find a topology where the sequence $\left(\frac1n\right)$ converges to $1$ can someone help me to find a topology where the sequence $\left(\frac1n\right)$ converges to an unique value which is $1$?
I was thinking of the Trivial topology $(X,T)$ but it's not the wanted topology because the sequence converges to every point of $X$.
 A: Paint a gigantic "$0$" symbol on $1$, and paint a gigantic "$1$" symbol on $0$. Now use the standard topology, except use the symbols you painted on $0$ and $1$ instead of the usual meanings of $0$ and $1$. 
To put this another way, let $f : \mathbb{R} \to \mathbb{R}$ be the function defined by $f(0)=1$, $f(1)=0$, and $f(x)=x$ if $x \ne 0,1$. Define $U \subset \mathbb{R}$ to be open in the topology $T$ if and only if $f(U)$ is open in the standard topology.
A: Just use the standard topology on $\mathbb{R}$, except switching $0$ and $1$ in the definition of the open sets.
A: Let $X=\{\frac{1}{n}:n\in\mathbb{N}\}$. Define a metric $d:X\times X\rightarrow\mathbb{R}$ by $d(x,y)=|x-y|$ if $x,y\neq 1$ otherwise $d\left(1,\frac{1}{n}\right)=\frac{1}{n}$. Also $d(1,1)=0$ of course. This is trivially a metric as all we have done is replaced the ``name" of $0$ with $1$. Then you get that $\left\{\frac{1}{n}\right\}$ converges to $1$ and only $1$ as you wanted.
A: In the topological space (*) induced by the metric $$d(x,y) = \inf_{n\in \mathbb{Z}} |x-y-n|, \qquad (x,y) \in (0,1]^2$$
then $$\lim_{n \to \infty} d(1,\frac1n) =0$$
(*) Called the circle $\mathbb{S}_1$ or $\mathbb{R}/\mathbb{Z}$
