# Homework Help: Properties of Open Sets (and what are Indexing Sets?)

1. Jan 4, 2007

### maverick280857

1. The problem statement, all variables and given/known data
Source: An Introduction to Complex Analysis (Classical and Modern Approaches) by Wolfgang Tutschke and Harkrishan L. Vasudeva (Champan and Hall/CRC)
Ref. page 17

Consider the metric space $(\mathbb{C}, d)$
1. If $U_{1}$, $U_{2}$,....,$U_{n}$ are open sets in $\mathbb{C}$, then so is $$\bigcap_{k = 1}^{n} U_{k}$$.
2. If {$U_{\alpha} : \alpha \in \Lambda$} is a collection of open sets in $\mathbb{C}$, where $\Lambda$ is any indexing set, then $$U = \bigcup_{\alpha \in \Lambda} U_{\alpha}$$ is also open.

2. Relevant equations

(None...all data stated in the question above)

3. The attempt at a solution

I am able to understand the proof of (1) but the proposition (2) is not clear to me.

What does an indexing set mean? Also, why does one need an indexing set in part (2) and not (1)?

Last edited: Jan 4, 2007
2. Jan 4, 2007

### cristo

Staff Emeritus
An index set is a set used to label the elements of another set, by means of introducing a surjective function from the index set, to the set being indexed.

An index set is not needed to describe the first example, since there are finitely many sets Ui. However, in the second example, there is no restriction on the set $\{U_\alpha:\alpha\in\Lambda\}$ to make it finite. For example, $\Lambda$ may be |N, in which case {Ua} would be countably infinite.

For (1) an intersection of finitely many open sets it open, whereas for (2) the union of any number of open sets is open.

3. Jan 4, 2007

### AKG

There's nothing really that special about the "indexing" set in this case. They could have just said "where $\Lambda$ is any set". Basically, they want you to prove that the union of $|\Lambda |$ open sets is open, where since $\Lambda$ can be any set, $|\Lambda |$ can be any cardinal number. More concisely, they want you to show that a union of any number of open sets is an open set.

In question 1, they could have talked about an indexing set. As you can see, they are using the indexing set {1, 2, ..., n}. If they were to talk about an indexing set, they would say "If {Ua : a is in L} is a collection of open sets in C, where L is any finite indexing set, then $\cap _{a \in L}U_a$ is also open."

4. Jan 4, 2007

### matt grime

Sorry. but how does the OP conclude that 1) does not have an indexing set but 2) does?

5. Jan 5, 2007

### maverick280857

$$U_{n} = D\left(0;\frac{1}{n}\right) = \left[z \in \mathbb{C} | |z| = \frac{1}{n}\right]$$

and the intersection of an arbitrarily large number of such sets, that is

$$\bigcap_{n = 1}^{\infty}U_{n} = \left[0\right]$$

which reduces to the singleton closed set.

(PS--How does one indicate parentheses { and } around sets?)

Last edited: Jan 5, 2007
6. Jan 5, 2007

### cristo

Staff Emeritus
I'm not sure that I understand your question. This example is showing that the intersection of infinitely many sets is not open.

When writing in tex use the commands \{ and \} to obtain { and } respectively.

7. Jan 5, 2007

### HallsofIvy

He doesn't. He should not have said that there is not indexing set. What he should have said was that the indexing set in 1) is {1, 2, 3, ..., n}, the set of positive integers less than or equal to n. His point is that that is given in the notation of the problem. In problem 2) the indexing set can be any set. Not only "infinite" but possibly "uncountable".

8. Jan 5, 2007

### maverick280857

Okay to make my query more explicit, I want to understand why an indexing set is invoked at all and why only in the second part.

9. Jan 5, 2007

### HallsofIvy

Because the proof for a finite collection of open sets is different from the proof for a infinite collection of sets. For a finite collection you could use proof by induction. Just show that the union of two open sets is open and induction does the rest for you. That's why you can prove that the intersection of a finite number of open sets is open but not the intersection of an infinite number of opens sets.

10. Jan 6, 2007

### maverick280857

Thank you all

11. Jan 7, 2007

### matt grime

An index set is invoked in the first one. It is the set {1,2,..,n}. Call that S. The following are identical

$$\cup_{s \in S}U_s$$

and

$$\cup_{i=1}^n U_i$$

However, the latter notation only works for a well ordered index set. i.e. the second one takes the FIRST set, intersects it with the SECOND, then the THIRD,.... . Not all index sets need to be well ordered so there is no concept of doing things first, second or third.

12. Jan 8, 2007

### maverick280857

Thank you! I think I see what you mean now. And I see that a similar theorem holds for closed subsets of $\mathbb{C}$ with the intersection and union interchanged.