# $(R, \tau_f)$

• 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, $(R, \tau_f)$ satisfies T1.

• T2~T4:

Since $(R, \tau_f)$ satisfies T1, if it also saitisfies T4, then it must satisfies T3, hence T2.

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

• C1:

If $(R, \tau_f)$ satisfies C1, then for any element x in $(R, \tau_f)$, there is a countable neighborhood base of it, say U = {$U_i$}. $\cup (U_i)^c$ is also countable. R is uncountable, so there exists one element y, $y\neq x$, $y\not\in \cup (U_i)^c$. That is, y is contained in any $U_i$ in U.

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

• C2: Any C2 space is C1 space.

So $(R, \tau_f)$ does not satisfy C2.

# $(R, \tau_c)$

The situation about $(R, \tau_c)$ is similarly.

• T1:

Let x and y be 2 distinct points. {x} is countable, so it is close. So $y\in {x}^c$. That is,$(R, \tau_c)$ satisfies T1.

• T2~T4:

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

• C1 & C2

We need only to show that $(R, \tau_c)$ does not satisfies C1.

If $(R, \tau_c)$ satisfies C1, then for any element x in v, there is a countable neighborhood base of it, say U = {$U_i$}. $\cup (U_i)^c$ is also countable. R is uncountable, so there exists one element y, $y\neq x$, $y\not\in \cup (U_i)^c$. That is, y is contained in any $U_i$ in U.

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

#### Other

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