I need to prove that there exists a homeomorphism f from R2 to itself with its standard topology such that:
when considering any pair of three distinct points (x1, x2, x3) and (y1, y2, y3), f maps (x1, x2, x3) -> (y1, y2, y3)
So far, I've shown that there exists a homeomorphism from the topological space with set X= {x1,x2,x3} and standard topology to Y={y1,y2,y3} with the same topology.
In order to extend this to R2, I'm considering the function
g = { f if x in X
{ Id if x in R\X
Is it correct to assume that, since f is on an open, it can be defined in a way that there is no point of discontinuity when g runs from R\X to X?
If not, how can I define a homeomorphism that satisfies the conditions in the problem?