Talk:Finite Union of Open Sets in Complex Plane is Open

Why limit this to finite unions?
The argument works just as well for infinite unions, and to state and prove it for infinite unions would be more general with barely any changes to the proof.

... In fact, on second thought, why do this proof at all? Since the open sets of the complex plane are the open sets induced by its metric, why not just refer to that proof? The proof given here for the complex plane is exactly the same as that proof, but the other proof is just stated for a general distance metric.


 * Short answer: because this is how it appears in the source work.
 * Longer, more informative answer: it is a feature of this site that a result couched in the language of a specific field of activity, e.g. real analysis, complex analysis, etc., should be accessible to those who are not familiar with deeper / more technical fields, e.g. (in this case) topology and metric space theory. While indeed such a proof falls immediately to more-or-less trivial results from metric space theory, we need to ensure that you can approach this from below, so to speak, as well as above.
 * This philosophy reverberates throughout the whole of, and we believe it enhances the learning experience for the student to be able to see multiple approaches to a particular problem; this gives insight, for example, into how a complex space is a metric space (and indeed a topological space), but it is completely predicated on the crucial fact that an open set in the complex plane is exactly an open set in a metric space whose metric is the modulus of the complex difference. It is the latter point which is often glossed over in texts in the fields of both metric spaces and complex analysis -- and I'm not 100 per cent sure we've got this completely watertight on . We are constrained by what texts are available for us to use as our source works, and the diligence by which these are used to construct pages from. --prime mover (talk) 19:40, 14 January 2022 (UTC)