- 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.

- T2~T4:

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.

- C1:

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.

- T1:

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

- T2~T4:

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.

- C1 & C2

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.