Prove that there is no continuous bijective map from the uni
Prove that there is no continuous bijective map from the unit circle in R^2 to the interval [0,1].
Solution
Assume on the contrary that there exists f: A (Unit Circle) -> B (=the interval [0,1]) , a continuous , bijective map.
Now f being a continuous map between compact Hausdorff spaces is a homeomorphism . PROOF: if f:X->Y is such a map, ,let Z be a closed subset of X. To show that f-1 is continuous, enough to show (f-1)-1 (Z) = f(Z) is closed in Y for every closed subset Z of X. But Z closed in a compact space X implies Z is compact implies f(Z) is compact in Y , which in turn implies f(Z) is closed in Y , as Y is Hausdorff, so we are done.
But the unit circle and the interval [0,1] cannot be homeomorphic : removing a point disconnects [0,1], whereas A-{point} remains connected.
Conclusion: there does not exist f: A (Unit Circle) -> B (=the interval [0,1]), f continuous and bijective
![Prove that there is no continuous bijective map from the unit circle in R^2 to the interval [0,1].SolutionAssume on the contrary that there exists f: A (Unit Ci Prove that there is no continuous bijective map from the unit circle in R^2 to the interval [0,1].SolutionAssume on the contrary that there exists f: A (Unit Ci](/WebImages/31/prove-that-there-is-no-continuous-bijective-map-from-the-uni-1090100-1761573841-0.webp)
