Complex Analysis: Open Set Intersection Proof

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SUMMARY

The discussion focuses on proving that the intersection of two open sets and the union of arbitrarily many open sets in the complex plane are both open sets. The proof for part (a) establishes that if U and V are open sets, then their intersection U ∩ V contains a disc of positive radius around any point z in the intersection. For part (b), it is shown that the union of n open sets U_1, U_2, ..., U_n also contains discs of positive radius around all points, confirming that it is an open set. The proofs rely on the definition of open sets in the context of complex analysis.

PREREQUISITES
  • Understanding of open sets in topology
  • Familiarity with the complex plane and complex numbers
  • Knowledge of basic set operations (union and intersection)
  • Ability to work with epsilon-delta definitions in analysis
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  • Study the properties of open sets in topology
  • Learn about the epsilon-delta definition of limits in complex analysis
  • Explore the concept of compactness in relation to open sets
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Homework Statement


An open set in the complex plane is, by definition, one which contains a disc of positive radius about each of its points. Prove that:
(a) the intersection of two open sets is an open set
(b) the union of arbitrarily many open sets is an open set


Homework Equations


If A, B are two sets, A = {a, b, c} and B = {c, d, e}, then:
AUB = {a, b, c, d, e}
A intersect B = {b}



The Attempt at a Solution


Here's what I've done:

(a) Let U, V be two non-empty open sets in the complex plane. Then, by definition, all the points of U and V have a positive radius about them.
Then, clearly, all the points in the intersection of U and V will be points that have a positive radius about them.
Thus the intersection of U and V is, by definition, an open set.

(b) Let U_1, U_2, ... , U_n be any n non-empty open sets in the complex plane. Then, by definition, all the points of U_1, U_2, ... , U_n have a positive radius about them.
Thus, clearly, all the points in the union of U_1, U_2, ... , U_n have a positive radius about them.
Thus the union of U_1, U_2, ... , U_n is, by definition, an open set.

---

I know heuristic proofs don't generally cut it in analysis, but I really don't see what else there is to say on this particular question. Am I missing something fundamental in my argument?
Thanks for any help
 
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I think you need to be a lot more explicit about the positive radius.

Here's how I would start (a):

Let U and V be two open sets in \mathbb{C}. If U \cap V is empty, then the claim is obviously true. So suppose that it is not empty, and choose z \in U \cap V.

Then z \in U, so there exists a radius r_u > 0 such that u \in U whenever |z - u| < r_u.

Similarly, z \in V, so there exists a radius r_v > 0 such that v \in V whenever |z - v| < r_v.

I now wish to show that U \cap V contains a disc of positive radius around the point z, which means that I must find a radius r > 0 such that x \in U \cap V whenever |z - x| < r.

Your job is now to find such an r. Hint: will one of r_u or r_v work? If so, which one? And you must prove that it works!
 

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