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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.

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  • $\begingroup$ More generally, you can prove that a topological space is completely regular iff it admits a uniform structure compatible with its topology; here, the metric gives such a uniform structure. $\endgroup$ – Seirios Oct 29 '12 at 12:36
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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.

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  • $\begingroup$ @ Matt,can you please help me to show this function is continuous? $\endgroup$ – ccc Nov 7 '12 at 14:48
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    $\begingroup$ @ccc Let's use that $d(.,.)$ is continuous in both arguments and that $1$ is also continuous. Then all we need to show is continuity at points $x$ such that $d(x,x_0) = \varepsilon$. Let $x$ be such a point. Then $x$ is in the closure of $B(x_0, \varepsilon)$ hence there is a sequence $x_n$ in $B(x_0, \varepsilon)$ converging to it, $x_n \to x$. Now since $d(.,.)$ is continuous, $d(x_0, x_n) \to d(x_0, x) = \varepsilon$. $\endgroup$ – Rudy the Reindeer Nov 7 '12 at 15:13
  • $\begingroup$ @Rudy There exist metric spaces and balls such that closure of an open ball is a proper subset of a closed ball. E.g. $\mathbb R^2 \backslash (-1,1) \times (-1,1)$ with the ball $B((1,0), 2)$. Its closure does not contain $(-1,0)$ but $B[(1,0), 2]$ does. So we don't know that $x$ is in the closure of $B(x_0, \epsilon)$. But actually it doesn't spoil your argument. If $x$ is a limit point, your argument is valid. If $x$ is not then in does not contains in $B(x_0, \epsilon)$ and has a neighborhood without points of $B(x_0, \epsilon)$. So the restriction of $f$ to that neighborhood is constant. $\endgroup$ – vanger Sep 3 '17 at 1:23
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$T_1$: 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 $p$.

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    $\begingroup$ If you see this, please try to register your account. It was just pointed out to me that your account has fragmented into 6 or 7. Registering would help prevent that. I've merged the fragments into this one. $\endgroup$ – Willie Wong Nov 28 '12 at 8:14
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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.

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