Hausdorff and Compact Space

• Jan 29th 2010, 02:51 AM
Cairo
Hausdorff and Compact Space
Let X be an infinite set and p a point in X, chosen once and for all. Let T be the collection of subsets V of X, for which either p is not a member of V, or p is a member of V and its complement ~V is finite.

I have proved that T is a topology on X, but I don't seem to be able to prove that;

a) (X,T) is a Hausdorff space.
b) (X,T) is a compact space.
• Jan 29th 2010, 04:13 AM
tonio
Quote:

Originally Posted by Cairo
Let X be an infinite set and p a point in X, chosen once and for all. Let T be the collection of subsets V of X, for which either p is not a member of V, or p is a member of V and its complement ~V is finite.

I have proved that T is a topology on X, but I don't seem to be able to prove that;

a) (X,T) is a Hausdorff space.
b) (X,T) is a compact space.

You don't say whether the sets V built as above are defined to be open or closed, but I'll assume they're open.

Let $x,y\in\left(X,T\right)\,,\,\,x\neq y$ . If both $x\,,\,y\neq p$ then $x\notin \{y\}\,,\,\,y\notin \{x\}\,,\,\,\{x\}\cap \{y\} = \emptyset$ and we're done, else: suppose $x\neq y=p$ , then $p\notin \{x\}\,,\,\,x\notin X\setminus \{x\}$ and

again we're done since of course $\{x\}\cap \left(X\setminus \{x\}\right)=\emptyset$ (Pay attention to the fact that in each case you MUST show the chosen subsets are open, each contains one

of the points but not the other one and their intersection indeed is empty)

Tonio