godel’s theorem has a kind of salacious, clickbait quality to it. Laymen will interpret all kinds of things into it. I’ve literally heard “so, math is useless then”. But, if you know algorithms, you know that saying they can’t determine if a loop ends is nowhere close to saying they’re going to stop working, or that they aren’t really good at what they can do.

    • mathemachristian[he]@lemmy.ml
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      1 day ago

      A banach space is just a vector space with a norm, there are many many weirder examples of spaces in topology. You’re thinking of the banach-tarski paradox I think

      • CanadaPlus@lemmy.sdf.org
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        23 hours ago

        It wasn’t a mistake. Usually you’re talking about an infinite-dimensional TVS when you say Banach space - as in it’s just Banach, that’s the most you can say about it. I don’t mean R3.

        Stuff like the axiom of choice has a way of coming up in functional analysis. Sure, there’s weirder spaces, like from general topology or TVS theory in general, but Banach spaces are an example that are pretty widely used and studied. It seems like going with some pathological object I have to search around for would make the point less clear.

        The Banach-Tarski paradox wouldn’t work either without the AoC, but it’s just a specific counterexample, and Banach and Tarski’s careers will be fine since they both died decades ago.

        • mathemachristian[he]@lemmy.ml
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          3 hours ago

          ah I gotcha, thats fair then.

          On another note, you’d propose a von Neumann hierarchy without AoC? How would that class be defined?

          Admittedly I’m not that big into class theory so I am likely missing something obvious, but to me it seems like the definition of the class requires such a selecting function

          • CanadaPlus@lemmy.sdf.org
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            5 minutes ago

            I mean, me neither. But if all sets are finite AoC just trivially holds, right? You can do it “manually”.

            If you back off to just ZF, parts of functional analysis will break. Linfinity space isn’t separable, and so isn’t necessarily Baire anymore, for example.

            If we go all the way to finitism or ultrafinitism it doesn’t really exist as a concept in the first place. But, whatever numerical engineering calculation will still work, and you can probably do something that looks like functional analysis to determine a mode of vibration, even if you’re actually just using a series of high-dimensional but finite spaces. Probably, anyway. Don’t ask me to prove it.