/sci/ - Science & Math

[Anonymous No.11643857]

How do you prove "topological continuity" (i.e. U open => f^{-1}(U) open) for f:C* -> R+, f(z)=|z|?

Both C* and R+ with the subspace topology of the euclidean topology.

I thought about rewriting z=x+iy and then looking at the pre-images in RxR

[Anonymous No.11643864]

Use the equivalent epsilon-delta definition. Then it becomes trivial.

[Anonymous No.11643875]

>>11643864
thats only equivalent to continuity in topology if the spaces are metrizable

[Anonymous No.11643878]

>>11643857
this is why using base for topology is useful. you don't need to check preimage of *every* open set, just the base ones. now, do you know some convenient base for usual topology on R ?

[Anonymous No.11643881]

>>11643875
>implying they aren't

[Anonymous No.11643888]

>>11643881
oops lol

[Anonymous No.11643889]

>>11643878
probably epsilon balls

[Anonymous No.11643895]

>>11643889
exactly. since we're talking R, these are just the open intervals (a,b). so what's the preimage of (a,b) ? it's definitely something tangible.

[Anonymous No.11643898]

>>11643895
in terms of \sqrt{x^2+y^2} it's still ugly

[Anonymous No.11643902]

>>11643898
so what if you square it ?

[Anonymous No.11643910]

>>11643902
>>11643898
i mean it would be (-\sqrt{b^2-y^2},-\sqrt{a^2-y^2}) x (\sqrt{a^2-x^2},\sqrt{b^2-x^2}) which is obviously open in the euclidean topology

[Anonymous No.11643911]

Literally just a simple application of the triangle inequality.

[Anonymous No.11643913]

>>11643895
It's an annulus between x^2+y^2 = a^2 and x^2+y^2 = b^2

[Anonymous No.11643915]

>>11643857
Just prove that sqrt(x) is continuous and use it as an auxiliary result

[Anonymous No.11643918]

>>11643913
exactly, so can you prove that any annulus in the complex plane is open ? it's basically >>11643911

[Anonymous No.11643922]

>>11643918
*any annulus given by strict inequalities obviously