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

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.

• Do you mean $C_b^\infty$ in the places you've written $C_c^\infty$? – Trevor Gunn Nov 28 '18 at 16:53
• Do you mean $f_i$ instead of $h_i$? – Paul Frost Nov 28 '18 at 16:55
• @PaulFrost Sorry, fixed that. – 0xbadf00d Nov 28 '18 at 16:56
• @TrevorGunn No, I mean $C_c^\infty$. – 0xbadf00d Nov 28 '18 at 16:56
• 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 $d(x,y)\geq\delta$. – 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 $$d(x,y)\geq\delta$$.