The sequence $(\frac 1n )$ of inverses of natural numbers converges to a limit other than $0$ Finding difficulty to construct a metric on $\Bbb R$ in which the sequence $(\frac 1n )$ of inverses of natural numbers that converges to a limit other than $0$.
 A: Given any metric, it is simple to construct a metric which essentially interchanges the role of two elements $a$ and $b$. Specifically, if $(X,d)$ is a metric space and $f:X\rightarrow X$ is a bijection then it is straightforward to show that
$$d_f(x,y):=d(f(x),f(y))$$
is also a metric on $X$. If we choose our bijection so that $f(a)=b$, $f(b)=a$ and $f(x)=x$ for $x\in X\setminus\{a,b\}$ then we obviously have $d_f(x,a)=d(x,b)$ and $d_f(x,b)=d(x,a)$ for any $x\in X$ with $x\neq a,b$.
For this particular problem, we take $X=\mathbb R$, $d$ as the Euclidean metric, $b=0$ and choose $a\in\mathbb R\setminus\{0\}$ arbitrarily. Then given the above construction, clearly $\frac1n\to a\neq0$. Explicitly, the metric in this case is given by
$$\begin{cases}
d(x,a)=d(a,x)=|x|,&x\neq0,a,\\
d(x,0)=d(0,x)=|x-a|,&x\ne0,a,\\
d(a,0)=d(0,a)=|a|,\\
d(x,y)=|x-y|,&x,y\ne0,a.
\end{cases}$$
A: The T-shaped set
$$T:=\{(x,0)\>|\>x>0\}\>\cup\>\{(0,1+x)\>|\>x\leq0\}\ \subset{\mathbb R}^2$$
inherits the euclidean metric from ${\mathbb R}^2$; whence is a metric space.
The map
$$\phi:\quad {\mathbb R}\to T,\qquad x\mapsto\cases{(x,0)\quad&$(x>0)$ \cr
(0,1+x)\quad&$(x\leq0)$\cr}$$
maps ${\mathbb R}$ bijectively onto $T$, wereby $\phi(-1)=(0,0)$. Consider now the metric
$$d(x,y):=\|\phi(x)-\phi(y)\|$$
on ${\mathbb R}$. With respect to this metric one has
 $$\lim_{n\to\infty}{1\over n}=-1\ .$$
