The point is not that ∀s∈S∃f(s) and g(s) | f(s)∈[0,1] and g(s)∈[0,1], it's that ∀y∈[0,1]∃h(y)∈S
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
replies:
The point is not that ∀s∈S∃f(s) and g(s) | f(s)∈[0,1] and g(s)∈[0,1], it's that ∀y∈[0,1]∃h(y)∈S