#

- T1: For any 2 distinct elements x, y, there is a neighborhood of x which doesn’t contain y, and a neighborhood of y which doesn’t contain x.

Since {x} is finite hence close, satisfies T1.

Since satisfies T1, if it also saitisfies T4, then it must satisfies T3, hence T2.

Therefore, we only need to show that does not satisfies T2. Say x, y are 2 distinct points, and U is a neighborhood of x, then is a close set contaning y. Now any open neighborhood of y is cannot be finite, so non of them is contained in . That is, does not saitisfies T2, hence T3 or T4.

If satisfies C1, then for any element x in , there is a countable neighborhood base of it, say U = {}. is also countable. R is uncountable, so there exists one element y, , . That is, y is contained in any in U.

Now we consider R\{y}. It is a open neighborhood of x, and does not contain any item of U. So does not satisfy C1.

- C2: Any C2 space is C1 space.

So does not satisfy C2.

The situation about is similarly.

Let x and y be 2 distinct points. {x} is countable, so it is close. So . That is, satisfies T1.

We need only to show that does not satisfy T2. Say U is an arbitrary open neiborhood of x, then is a neighborhood of y, and it’s countable. So any subset of is countable, too. However, any open neighborhood of y must be uncountable. So does not satisfies T2.

We need only to show that does not satisfies C1.

If satisfies C1, then for any element x in v, there is a countable neighborhood base of it, say U = {}. is also countable. R is uncountable, so there exists one element y, , . That is, y is contained in any in U.

Now we consider R\{y}. It is a open neighborhood of x, and does not contain any item of U. So does not satisfy C1.

#### Other

Notice that the same proof is valid when R is substituted with any uncountable space.