winter85
Nov11-09, 01:21 AM
1. The problem statement, all variables and given/known data
This problem is from Schaum's Outline, chapter 7 #38 i believe.
Let f: (0, inf) -> [-1,1] be given as f(x) = sin(1/x), where R is given the usual euclidean metric topology and (0,inf) and [-1,1] are given the relative subspace topology.
Show that f is not an open map.
3. The attempt at a solution
I thought about this for so long but I can't see how this function is not an open map. Let A be an open subset of (0,inf) (we can take A to be an interval of the form (a,b) since these form a base for the topology). Pick any point x in A. Then there is an open neighborhood G of x that lies in A. If f is monotone in an open neighborhood of x, I can intersect that neighborhood with G to get a neighborhood H of x, lying inside A, on which f is monotone. Clearly in that case f[H] would be of the form (a,b) subset of [-1,1].
For the case when f is not monotone on any neighborhood of x, this only happens when f[x] = 1 or f[x] = -1. in this case, any open neighborhood of x would map to an interval of the form (a,1] or [-1, a). This is open in [-1,1] under the relative subspace topology.
Doesn't this show that f is open?
Any help would be greatly appreciated, thanks.
This problem is from Schaum's Outline, chapter 7 #38 i believe.
Let f: (0, inf) -> [-1,1] be given as f(x) = sin(1/x), where R is given the usual euclidean metric topology and (0,inf) and [-1,1] are given the relative subspace topology.
Show that f is not an open map.
3. The attempt at a solution
I thought about this for so long but I can't see how this function is not an open map. Let A be an open subset of (0,inf) (we can take A to be an interval of the form (a,b) since these form a base for the topology). Pick any point x in A. Then there is an open neighborhood G of x that lies in A. If f is monotone in an open neighborhood of x, I can intersect that neighborhood with G to get a neighborhood H of x, lying inside A, on which f is monotone. Clearly in that case f[H] would be of the form (a,b) subset of [-1,1].
For the case when f is not monotone on any neighborhood of x, this only happens when f[x] = 1 or f[x] = -1. in this case, any open neighborhood of x would map to an interval of the form (a,1] or [-1, a). This is open in [-1,1] under the relative subspace topology.
Doesn't this show that f is open?
Any help would be greatly appreciated, thanks.