Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

At first, I thought the answer is “yes.” I reasoned like this: 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.

After more thought, I changed my mind. Here's my new 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

share|improve this question
1  
You’re right: they are different topologies, and your example with $\{3\}$ correctly shows this. –  Brian M. Scott Dec 10 '13 at 0:02
1  
You are right. The final segment topology is strictly coarser than the cofinite topology. –  Daniel Fischer Dec 10 '13 at 0:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.