Show this is a topology?

[thedarjeeling]

Let the following be a topology on the natural numbers: \(\displaystyle \tau = \{ \mathbb{N}, \emptyset, \{1\}, \{1,3\}, \{1,3,5\} \ldots \}\)

Intuitively I know this is a topology, but I'm having trouble writing the proof nicely

1. Clearly, by construction of the set, the emptyset and the universe (N) are in the topology.

2. Take 2 arbitrary open sets, and consider the intersection. There are three cases

a) one of them is N - the other one must be a subset of N, hence the intersection is just the subset, which is open to begin with
b) one of them is the emptyset - then the intersection regardless is the emptyset ,and the emptyset is in tau
c) both are not n and both are not the emptyset - by construction of the set, then they are both finite sets. Take the maximum of the two; it has the form {1,3,5....2n+1}. Clearly both sets must be subsets of the maximum by the construction of the set, for the other one must have the elements {1,3,5...2m+1} where m<=n. Since one is the subset of the other, the intersection is just the subset which is again, open to begin with.

3. Take an arbitrary union of open sets. There are some cases:

a) suppose that they are all empty, the intersection is empty, and the emptyset is in tau
b) suppose that one of them is not empty, the union is either infinite or finite. If it's infinite, then it has to be equal to the natural numbers, and the natural numbers is in tau. If it's finite, then by construction of the set, all the sets are subsets of the largest finite set since they all have elements in the form {1,3,5...2n+1}. Since they are all subsets of the largest set, then the union is just the largest set which is open to begin with.

Does that sound formal/logical enough?

Edit: part 3 cannot be right where I say "has to be equal to the natural numbers" because you don't have the even numbers, but the natural numbers has even integers.

Maybe it's not a topology, since the arbitrary union could be equal to all the odd natural numbers, but there is no such infinite open set in the topology to begin with?

 

[pka]

Let the following be a topology on the natural numbers: \(\displaystyle \tau = \{ \mathbb{N}, \emptyset, \{1\}, \{1,3\}, \{1,3,5\} \ldots \}\)
Intuitively I know this is a topology, but I'm having trouble writing the proof nicely
1. Clearly, by construction of the set, the emptyset and the universe (N) are in the topology.
2. Take 2 arbitrary open sets, and consider the intersection. There are three cases
a) one of them is N - the other one must be a subset of N, hence the intersection is just the subset, which is open to begin with
b) one of them is the emptyset - then the intersection regardless is the emptyset ,and the emptyset is in tau
c) both are not n and both are not the emptyset - by construction of the set, then they are both finite sets. Take the maximum of the two; it has the form {1,3,5....2n+1}. Clearly both sets must be subsets of the maximum by the construction of the set, for the other one must have the elements {1,3,5...2m+1} where m<=n. Since one is the subset of the other, the intersection is just the subset which is again, open to begin with.

Well of course that is a topology
Each of those sets except \(\displaystyle \mathbb{N}~\&~\emptyset\) are what is known as an initial segment in the odd positive integers. So clearly that collection is closed with respect to finite intersection and arbitrary union.

 

[thedarjeeling]

Each of those sets except [FONT=MathJax_AMS]N[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]&[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]∅[/FONT] are what is known as an initial segment in the odd positive integers. So clearly that collection is closed with respect to finite intersection and arbitrary union.

But is it? Like I noted in the original post, if you take an arbitrary infinite union, you will get all the odd positive integers, but there is no such open set in the topology which consists of all the odd positive integers; all of these initial segments are finite.

Plus, the challenge is in writing it out, so while I agree that it may seem intuitive, the challenge is in writing it nicely and not just "clearly it's a topology!" statements.

 

[pka]

But is it? Like I noted in the original post, if you take an arbitrary infinite union, you will get all the odd positive integers, but there is no such open set in the topology which consists of all the odd positive integers; all of these initial segments are finite.

Plus, the challenge is in writing it out, so while I agree that it may seem intuitive, the challenge is in writing it nicely and not just "clearly it's a topology!" statements.

Well if you understand that the only infinite set in that collection is \(\displaystyle \mathbb{N}\) then you do not have a topology.
I assumed the the set of all odds was included in the collection. If not then you are correct,