Let $C_b(\mathbb R)$ denote the set of bounded continuous function from $\mathbb R$ to $\mathbb R$. We say that $M\subseteq C_b(\mathbb R)$

  1. separates points $:\Leftrightarrow$ $$\forall x,y\in\mathbb R:x\ne y\Rightarrow\exists f\in M:f(x)\ne f(y)\tag1$$
  2. strongly separates points $:\Leftrightarrow$ $$\forall x\in\mathbb R,\delta>0:\exists k\in\mathbb N,\left\{f_1,\ldots,f_k\right\}\subseteq M:\inf_{y\::\:d(x,y)\:\ge\:\delta}\max_{1\le i\le k}|f_i(x)-f_i(y)|>0\tag2$$

How can we show that $C_c^\infty(\mathbb R)$ strongly separates points?

It's clear that $C_c^\infty(\mathbb R)$ separates points.

  • $\begingroup$ Do you mean $C_b^\infty$ in the places you've written $C_c^\infty$? $\endgroup$ – Trevor Gunn Nov 28 '18 at 16:53
  • $\begingroup$ Do you mean $f_i$ instead of $h_i$? $\endgroup$ – Paul Frost Nov 28 '18 at 16:55
  • $\begingroup$ @PaulFrost Sorry, fixed that. $\endgroup$ – 0xbadf00d Nov 28 '18 at 16:56
  • $\begingroup$ @TrevorGunn No, I mean $C_c^\infty$. $\endgroup$ – 0xbadf00d Nov 28 '18 at 16:56
  • 2
    $\begingroup$ Cant you just take a single $f\in C^\infty_c(\mathbb R)$ with $f(x)=1$, $\mathrm{supp}(f)\subset B(x,\delta)$? Then you would have $f(x)-f(y)=1$ for $d(x,y)\geq\delta$. $\endgroup$ – Federico Nov 28 '18 at 16:58

Given $x\in\mathbb R$ and $\delta>0$, take a single function $f\in C^\infty_c(\mathbb R)$ with $f(x)=1$ and $\mathrm{supp}(f)\subset B(x,\delta)$.

Then $f(x)-f(y)=1$ for $d(x,y)\geq\delta$.

  • $\begingroup$ Oops, way to simple. $\endgroup$ – 0xbadf00d Nov 28 '18 at 17:10
  • $\begingroup$ Ah those lucky days when we can actually solve something! So satisfying... :-) $\endgroup$ – Federico Nov 28 '18 at 17:11

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