Exercise 24.7

Proof of $(a)$. $f$ is injective since if $f(a)=f(b)$ but $a\ne b$, then (with possible swapping) $a<b$, and so $f(a)<f(b)$, a contradiction. We thus must show $f$ and $f^{-1}$ are continuous. But $f$ is continuous since $f^{-1}((a,b))=(f^{-1}(a),f^{-1}(b))$ is open (apply the same argument to the intervals of the form $[a_0,b),(a,b_0]$ for $a_0,b_0$ minimal and maximal respectively); the same argument applies for $f^{-1}$ as well. □

Proof of $(b)$. $f(x)=x^n$ is order preserving since $a<b⟹a∕b<1⟹a^n∕b^n<1⟹a^n<b^n⟹f(a)<f(b)$. $f$ is continuous since it is the product of $n$ copies of the identity function, which is continuous. We want to show $f$ is surjective. Letting $N=\{x^n\mid x\in {\mathbb{Z}}_{\ge 0}\}$, we see that every real number $y\in Y$ is between two consecutive members of $N$, or it is already an $n$th power of an integer, in which case it is trivially mapped to by its $n$th root. In the case $y\in Y$ is not an $n$th power, we have $f(n)<y<f(n+1)$, and so by the Intermediate value theorem (Theorem 24.3), we see that there exists a point $c\in X$ such that $f(c)=r$, i.e., $f$ is surjective.

Since $f$ is order preserving and surjective, by $(a)$ it is then a homeomorphism, and so $f^{-1}$, the $n$th root function, is also continuous. □

Proof of $(c)$. $f$ is order-preserving on $(-\infty ,-1)$ since $a<b⟹f(a)=a+1<b+1=f(b)$, and on $[0,\infty )$ since it is the identity. We check that it is order preserving around the boundary. So, suppose $a<-1$ and $b\ge 0$. Then, $a<b$ but also $a+1<0\le b$, and so $f$ is order-preserving. $f$ is surjective since if $x\in ℝ$, if $x\ge 0$ its preimage is itself, and if $x<0$, its preimage is $x-1$. $f$ is not a homeomorphism by Theorem $23.6$ since $ℝ$ is connected but $X$ is not, by considering $f^{-1}(ℝ)$.

This does not contradict $(a)$ since $X$ is not in the order topology. Even if $ℝ$ is in the order topology, the subspace topology induced on $X$ is not the order topology. For, $(-1∕2,1)$ is open in $ℝ$, and so $(-1∕2,1)\cap X=[0,1)$ is open in $X$, but not open in the order topology on $X$. □