Woohoo proving topologies o_o

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thedarjeeling

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I know it looks like I'm listing a lot of homework problems, but please bear with me; I'm just giving a sample and if you could help on any of them it might be enough to let me proceed through the rest. I know that in order to have a topology, that the collection of sets must satisfy three conditions, which are

1) X, emptyset are elements of tau (X is the universal set). This one is generally easy to show, except on number 4 I can't even show that X is in tau!
2) the intersection of any two arbitrary open sets is open. I have no idea how to intersect two open sets in the formats below
3) the arbitrary union of open sets is open. I have no idea how to arbitrary union the sets below.

1. Let X and Y be any two sets and let \(\displaystyle \tau\) be a topology for X. If f:Y --> X is any function, show that \(\displaystyle \hat{\tau} = \{B \subset Y | \exists A \in \tau : B = f^{-1}(A)\}\) is a topology for Y. SOLVED

2. Let (X, \(\displaystyle \tau\)) be a topological space and let Y \(\displaystyle \subset\) X. Show that \(\displaystyle \tau_Y = \{B \subset X: \exists A \in \tau : B = Y \cap A\}\) is a topology for Y. SOLVED

3. Show that \(\displaystyle \tau = \{A \subset \mathbb{R}: A\) is finite or its complement is finite} U emptyset is a topology for R. Counterexampled, going to ask professor if it's really a topology, since I think I showed under arbitrary union that it isn't

4. Show that \(\displaystyle \tau = \{A \subset \mathbb{R} : \exists N \in \mathbb{N} \ \forall n \ge N \ \frac{1}{n} \in A\} \cup \emptyset\) is a topology for R.
 
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thedarjeeling said:
3. Show that \(\displaystyle \tau = \{A \subset \mathbb{R}: A\) is finite or its complement is finite} U emptyset is a topology for R.
Click to expand...

Here's my attempt so far at number 3, tell me what you think.

For condition 1, I need to show that R and the emptyset are elements of tau. Since the empty set is by definition finite, then it is in tau. Since R's complement, the empty set, is finite, then it also belongs in tau.

For condition 2, Let A, B arbitrary sets in tau. There are 4 possible cases:

A is finite, B is finite --> clearly, the finite intersection of finite sets is finite, so A intersection B is in tau
A is finite, B is infinite --> clearly, the finite intersection of a finite set an an infinite set is finite, so A intersection B is in tau
A is infinite, B is finite --> clearly, the finite intersection of a finite set an an infinite set is finite, so A intersection B is in tau
A is infinite, B is infinite --> in this case, we use the other property of the sets in tau; since neither A nor B are finite, then both of their complements are finite. We need to show that A intersection B is open, or in other words that its complement is finite.

Thus \(\displaystyle (A \cap B)^C = A^C \cup B^C\)

Since the finite union of finite sets is finite, then \(\displaystyle A^C \cup B^C\) is finite, so therefore \(\displaystyle A \cap B\) is open since its complement is finite.

For condition 3, I have no idea what to do with an arbitrary union :oops:

Edit: I tried arbitrary union. I said there are two cases,

1) the arbirary union is finite - this trivially is in tau
2) the arbitrary union is infinite - I cannot prove that the complement of the arbitrary union is finite. In fact, I've come up with a counterexample. If you have the sets {1}, {2}, {3}, etc which are open sets in tau, their infinite union is clearly an infinite set. But also, the complement of this infinite set is also infinite, so I've run into a roadblock on how to prove that the arbitrary union is closed.

Also again on number 4, I can't even show that X = R is in tau. To me it looks like the collection of sets are all positive, but R contains negative numbers and it's not explicitly in the collection so I can't maneuver R into tau.

Edit: Ok, figured number 1 and 2 out! Woohoo. Brick wall on 3 due to counterexample. I actually figured out why R is in number 4, so I'll try again on the other conditions.
 
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Thanks, I will ask my professor on that one.

Update: 1 and 2 are solved, 3 the counterexample seems to deny that it is a topology, still working on 4. I actually realized what I was doing wrong and R is clearly in tau, so the first condition is met and I'm trying the other two.
 
4. Show that \(\displaystyle \tau = \{A \subset \mathbb{R} : \exists N \in \mathbb{N} \ \forall n \ge N \ \frac{1}{n} \in A\} \cup \emptyset\) is a topology for R.
Click to expand...

Ok here's what I have, it looks like the collection of sets has the form ([a, b]), where a <= 0, and b > 0. Just plug in some numbers to see that this true. In short, the left bound has to be at most 0 inclusive, since if it's even a tiny bit over 0, then it will fail the condition that for all natural numbers, 1/n is in the set. The right bound then has to be something above 0, since if it is zero or below then it will either result in the empty set or the single number 0, which doesn't contain any numbers of them form 1/n. I don't know how to notate that the boundaries can be either [] or (), because both work (try (-1, 1) and [-1, 1], they are both open sets in tau).

Condition 1: Trivially the emptyset is in tau by the very construction of the set. The real numbers can be said to have the form (a,b) for arbitrarily large a, b, so the real numbers is in tau.

Condition 2: Take 2 arbitrary sets ([a,b]) ([c,d]). Their intersection is ([e,f]) where e is either a or c whichever is larger, and f is either b or d, whichever is smaller. The intersection has the form of tau, so it's an open set.

Condition 3: Take an arbitrary union of the sets. Whoever has the largest negative number is the lower bound, and whoever has the largest positive number is the upper bound, and it's still in the form ([a,b]) so the arbitrary union is open.

This seems like a highly suspicious problem to me, what do you guys think? My proof looks very sloppy and informal, but I don't know how else to word it or notate it.
 
I don't understand your proof. Use the definition you are given.

If A and B belong to T then there exist N and M such that for all n>=N, m>=M, 1/n belongs to A, 1/m belongs to B.

For intersections you want (I'll do the case for n=2), There is a K such that for all k >= K, 1/k belongs to A intersect B.

Let K=max{M,N}. Then if k >= K, we know k>= M, k>=N, so 1/k belongs to A and 1/k belongs to B. Hence 1/k belongs to their intersection.
 
Ah, thank you for clearing my head, much better way of looking at it without resorting to my sucky notation and trying to intersect the entire set.

How might you go about showing the arbitrary union part? Would it simply be K = min(M,N,O,.....) where M,N,O.... represent the lower bound such that for all m >= M, n>= N, o>=O....that 1/m belongs in set A, 1/n belongs in set B, 1/o belongs in set C, etc, etc.

Because for arbitrary union we're looking for the largest 1/n, so we need the smallest n, right?
 
Arbitrary union is immediate, you are over-thinking it...
 
Wait wait wait, I think I might be seeing it, just let K = 1, since that would be the minimum possible natural number?

Edit: no that can't be it, because if I union 2 sets A and B where M = 10, N = 11, then K =1 will not work. Hmm, it's immediate?

I know this sounds silly, but how do I sort of word it formally? This is how I've started it:

Let \(\displaystyle A_{\alpha} \in \tau\), where \(\displaystyle $\alpha \in a$\). Consider \(\displaystyle $\bigcup_{\alpha \in a}{A_{\alpha}}$\).

The definition of union I understand intuitively but I can't seem to push it onto this example formally. I've been taught that the union of sets is the set of points in X such that there exists a set which contains that point.
 
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Still over thinking it! Let {A_i} be an infinite collection in T and let U be their union. Then consider any set A in this collection. By definition there is an N such that n>=N implies 1/n is in A. Hence for all n>=N, 1/n is in U.
 
Oh wow, my brain just unraveled after I saw what you did there. Very sneaky. Thanks man. How'd you get so good at this o.o
 
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