Fort Space Topology

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February 7th 2010, 08:24 AM
Cairo

Fort Space Topology

Does anybody know which sets are both open and closed, and which sets are neither open nor closed for the Fort Space Topology?
February 7th 2010, 08:46 AM
Moo

Hello,

Just write down what open and closed sets are in this topology.

Let X be the infinite set. Fix $p\in X$ .
Open sets A :
1.a) $p\not\in A$
1.b) or $A'=X\setminus A$ is finite

Closed sets B :
2.a) $p\in B$ (complement of a set in which p is not)
2.b) or $B$ is finite.

So now try to combine...
It's obvious that a set can't satisfy both 1.a) and 2.a), and can't satisfy both 1.b) and 2.b)

Now is it possible for a set to satisfy 1.a) and 2.b) ? Of course ! A finite subset of X which doesn't contain p will satisfy the two conditions. Thus it's both closed and open.
Similarly, it's possible to find a set that satisfies 1.b) and 2.a)...

A set that is neither open nor closed ? No it's not possible, because either p is in the set, either it's not (that's the law of excluded middle :D). If it is, then the set is closed, and if it's not, then the set is open !
February 7th 2010, 10:24 AM
Cairo

Thanks for this Moo.

Am I right in thinking that the neigbourhoods of the point p, are just the open sets U of p whose complement is finite?
February 7th 2010, 10:30 AM
Moo

Quote:

Originally Posted by Cairo

Thanks for this Moo.

Am I right in thinking that the neigbourhoods of the point p, are just the open sets U of p whose complement is finite?

Yes, the neighbourhood U has to contain an open set, say A, which contains p. So the complement of A will be finite (since it doesn't satisfy 1.a), it has to satisfy 1.b)).
And since $A \subset U$ , $X\setminus U \subset X\setminus A$ . Thus $X\setminus U$ is finite...
February 8th 2010, 09:46 AM
Cairo

After a sleepless night and much scribbling, I am still unable to prove this result for the Fort space topology.

Let (a_n) be a sequence in X, such that the set of the sequence, {a_n : n in N} is infinite. Using the definition of convergence in topological spaces, prove that (a_n) has a subsequence which converges to p.

If it were a metric space, I think i could do it, but as it stands I can't.

Could anybody offer a proof please?

(Headbang)