2 ms·

I don’t think it’s fair to call {}-> injective just because no two inputs map to the same output. That’s vacuous.

BeetleB 2mo 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 2mo 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 2mo 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 2mo 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.
gpm 2mo ago
Edit: Removed incorrect claim that |B| > |A| sufficed for the counter example. It's also the definitions the book supplies though (and the standard ones). Mathematics works over definitions. Everyone is free to do math over whatever definitions they want - but what is or isn't true follows from them. Lots of definitions and theorems exclude things like empty-set cases because they're weird, but that has to be explicit (otherwise someone will apply a theorem to the empty set and it will lead them to incorrect conclusions).
- ndriscoll 2mo ago
  No, empty A is critical to the counterexample. In your example, g(x) = 1 is a left inverse. The point is you either send an element of the codomain to its (unique by injectivity) preimage if it's in the image, or to an arbitrary element of A if it's not, and that's a left inverse. But then if B has an element, A needs one for you to pick your arbitrary target. In a sense, your claim that the problem is a smaller domain than codomain does contribute though; if f is also surjective, then this case can't happen, so bijective iff invertible (the empty function is vacuously bijective and its own inverse).
  - gpm 2mo ago
    Oh, oops, you're right. Sorry.
mitxela 2mo ago
But that is the definition of injective.

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 2mo 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 2mo 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 2mo 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.

gpm 2mo ago

Edit: Removed incorrect claim that |B| > |A| sufficed for the counter example. It's also the definitions the book supplies though (and the standard ones). Mathematics works over definitions. Everyone is free to do math over whatever definitions they want - but what is or isn't true follows from them. Lots of definitions and theorems exclude things like empty-set cases because they're weird, but that has to be explicit (otherwise someone will apply a theorem to the empty set and it will lead them to incorrect conclusions).

ndriscoll 2mo ago
No, empty A is critical to the counterexample. In your example, g(x) = 1 is a left inverse. The point is you either send an element of the codomain to its (unique by injectivity) preimage if it's in the image, or to an arbitrary element of A if it's not, and that's a left inverse. But then if B has an element, A needs one for you to pick your arbitrary target. In a sense, your claim that the problem is a smaller domain than codomain does contribute though; if f is also surjective, then this case can't happen, so bijective iff invertible (the empty function is vacuously bijective and its own inverse).
- gpm 2mo ago
  Oh, oops, you're right. Sorry.

mitxela 2mo ago

But that is the definition of injective.