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dourtuntellorvl 2022-07-01 Answered
Let A 1 , A 2 , P be CPOs and let ψ : A 1 × A 2 P be a map, then ψ is continuous it is so in each variable separately
Proving this direction is quite easy, but I am not sure how to conclude the opposite direction:
If D = ( d 1 , d 2 ) = ( D 1 × D 2 ) , we have:
ψ ( D ) = ψ ( d 1 , d 2 ) = ψ d 1 ( d 2 ) = ψ d 2 ( d 1 ) , where
ψ d 1 ( d 2 ) = ψ D 1 ( D 2 ) = ψ D 1 ( D 2 ) = ψ ( { D 1 } × D 2 )
and, similarly,
ψ d 2 ( d 1 ) = ψ D 2 ( D 1 ) = ψ D 2 ( D 1 ) = ψ ( D 1 × { D 2 } )
how to conclude that ψ ( D ) = ψ ( D ) ?
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Answers (1)

Trey Ross
Answered 2022-07-02 Author has 30 answers
Step 1
Anyway, I'll start in a point you have already reached.
ψ ( D ) = ψ d 2 ( d 1 ) = ψ d 2 ( D 1 ) = ψ d 2 ( D 1 ) = { ψ d 2 ( x ) : x D 1 } = { ψ ( x , d 2 ) : x D 1 } = { ψ x ( D 2 ) : x D 1 } = { { ψ x ( y ) : y D 2 } : x D 1 } = { ψ ( x , y ) : x D 1 , y D 2 } = ψ ( D ) .
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