# The metric spaces are Tychonoff spaces

Following a reference from "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.

Since the sum and the product of real continous functions, defined in any topological space, are continuous we can easily prove that every metric space is a Tychonoff spaces. So we suppose that $$(X,d)$$ is a metric space and we consider a closed set $$F$$ a point $$p\notin F$$ and so we define a function $$f:X\rightarrow[0,1]$$ by the condiction $$f(x)=\frac{d(x,p)}{d(x,p)+d(x,F)}$$ where $$d(x,F):=\mathscr{inf}\{d(x,a):a\in F\}$$. So we observe that $$f(p)=0$$ and $$f(x)=1$$ for any $$x\in F$$; then $$f$$ is continuous, since any function $$g:X\rightarrow\mathbb{R}$$ defined as $$g(x)=d(x,A)$$

for any $$x\in X$$ and for any fixed $$A\subseteq X$$, is a countinuous function and so $$f$$ is continuous function, since the sum and the ratio of continuous function is a continuous function.

Well by definition of Tychonoff space we can claim that the metric space $$(X,d)$$ is a Tychonoff space.

Well unfortunately I don't be able to prove that the function $$g$$ is continuous: could someone help me, please?

Not only $$g$$ is continuous, $$g$$ is, in fact, non-expansive: $$|g(x) - g(y)| \le d(x, y),$$ for all $$x, y$$. To prove this, suppose $$x, y \in X$$ and $$a \in A$$. Then, by triangle inequality, $$d(x, a) \le d(x, y) + d(y, a).$$ By definition of $$d(x, A)$$, we have $$d(x, A) \le d(x, a)$$, so, $$d(x, A) \le d(x, y) + d(y, a)$$ for all $$x, y \in X$$ and $$a \in A$$. Because this holds for all $$a \in A$$, we therefore see that $$d(x, A) - d(x, y)$$ is a lower bound for $$d(y, a)$$ when $$a \in A$$. So, the infimum must be larger than this lower bound, i.e. $$d(x, A) - d(x, y) \le \inf_{a \in A} d(y, a) = d(y, A).$$ Thus, $$d(x, A) - d(y, A) \le d(x, y)$$. Symmetrically, $$d(y, A) - d(x, A) \le d(y, x) = d(x, y)$$, hence $$|d(x, A) - d(y, A)| \le d(x, y),$$ as promised.
• Sorry, but I don't understand why if $|g(x)-g(y)|\le d(x,y)$ then g is continuous? Could you explain better, please? – Antonio Maria Di Mauro Mar 4 '20 at 10:34
• For the first question, recall that $|a| = \max\{a, -a\}$, so $|a| \le b$ if and only if $a \le b$ and $-a \le b$. That's what I'm doing here, with $b = d(x, y)$ and $a = d(x, A) - d(y, A)$. For the second question, fix $\varepsilon > 0$, and let $\delta = \varepsilon$. To show $g$ is continuous at $y$, then observe $$d(x, y) < \delta \implies |g(x) - g(y)| \le d(x, y) < \delta = \varepsilon,$$ proving continuity at $y$ (which is also uniform continuity!). – user754697 Mar 4 '20 at 10:37
• @AntonioMariaDiMauro We can take $\delta=\varepsilon$ at any point! – Henno Brandsma Mar 4 '20 at 10:46
• @user754697 So to prove the continuity of $g$ I must prove that for every neighborhood $U_{g(x)}$ of $g(x)$ there exist a neighborhood $V_x$ of $x$ such that $g(V_x)\subseteq U_{g(x)}$: so in your proof the neighborhood of $x$ is the ball $B(x,\delta)$ but I don't understand what is the neighborhood of $g(x)$. Could you explain to me this? – Antonio Maria Di Mauro Mar 4 '20 at 10:49
• For any neighbourhood $U_{g(x)}$, by definition of the topology induced by a metric, there must exist an open ball $B(g(x), \varepsilon) \subseteq U_{g(x)}$. Then, $B(x; \varepsilon)$ is a neighbourhood of $x$, and $g(B(x; \varepsilon)) \subseteq B(g(x), \varepsilon) \subseteq U_{g(x)}$. Note that if $y \in B(x; \varepsilon)$, then $d(x, y) < \varepsilon$, so $|g(y) - g(x)| < \varepsilon$, and so $g(y) \in B(g(x); \varepsilon)$. – user754697 Mar 4 '20 at 11:44