1

In Goldblatt's Topoi he discusses nine equivalent claims in his Theorems 7.3 and 8.3, and shows they are equivalent by proving two cycles of implications.

I am wondering if some of results have neater proofs instead going on a circular tour. For example, in answering Showing that, in a topos where $[\top, \bot]: 1 \oplus 1 \to \Omega$ is iso, $\neg\circ\neg = 1_\Omega$ Daniel Schepler gives a very neat proof that the implication in that question title holds.

Question: is there another neat and elementary proof that shows that in a topos where $\neg\circ\neg = 1_\Omega$, $\top \cup \neg\top \equiv 1_\Omega$ so $\top$ and $\bot$ are complements?

(Or, if neater, an elementary proof that shows that if $[\top, \bot]$ is an isomorphism, then $\top \cup \bot \equiv 1_\Omega$.)

Late starter
  • 242
  • 1
    I think the last sentence ("if $[\top, \bot]$ is iso, then $\top \cup \bot \equiv 1_\Omega$") is trivial to prove: the union, $\top \cup \bot$, is (defined as) the monomorphic part of $[\top, \bot]$, so if the latter arrow is iso, then it is its own monomorphic part, $\top \cup \bot = [\top, \bot]$ (i.e., $\top \cup \bot$ is iso). So then $1_\Omega = (\top \cup \bot) \circ (\top \cup \bot)^{-1}$ and obviously $\top \cup \bot = 1_\Omega \circ (\top \cup \bot)$ which means that $1_\Omega \equiv \top \cup \bot$ since each factors thorugh the other. – Tony Dolezal Mar 03 '25 at 01:14
  • @TonyDolezal That looks very neat! – Late starter Mar 04 '25 at 07:58

1 Answers1

3

Here's an argument in the internal language of the topos. $$ \top\cup\neg\top=\neg\neg(\top\cup\neg\top) =\neg(\neg\top\cap\neg\neg\top) =\neg\bot=\neg\neg\top=\top. $$ I've used the assumption $\neg\neg=1_\Omega$ at the first and last step. The second step uses the de Morgan law $\neg(A\cup B)=(\neg A)\cap(\neg B)$, which is (unlike its dual de Morgan law) intuitionistically valid.

Andreas Blass
  • 75,557
  • Neat thanks (and perhaps also the idea that Goldblatt had in mind in his "hint" to ex. 30 on his page 184). But elementary? I wasn't too clear what I meant by "elementary", sorry! (I mean I wasn't clear in my own mind, not just in what I wrote). But I guess I was thinking of a proof that came more or less immediately from the definitions and didn't call on a more general/sophisticated background claim e.g. about the internal logic being intitioniistic. – Late starter Mar 04 '25 at 08:14