Let $\alpha\in\mathbb{R}^{d}$. Construct a continuous function $f:\mathbb{R}^{d}\rightarrow\mathbb{R}$ such that $0\leq f(x)\leq1$ for all $x\in\mathbb{R}^{d}$; $f(\alpha)=0$ and $f(x)=1$ if $|x-\alpha|\geq\epsilon$ for a given $\epsilon>0$.
My initial reaction is that $f$ is not continuous. There seems to be a jump at $\alpha$. I use the tag ``probability theory'' because this is from a probability class. Given this context, I wonder whether $f$ is a cumulative distribution function. I also considered constructing this directly by $f(x)={\displaystyle \frac{|x-\alpha|}{1+|x-\alpha|}}$. But it does not satisfy $f(x)=1$ if $|x-\alpha|\geq\epsilon$ for a given $\epsilon>0$.
Thank you!