Show that $f$ is a continuous function.

Question

Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$

Let $f(x) = \lim_{n\to \infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). We want to show that $f(x)$ is continous.

For $x,y\in X$ consider $$|f(x)-f(y)|\leq |f(x)-d(a_n,x)|+|d(a_n,x)-d(a_n,y)|+|f(y)-d(a_n,y)|$$ then for large enough $n$ we can make $$|f(x)-d(a_n,x)|<\epsilon/3, |f(y)-d(a_n,y)|<\epsilon/3.$$ And since $|d(a_n,x)-d(a_n,y)|<d(x,y)$ we have that

$$|f(x)-f(y)|<2\epsilon/3+d(x,y)$$ If we choose $\delta = \epsilon/3$ then we are done.

Is this proof correct?

Answer 1

There is a faster way: $$ |\,f(x)-f(y)|=\lim_{n\to\infty}|d(x,a_n)-d(y,a_n)|\le d(x,y). $$