Does the set of Continuous functions always separate points and vanish nowhere?

Question

Let $X$ be an arbitrary metric space. Does $C(X;\mathbb{R})$ always separate points and vanish nowhere?

I think this is true because the functions $f(x) = x$ and $g(x) = 1$ are both in $C(X;\mathbb{R})$ and separate points and does not vanish.

Thank you for the help.

$f(x)=x$ is not an element of $C(X;\mathbb R)$ for a general metric space $X$. — Kavi Rama Murthy, Feb 01 '19 at 06:04

Henno Brandsma · Accepted Answer · 2019-02-01T09:01:54.830

3

Yes, these two functions do suffice to show that for $X\subseteq \mathbb{R}$. Or use $f$ and $g(x)=x+1$, as $f$ and $g$ never simultaneously are $0$. But the definition of $f$ assumes all points of $X$ are reals.

For general metric $X$ consider using $f_p(x) = d(x,p)$ for $p \in X$ as continuous functions, and also $g$.

edited Feb 01 '19 at 09:01

answered Feb 01 '19 at 05:30

Henno Brandsma

242,131

would you mind elaborating on your last statement for a general metric? – Tomislav Feb 01 '19 at 05:33
@MatteoLepur If $x \neq y$, what are $f_x(x)$ and $f_x(y)$? – Henno Brandsma Feb 01 '19 at 05:34
$f_x(x) = 0$ and $f_x(y)$ $\not=$ $0$. I don't see how this helps us. Thanks for the help. – Tomislav Feb 01 '19 at 05:38
@hennobrandsma you're saying since there are at least two different points in the metric space then $ f_p(x)= d(x,y) $ is not equal to zero for some x in X, do I understand that correctly? – Kaan Yolsever Feb 01 '19 at 05:39
@KaanYolsever If we have at least two points we can separate them by some $f_x$ and the constant function always shows non-vanishingness. – Henno Brandsma Feb 01 '19 at 05:49
@MatteoLepur The definition of separating points, is that for $x \neq y$ some $f$ has $f(x) \neq f(y)$. – Henno Brandsma Feb 01 '19 at 05:50
@MatteoLepur The fucntions $f_p$ I defined are just some of the continuous functions on $X$ but they suffice to show that the real-valued continuous functions separate points. – Henno Brandsma Feb 01 '19 at 05:54
@HennoBrandsma thank you very much. Perfect. – Tomislav Feb 01 '19 at 05:57
@HennoBrandsma sorry for bothering but why are metrics continuous in every metric space – Kaan Yolsever Feb 01 '19 at 06:00
2

@KaanYolsever search for “metric continuous” on this site. Plenty of answers. But basically the triangle inequality implies $|f_p(x) - f_p(y) \le d(x,y)$ which implies continuity right away. – Henno Brandsma Feb 01 '19 at 06:04

Does the set of Continuous functions always separate points and vanish nowhere?

1 Answers1