Does anybody know which sets are both open and closed, and which sets are neither open nor closed for the Fort Space Topology?

Printable View

- Feb 7th 2010, 08:24 AMCairoFort 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?

- Feb 7th 2010, 08:46 AMMoo
Hello,

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

Let X be the infinite set. Fix $\displaystyle p\in X$.

Open sets A :

1.a) $\displaystyle p\not\in A$

1.b) or $\displaystyle A'=X\setminus A$ is finite

Closed sets B :

2.a) $\displaystyle p\in B$ (complement of a set in which p is not)

2.b) or $\displaystyle B$ 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 ! - Feb 7th 2010, 10:24 AMCairo
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? - Feb 7th 2010, 10:30 AMMoo
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 $\displaystyle A \subset U$, $\displaystyle X\setminus U \subset X\setminus A$. Thus $\displaystyle X\setminus U$ is finite... - Feb 8th 2010, 09:46 AMCairo
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)