Originally Posted by

**zhandele** 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.