How to show the given function is continuous

Question

While solving a problem on compactness

, I just came across this function, I need to know whether it is well defined or not and if is well defined then need to prove that this function is continuous.

Let $(X,d)$ be a non-compact metric space and $\{x_n\}$ be a sequence which has no accumulation points. Then show that there exits a continuous real-valued unbounded function.

In MathStackExchange, one defined (Existence of a continuous function which does not achieve a maximum. ) $g(x)=\inf_nd(x,a)+1/n$ and then asked to take $f(x)=\frac{1}{g(x)}$ now $g(x)$ is not zero anywhere and he said $g(x)$ is continuous and $f(x_n)=n $ and so the function $f$ is continuous and unbounded.

My doubt is this function $g$ well defined? If yes what does it really mean? what is that 1/n? how does it varies? If this function is well defined, kindly help me to prove how that it is continuous.

Answer 1

$g$ is continuous because:

Let $\overline{x}\in X$, $\epsilon>0$ and choose $\delta=\frac{\epsilon}{2}$. Now by definition of $g(x)$ there exists $N\geq1$ such that $g(x)+\frac{\epsilon}{2}>d(x,x_{N})+\frac{1}{N}$ and also by definition of $g(\overline{x})$ we have $g(\overline{x})\leq d(\overline{x},x_{N})+\frac{1}{N}$, therefore: \begin{align*} g(\overline{x})-g(x)&< g(\overline{x})-d(x,x_{N})-\frac{1}{N}+\frac{\epsilon}{2} \\ &\leq d(\overline{x},x_{N})+\frac{1}{N}-d(x,x_{N})-\frac{1}{N}+\frac{\epsilon}{2} \\ &=d(\overline{x},x_{N})-d(x,x_{N})+\frac{\epsilon}{2} \\ &\leq d(\overline{x},x)+\frac{\epsilon}{2} \\ &<\epsilon,\ \text{for every}\ x\in B(\overline{x},\delta). \end{align*} To get $g(x)-g(\overline{x})<\epsilon$ repeat the above procedure but considering $g(\overline{x})+\frac{\epsilon}{2}>d(\overline{x},x_{N})+\frac{1}{N}$ and $g(x)\leq d(x,x_{N})+\frac{1}{N}$, thus $g$ is continuous in $X$ by definition of continuity.

$g$ being well define means that $g$ is always a real number and that is true because $g\geq0$ (is the infimum of non negative numbers), in fact, we can prove that $g>0$, since $(x_{n})_{n\geq1}$ does not have any accumulation point there exists $M\geq1$ and $c>0$ such $d(x,x_{n})\geq c$ for all $n\geq M$ (just by negating what to be an accumulation point means) and for $n<M$ set $b=\min_{n<M}(d(x,x_{n})+\frac{1}{n})>0$, therefore, $0<\min(b,c)\leq\inf_{n\geq1}(d(x,x_{n})+\frac{1}{n})=g(x)$.

Answer 2

Given a sequence $\{x_n\}$ in $X$. Consider the following sequence in $\mathbb R$: $$a_{n,x}:=d(x,x_n)+\frac{1}{n}$$ where $x\in X$ is chosen to be fixed, and $n\in\mathbb N$. This is a well-defined sequence (in $\mathbb R$) without any doubt. And then in his construction he defined a new mapping $g:X\to\mathbb R$ by $$g(x)=\inf_{n\in\mathbb N}(a_{n,x}).$$ Since the sequence $a_{n,x}$ is bounded below (clearly) by $0$, we can always take $\inf$ and get a finite number. Thus the mapping $g$ is well-defined as well.

Edit: as for the continuity of $g$, I think you can mimic the proof in this answer Infimum of a metric is continuous. Maybe you will use the fact that $\inf(A+B)=\inf A+\inf(B)$ for two bounded below subsets of $\mathbb R$.