Does there exist a metric space $(\mathbb{R}, d)$ such that the sequence $\{\frac{1}{n}\}$ converges to $3?$
Here is an answer Sequence converging to different limits
I have tried this: Let $A=\{\frac{1}{n}: n\in \mathbb{N}\}, B=3+A$. Also let
$f(x)= \begin{array}{cc} \{ & \begin{array}{cc} x & x\notin A\cup B \\ x+3 & x\in A \\ x-3 & x\in B \end{array} \end{array} $
and $d(x,y)=|f(x)-f(y)|$. Then the sequence converges to 3
Yes, this is a successful construction!
In general, if $(M,\delta)$ is any metric space and $f\colon \Bbb R\to M$ is any injective function, then $d(x,y)=\delta(f(x),f(y))$ defines a metric on $\Bbb R$, called the "pullback" of $\delta$ to $\Bbb R$. (The idea being, just measure distances "over here" by seeing what the distances of the images are "over there".) Your construction is a special case of this, with $M =\Bbb R$ and $\delta$ being the usual metric on $\Bbb R$.
(Indeed, pullbacks are defined for any set, not just $\Bbb R$.)
