Complex Analysis: Open Set Intersection Proof

[Pyroadept]

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.

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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

 

[jbunniii]

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!