Okay, here's an example. Note that the half-open interval [0,1) maps bijectively to the unit circle S^1 in the complex plane by a continuous map with no continuous inverse: f : [0,1) —> S^1 via f(x) = exp(2πix). Now define the space X as a subset of C as follows: For each integer n <= 0, include a half-open interval X_n. For each integer n > 0, include a circle X_n. Let X denote the union of all H_n, n <= 0, and all C_n, n > 0. Then move X_n to X_(n+1) by a homeomorphism for n nonzero, and use a function like f above to map the half-interval X_0 bijectively onto the circle X_1 by a continuous map with no continuous inverse. This is clearly a continuous bijection of X to itself with no continuous inverse. If a connected set is desired, just arrange each X_n consecutively in the upper half plane intersecting the x-axis in one point (e.g., the point 4n so the X_n will be disjoint). Then add the x-axis to X, and we're done. —Dan For each integer n <= 0, include the half-open interval H_n = {(x,4n) in R^2 | 0 <= x < 1} in X. For each integer n > 0, include the circle C_n = {(1-exp(2πix),4n) ----- There do exist spaces X that are subsets of the plane possessing a continuous bijection X —> X without continuous inverse. Consider it a puzzle if you like. (But not the Klein bottle.) —Dan Bill Gosper wrote: ----- DanA: Consider the space Homeo(K) of self-homeomorphisms of the Klein bottle. I.e., Homeo(K) consists of all continuous bijections h : K —> K having a continuous inverse. __________ Umm, can somebody show me a continuous bijection with a *dis*continuous inverse? -----