Interesting topologies on $\mathbb{R}$? It is evident in with respect to the usual topology, the sequence $0.9,0.99,0.999,0.9999,\ldots$ converges to $1$.
I am also aware that the same sequence does not converge to $1$ with respect to the lower limit topology.
My question is: Does there exist a topology on $\mathbb{R}$ with respect to which the aforementioned sequence converges to a real number that is NOT $1$?
 A: Yes. Take any injective function $f\colon\Bbb R\longrightarrow\Bbb R$ such that $f(x)=x$ for any $x$ from your sequence, but such that $f(1)\neq1$; for instance, you can take$$f(x)=\begin{cases}x&\text{ if }x<1\\x+1&\text{ if }x\geqslant1.\end{cases}$$Then consider the topology induced by the distance $d$ defined by$$d(x,y)=\bigl|f(x)-f(y)\bigr|.$$With repect to this topology, your sequence converges to $f(1)$.
A: Yes, of course. A couple of uninteresting examples:

*

*in the indiscrete topology that sequence converges to $2$. And to $1$, and to $0$, and to every other point.


*if $f(x)=\begin{cases}0&\text{if }x=1\\ 1&\text{if }x=0\\ x&\text{if }x\ne 0\land x\ne 1\end{cases}$ and $\tau=\{f^{-1}[\Omega]\,:\, \Omega\text{ open in the Euclidean topology}\}$, then that sequence $\tau$-converges to $0$ and to no other point.
A: Take into account that a topology on a set defines its structure, without it $\mathbb R$ is the same as any other set with $\aleph_1$ cardinality. Take any real point $a$ and define the distance in $\mathbb R$ as $d(x,y)=|x-y|$ if $x,y\not\in\{1,a\}$, $d(1,y)=d(y,1)=|a-y|$ and $d(a,y)=d(y,a)=|1-y|$, it is clear that under this topology any sequence that converges to $1$ in the usual topology converges to $a$ in this one.
