The final segment topology of the naturals

Question

We define the topological space $(\mathbb{N},\tau)$. The open sets of this topological spaces are the sets of the form $\{n+1,n+2,\dotsc\}$, for some $n\in\mathbb{N}$. This topology is called the final segment topology.

My question is: is this topology defined above the finite complement topology on the naturals?

Ι know that the only closed sets in the finite complement topology are the finite sets.

So if $m\in\mathbb{N}$ the set $A=\{1,\dots,m-1,m+1,\dotsc\}$ is not open in the final segment topology, therefore $\mathbb{N}\setminus\{A\}$ is not closed in this topology. But is closed in the finite complement topology so the two topologies are different.

Is this example right?

You are correct: the two topologies are not the same. Your example is correct if you specify that $m>1$. (By the way, the word that you want is complement, not compliment.) — Brian M. Scott, Sep 01 '16 at 16:53

score 1 · Accepted Answer · answered Sep 01 '16 at 17:23

Clearly every open set in the final segment topology is open in the finite complement topology.

However, there are open sets in the finite complement topology that are not open in the final segment topology, for instance $$ \mathbb{N}\setminus\{2\} $$

So, yes, your example is correct.

The only closed point in the final segment topology is $1$ (it would be $0$, if you consider it in the natural numbers), whereas all points are closed in the finite complement topology.

The final segment topology of the naturals

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