5 comments

  • generationP 39 minutes ago
    This one is not just in Dummit and Foote; it's just too easy to miss. I'd guess it appears in half the places that state this result. Fixed it in my own lecture notes a few months ago.
  • Paracompact 58 minutes ago
    It warms my heart every time I see an interactive proof assistant being used to improve rather than simply slow down mathematical thinking.

    After years of using the things, I believe not enough focus is given to high-velocity uses of proof assistants for prototyping. They can altogether replace scratch paper for fumbling around with new concepts.

    • dnautics 45 minutes ago
      WIP, but that is the target ethos in the prover I'm building:

      https://github.com/ityonemo/bpa

      Its painfully verbose and explicit but its designed to let you cut down to the structure of the proof with a query language

  • hyperhello 23 minutes ago
    I don’t think it’s fair to call {}-> injective just because no two inputs map to the same output. That’s vacuous.
    • BeetleB 17 minutes ago
      Generally mathematicians treat vacuous statements as true.

      I believe it doesn't make any difference to any meaningful result. It merely makes it easier to write theorems without specifying exceptions.

    • tim-kt 13 minutes ago
      It's true precisely because it's vacuous. If you quantify over the empty set, anything is true.

      In other words, the statement "for every x in {} it holds that <anything>" is always true.

      • layer8 8 minutes ago
        What can be confusing is that the statement "for every x in {}, it doesn’t hold that <anything>" is always true as well.
        • tim-kt 6 minutes ago
          I mean, yes. But "it doesn't hold that <anything>" is equivalent to "it holds that <not anything>" and since not anything is also anything... Ah, I see.
  • psYchotic 17 minutes ago
    Help me out, I feel dumb.

    The first criterion for a function is stated as:

    > The first item in each pair comes from A.

    The counter-evidence for the proposition says:

    > Let A = {}, and B = {1}. Let f: A -> B = {}

    How does this f satisfy the first criterion, if A is uninhabited? It feels like this function can't be invoked. Am I thinking too much in terms of types here?

    • changoplatanero 13 minutes ago
      When there are no pairs, its certainly true that the first element of each pair comes from A. Just like if there are no living dinosaurs its true that all living dinosaurs speak English.
  • shmoil 25 minutes ago
    I asked AI to formalize an old important paper in analysis. In the paper there is a sequence of epsilon_n > 0, epsilon_n -> 0. It came back, and said: "I formalized it, it is all good, but the assumption that epsilons > 0 is not used anywhere. Shall we remove it, you a get a stronger result this way?"

    LOL