Completely regular-Topology Prove that every metric space is a Tychonoff space.
Can somebody please help me to show this space satisfies the completely regular axiom and the $T_1$ axiom.
 A: According to Wikipedia, "$X$ is a Tychonoff space, ... if it is both completely regular and Hausdorff."
Every metric space is Hausdorff (but perhaps you might want to show that). 
And, again according to Wikipedia, "$X$ is a completely regular space if given any closed set $F$ and any point $x$ that does not belong to $F$, then there is a continuous function $f$ from $X$ to the real line $\mathbb R$ such that $f(x)$ is $0$ and, for every $y$ in $F$, $f(y)$ is $1$."
Let $F \subset X$ be closed. Let $x_0 \in F^c $. Since $F^c$ is open there is $\varepsilon > 0$ such that $B(x_0, \varepsilon) \subset F^c$. Define $$ f(x) = \begin{cases} \frac{d(x,x_0)}{\varepsilon} & x \in B(x_0, \varepsilon) \\ 1 & x \in B(x_0, \varepsilon)^c \end{cases}$$
Then $f(x_0) = 0$ and $f\mid_F = 1$. Now show that $f$ is continuous.
A: $T_1$: By the triangle inequality $\delta_p \colon X \to [0,\infty), x \mapsto d(p,x)$ is continuous and it is zero iff $x = p$. Therefore $\{p\} = \delta_p^{-1}\{0\}$ is closed. 
$T_{3\frac{1}{2}}$: Let $A$ be closed and set $\delta_{A}(x) = \inf_{a \in A} d(x,a)$. Again the triangle inequality shows that $\delta_A$ is continuous. Moreover, $\delta_A(x) = 0$ iff $x \in A$, so if $p \notin A$ the function $\frac{\delta_A}{\delta_{A}(p)}$ is zero on $A$ and $1$ on $p$.
A: Every metric space is normal Hausdorff, hence for any two disjoint nonempty closed sets $A$ and $B$, there is a function that sends $A$ to $0$ and $B$ to $1$. This is true if $A$ is a singleton, so the space is Tychonoff.
