Final segment topology and finite-closed topology?

Nov 2012
197
30
Normal, IL USA
I'm trying to answer a question I found in an elementary topology textbook. I keep feeling I'm missing something.

Suppose N is the set of all positive integers and t consists of N, the empty set, and every set { n, n+1, ... } for n any positive integer. This is a topology and is called the “final segment topology.”

Is it the finite-closed topology? The finite-closed topology contains X (the "whole set"), the empty set, and all sets that have finite complements. These complements are the ONLY closed sets.

I think the answer is “yes.” If n = 1, then the “final segment” formed is N. If n = 4 (for example), then the complement is { 1, 2, 3 } which is a finite set, so all these “final segments” belong to this topology. There are other finite subsets, such as { 2, 4, 6 }, but they will be neither open nor closed, since they are not the complement of any “final segment” set.

Can anyone tell me if I'm right? Thanks in advance for any help.
 
Nov 2012
197
30
Normal, IL USA
No answer yet?

I rethought this problem, and changed my mind. I now think the answer is "no."

Here's my reason: consider a singleton set in N such as { 3 }. It is finite, and its complement in N is infinite, so it is a closed set in the "finite closed" topology. It's complement is { 1 , 2, 4 , 5 ... } which is open in the "finite closed" topology, but it is not a "final segment," so it's not open in the "final segment" topology. If I'm right, this means the two topologies have different set of open sets, and so are different.

Am I right about that? Thanks. R
 

Plato

MHF Helper
Aug 2006
22,490
8,653
No answer yet?
I rethought this problem, and changed my mind. I now think the answer is "no."
Here's my reason: consider a singleton set in N such as { 3 }. It is finite, and its complement in N is infinite, so it is a closed set in the "finite closed" topology. It's complement is { 1 , 2, 4 , 5 ... } which is open in the "finite closed" topology, but it is not a "final segment," so it's not open in the "final segment" topology. If I'm right, this means the two topologies have different set of open sets, and so are different.
That is correct. I did not answer before because I was not sure I understood your description.
I have seen this topology called long line topology and yours is restricted to the naturals.
And no it is not the cofinite topology. I was just not sure that is what you were asking.