We say two topologies $\tau$ and $\rho$ on $X$ are similar if the set of continuous functions $f:(X,\tau) \rightarrow (X,\tau)$ is the same as the set of continuous functions $f:(X,\rho)\rightarrow (X,\rho)$.


Does there exist a topology $\tau$ that is similar to the euclidean topology on $\mathbb R$?


This was asked [here](https://math.stackexchange.com/questions/1890383/is-there-another-topology-on-mathbbr-that-gives-the-same-continuous-functio/1890409#1890409) but all we could prove is that $\tau$ must be a refinement of the euclidean topology.


Regards.



can another topology be given to $\mathbb R$ so it has the same continuous maps $\mathbb R\rightarrow \mathbb R$?
The only topology similar to the Euclidean topology on $\mathbb{R}$ is the Euclidean topology. 




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Suppose there is such a topology $\tau$. I'll use "open," "continuous," etc. to mean with respect to the Euclidean topology and "$\tau$-open" etc. for $\tau$.


Since $\tau$ is a refinement of the Euclidean topology, there exists a point (without loss of generality $0$) and a $\tau$-open neighborhood $U$ which does not contain any open interval containing $0$. Then there is a sequence $x\_i$ in the complement of $U$ converging to $0$. For convenience choose a monotonic subsequence $y\_i$, without loss of generality positive, and suppose $y\_1=1$.


Now consider the function $\mathbb{R}\to\mathbb{R}$ which fixes the complement of $(0,1)$, takes $[\frac{1}{2n},\frac{1}{2n-1}]$ to $y\_n$, and takes $[\frac{1}{2n+1},\frac{1}{2n}]$ to $[y\_{n+1},y\_n]$ linearly. This function is continuous, hence $\tau$-continuous. 


Then the preimage of $(-1,1)\cap U$ is a $\tau$-open subset $V$ of $(-1,0] \cup \bigcup\_n (\frac{1}{2n+1},\frac{1}{2n})$ which contains $0$. 


There is a homeomorphism (hence a $\tau$-homeomorphism) $\phi$ of $\mathbb{R}$ which fixes the complement of $(0,2)$, takes $[1,2]$ to $[\frac{1}{2},2]$ linearly, and takes $[\frac{1}{k+1},\frac{1}{k}]$ to $[\frac{1}{k+2},\frac{1}{k+1}]$ linearly. 


Then $\phi(V)\cap V$ is a $\tau$-open subset of $(-1,0]$ containing $0$. 


But then $(-\infty,0]$ is $\tau$-open, so that $\mathbb{R}$ is not $\tau$-connected, a contradiction since then there is a $\tau$-continuous map sending $(-\infty,0]$ to one point and $(0,\infty)$ to another.


