Can someone help me to show that Q is not homeomorphic to N.
We know that there exists a bijection between Q and N. How can I show that f:Q onto N is not continuous ?
I also have trouble with showing any isometry is continuous and is a homeomorphism.
We know that f: M onto N is an isometry if d_N (fp, fq) = d_M (p,q) for each p,q in M. So, f preserves distance. We now want to show that f is continuous. Assume that f is not continuous, then there exists x in M such that f is not continuous at x. This implies that there exists epsilon greater than 0 st for each delta greater than 0 and for each y in M, we have d_M (x,y) less than delta implies d_N (f(x), f(y)) greater than or equal to 0.
I am not able to see a contradiction here. Anyone can help pls?
We want to show that f is a bijection, but this is given in the definition of isometry. We need to show that f is continuous, this will be given if I can prove part 1). The last proof is to prove that inverse of f is continuous. Anyone can help?
There is no need to use contradiction for Part (1). Itís totally straightforward.
Let . You want to show that f is continuous at x. Given , simply set and you have that , .
f is a function from M onto N; in other words, it is surjective.
Let . Then
So f is injective as well. Hence it is a homeomorphism.
Thanks a lot for your help, but I don't know why in part 2) showing f is bijective, then it is homemorphism. Since from what I learned, to show f is homeomorphism, I need to check 3 conditions: f is a bijection, f is continuous, and inverse of f is continuous.
Originally Posted by JaneBennet
However, I now see how to approach part 2) . Since f is a bijection, the inverse of f is also a bijection, and inverse of f also preserves distance. Hence, inverse of f is isometry. By applying part 1), inverse of f is continuous
Youíre right, one has to check that is continuous as well.
Since f is a bijection, we can write .
Let . You want to show that is continuous at . Given , simply set and you have that , .
Do you have any suggestion for me on how to show Q is not homeomorphic to N. Where Q is the set of rationals, and N is the set of naturals.
Show that there are no isometries between and . Make use of the fact that distances in can be arbitrarily small whereas distances in can’t be arbitrarily small. (I mean of course distances between distinct elements.)
I thought that isometry is a stronger property than homeomorphism. So, if f is an isometry, then f is a homeomorphism. However, if f is not an isometry, then it doesn't guarantee that f is not a homeomorphism. I think this way because I take the contrapositive statement: if f is not homeomorphism, then f is not isometry.
Originally Posted by JaneBennet
The correct contrapositive statement is: If f is not an isometry, then f is not a homeomorphism.
In any case, f can be an isometry without being a homeomorphism. For example, . Hence the statement “if f is not a homeomorphism, then f is not an isometry” is false.
Try this: Let be an isometry, be a fixed point and be a sequence in M such that for any . Then if as , as .
Now suppose is an isometry. Let and . We have shown above that any isometry is injective; hence for any n. But , whereas can never tend to zero if since the set has a lower bound of 1. This is a contradiction – hence f cannot be an isometry.