# Property for X 3 spaces

#### Amer

##### Active member
if X is T3 space and A is infinite subset of X there exist $$U_1,U_2 , ...$$
open sets such that $$\overline{U_n} \bigcap \overline{U_m}$$ for $$n \neq m$$
and $$U_m \bigcap A \neq \phi$$ for all m (Use induction )

My work
I will work on $$x_1,x_2 ,...$$ subset of A

for k=2 let $$x_1 , x_2 \in A$$ there exist two disjoint open sets $$U_1 ,U_2$$ such that $$x_1 \in U_1 \; , \; x_2 \in U_2$$
from regularity we can find two open sets $$V_1 , V_2$$ such that
$$x_1 \in V_1 \subseteq \overline{V_1 } \subseteq U_1$$
$$x_2\in V_2 \subseteq \overline{V_2 } \subseteq U_2$$

and $$V_1 , V_2$$ have the properties we are searching for

now suppose it is true for k now want to test if it is true for k+1

$$x_1 , x_2 ,...,x_{k+1}$$
there exist two disjoint open sets $$U_n , U_m$$ such that $$x_n \in U_n$$ and $$x_m \in U_m$$ fixed these open sets
we have now for any $$x_i$$ of these x's k+1 open sets containing $$x_i$$ wihtout containing other x, Let
$$V_1 = \bigcap_i^{k+1} U_{i}$$ such that $$U_1 ,...,U_{k+1}$$ containing $$x_1$$
do the same for all x's we have now we use the regularity property we have $$H_1 \subseteq \overline {H_1} \subseteq V_1$$

these $$H_1,H_2 ,...,H_{k+1}$$ are the U's we are looking for

how about my proof any notes
Thanks