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Closure, Interior, Boundary and Exterior of a Set in different topologies...

This is something I seem to be majorly struggling with I am looking at the set of all $A = \mathbb R - \mathbb Q $. I need to determine the closure, interior, boundary and exterior in different topologies.

I need to do this for the usual topology, the finite complement topology and the countable complement topology.

The usual topology is defined to be $X$ and the empty set are in the topology, the infinite collection of unions is in the topology and the finite number of intersections is in the topology.

I think the $Cl(A)= \mathbb R $, $Int(A) = \emptyset$, $Bd(A) = \mathbb R$ and I am not sure what the exterior of A is.

The finite complement topology is defined to be "Let $X$ be an set. The collection $\mathfrak T$ of all subsets $U$ of $X$ such that either $U = \emptyset$ or $ X-U$ is a finite set."

The countable complement topology is defined to be $\{ U : \mathbb R - U$ is countable$\}$ $\cup \{\emptyset, \mathbb R\}$ is a topology on R.

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  • $\begingroup$ The exterior of $A$ is just the complement of Cl$(A)$, which in this case is empty. It can also be realized as the interior of the complement of $A$. $\endgroup$ – Shalop Apr 22 '15 at 22:41
  • $\begingroup$ Thank you. I thought I might be making that too complicated. Any insight on the other topologies? I would even love some sites to go to for further reading. $\endgroup$ – user219081 Apr 22 '15 at 23:19
  • $\begingroup$ For the finite (resp. countable) complement topology, here's a start: a set is closed if and only if it is either $\mathbb{R}$ or finite (resp. countable). So the closure of an uncountable set in the finite (resp. countable) complement topology must be $\mathbb{R}$. So now all that is left is to find the interior, and the rest follows by the standard procedure (e.g. the boundary is the closure minus the interior, etc.) A hint for the interior in the two complement topologies: the interior of an open set is the set itself. $\endgroup$ – Ian Apr 22 '15 at 23:26
  • $\begingroup$ Oh sorry, I was only talking about the usual topology. I didn't read the entire question. Sorry! $\endgroup$ – Shalop Apr 23 '15 at 1:31

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