# A countable Tychonoff space is normal?

I am trying to prove a countable tychonoff space must be normal but I cannot.

Here is my work so far: We take two disjoint closed sets $F_1$ and $F_2$. Since $F_1$ is countable and $F_2$ is closed and $X$ is Tychonoff we can let $F_1=\{x_1,x_2\dots\}$ and for each $i$ take a continuous functions from $X$ to $[0,1]$ so that $f(x_i)=1$ and $f(F_2)\subseteq \{0\}$. Taking the series $\sum\limits_{i=1}^\infty \frac{f(x)}{2^n}$ gives us a continuous function satisfying $f(x)>0$ if $x\in F_1$ and $f(x)=0$ if $x\in F_2$ (Continuity is because of the Weierstrass M test).

I have no idea how to proceed, I want my function to be $1$ in $F_2$ without losing the fact that it is $0$ in $F_2$.

Things that may be useful:

The maximum of two continuous functions is continuous

The product of two continuous functions is continuous.

$T_{3}$ and Lindelöf imply $T_{4}$. Countable implies Lindelöf, Tychonoff implies $T_{3}$.

• Thank you, how do you prove this? – Onir May 25 '15 at 0:35
• Countable implies Lindelöf trivially. You can find a proof for second countable + regular imply normal, on chapter 32 (page 200) of Topology by Munkres that is easily adapted to a proof for Lindelöf + regular imply normal. – JKEG May 25 '15 at 0:43
• oh ok, thank you kindly – Onir May 25 '15 at 0:44
• I found a solution using $T_{3.5}$ and not just $T_3$, what do you think? – Onir May 25 '15 at 16:55

Every countable space is Lindelöf, and every Tikhonov space is $T_3$. A Lindelöf $T_3$-space is paracompact, and a paracompact Hausdorff space is $T_4$.

We take two disjoint closed sets $F_1$ and $F_2$. Since $X\setminus F_2$ is countable and $F_2$ is closed and $X$ is Tychonoff we can let $X\setminus F_2=\{x_1,x_2\dots\}$ and for each $i$ take a continuous functions from $X$ to $[0,1]$ so that $f(x_i)=1$ and $f(F_2)\subseteq \{0\}$. Taking the series $\sum\limits_{i=1}^\infty \frac{f(x)}{2^n}$ gives us a continuous function satisfying $f(x)>0$ if $x\not \in F_2$ and $f(x)=0$ if $x\in F_2$ (Continuity is because of the Weierstrass M test).

This gives us a function $f$ satisfying $f(F_1)=0$ and $f(X\setminus F_2)\cap\{0\}=\emptyset$. We can use the same argument again to get a function $g$ satisfying $g(x\setminus F_1)\cap \{0\}=\emptyset$ and $g(F_1)\subseteq \{0\}$

We claim the function $h(x)=\frac{f(x)}{f(x)+g(x)}$ separates $F_1$ and $F_2$, It is clear that $f+g$ is continuous and because $F_1$ and $F_2$ are disjoint $f(x)>0$ or $g(x)>0$ and so $f+g$ is a function that is never zero, because of this $\frac{f(x)}{f(x)+g(x)}$ is continuous. We now prove $h$ separates $F_1$ and $F_2$ Proof: take $x\in F_1$, then $f(x)=0$ and so $h(x)=0$. Take $x\in F_2$. Then $g(x)=0$ and so $\frac{f(x)}{f(x)+g(x)}=\frac{f(x)}{f(x)}=1$ as desired.