Exercise 22.4

Solution for $(a)$. Set $g(x\times y)=x+y^2\in ℝ$. We see it is a surjection onto $ℝ$ since $ℝ\times \{0\}\mapsto ℝ$. It is continuous since for $x_0\times y_0\in X$, given $𝜖>0$, letting $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->(1,𝜖∕2(\left|y_0\right|+1))$, $\rho (x_0\times y_0,x\times y)<\delta$ implies

If we define $f:ℝ\to X$ by $x\mapsto x\times 0$, which is continuous since $(a,b)\times (c,d)$ maps back to $(a,b)$ which is open in $ℝ$, we see $g\circ f$ is the identity on $ℝ$, and so $g$ is a quotient map by the lemma above. Since $x_0\times y_0\sim x_1\times y_1⟺g(x_0\times y_0)=g(x_1\times y_1)$, by Corollary $22.3$, this induces a bijective continuous map $g^{\prime }:X^{\text{∗}}\to ℝ$, which is a homeomorphism since $g$ was a quotient map. □

Solution for $(b)$. Set $g(x\times y)=x^2+y^2\in ℝ$. We see it is a surjection onto ${ℝ}_{\ge 0}$, since $ℝ\times \{0\}\mapsto {ℝ}_{\ge 0}$, and it does not map to anything else since $x^2+y^2\ge 0$ for all $x,y$. It is continuous since for $x_0\times y_0\in X$, given $𝜖>0$, letting $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->(1,𝜖∕2(\left|y_0\right|+\left|x_0\right|+1))$, $\rho (x_0\times y_0,x\times y)<\delta$ implies

We define $f:{ℝ}_{\ge 0}\to X$ by $x\mapsto \sqrt{x}\times 0$, which is continuous since the preimage of $(a,b)\times (c,d)$, if $(c,d)\ni 0$, is the open set ${ℝ}_{\ge 0}\cap (a^{\prime },b^{\prime })$, where $a^{\prime }=a^2$ if $a\ge 0$, and $-1$ otherwise, and similarly for $b^{\prime }$ (we chose $-1$ out of convenience; we really only have to make sure the preimage is a half-open set $[0,b^{\prime })$ or the empty set in these cases); if $(c,d)\not\ni 0$, then the preimage would be empty. We then see $g\circ f$ is the identity on ${ℝ}_{\ge 0}$, and so $g$ is a quotient map by the lemma above. Since $x_0\times y_0\sim x_1\times y_1⟺g(x_0\times y_0)=g(x_1\times y_1)$, by Corollary $22.3$, this induces a bijective continuous map $g^{\prime }:X^{\text{∗}}\to {ℝ}_{\ge 0}$, which is a homeomorphism since $g$ was a quotient map. □

Proof. So define $f:X^{\text{∗}}\to ℝ$ as follows: if $U$ is an equivalence class (so that $U\in X^{\text{∗}}$) containing $x_{}\times y_{}$ then set $f(U)=x_{}+y_{}^2$, noting that clearly $f(U)\in ℝ$. We also note that if $x_0\times y_0$ and $x_1\times y_1$ are both in the equivalence class $X$ then $x_0+y_0^2=x_1+y_1^2$ so that $f(X)=x_0+y_0^2=x_1+y_1^2$ is well defined.

Now suppose that $U_0$ and $U_1$ are two distinct equivalence classes and that $x_0\times y_0\in U_0$ and $x_1\times y_1\in U_1$. It then follows that it is not true that $x_0\times y_0\sim x_1\times y_1$ so that $x_0+y_0^2\ne x_1+y_1^2$. Thus we have $f(U_0)=x_0+y_0^2\ne x_1+y_1^2=f(U_1)$, which shows that $f$ is injective. Next consider any real $c$ and, the point $c\times 0$ in the plane, and let $U$ be the equivalence class containing $c\times 0$, which must exist since the equivalence classes form a partition of the plane. Then we have $f(U)=c+0^2=c$, which shows that $f$ is surjective since $c$ was arbitrary. This completes the proof that $f$ is a bijection, noting that of course this means that $f^{-1}$ is also a bijective function.

Now, the standard topology of $ℝ$ of course can have different bases, but we will concern ourselves with the order topology basis first. So consider any basis element $B=(a,b)$ of $ℝ$ in the order topology. Clearly then the inverse image of $B$ is the collection $U$ of equivalence classes $U$ where $a<f(U)<b$. Then the union of all the sets in $U$ is the set of points in the plane $x_{}\times y_{}$ where $a<x_{}+y_{}^2<b$. An illustration of such a subset of the plane is illustrated below for $a=-2$ and $b=3$:

It is easy to see that such a set is open in ${ℝ}^2$ for any $a$ and $b$, which we shall not show formally. Thus $U$ is an open subset of $X^{\text{∗}}$, which suffices to show that $f$ is continuous.

Now, to show that $f^{-1}$ is continuous we utilize metric topology bases for both ${ℝ}^2$ and $ℝ$, which we know induce the standard topologies on those sets. In particular, we use the standard metric $d$ on $ℝ$ but use the square metric $\rho$ on ${ℝ}^2$, which induces the standard topology on ${ℝ}^2$ by Theorem 20.3. Recall that the square metric is defined by

So consider any $x\in ℝ$, let $U$ be the equivalence class in $X^{\text{∗}}$ such that $U=f^{-1}(x)$ so that $f(U)=x$, and let $U$ be a neighborhood of $U=f^{-1}(x)$ in the quotient topology on $X^{\text{∗}}$. Then we have that $U$ is a collection of equivalence classes and that $x\times 0$ must be in $U$ since $x+0^2=x=f(U)$. Clearly also the union $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{U^{\prime }\in U}{U}^{\prime }$ is then open in ${ℝ}^2$ since $U$ is open in the quotient space, and $x\times 0\in A$ since $x\times 0\in U$ and $U\in U$. Thus by Lemma 20.4.1 there is a $\delta >0$ where $B_{\rho }(x\times 0,\delta )\subset A$.

Now, consider the set $B_d(x,\delta )$, which is clearly a neighborhood of $x$ in $ℝ$. Consider any $V\in f^{-1}(B_d(x,\delta ))$ and let $v=f(V)$ so that it must be that $v\times 0\in V$ since $v+0^2=v=f(V)$. Then $v=f(V)\in B_d(x,\delta )$ so that $d(v,x)=\left|v-x\right|<\delta$. Since we have $\left|v-x\right|\ge 0=\left|0-0\right|$, it follows that

and hence $v\times 0\in B_{\rho }(x\times 0,\delta )\subset A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{U^{\prime }\in U}{U}^{\prime }$. Therefore there is an equivalence class $W$ in $U$ such that $v\times 0\in W$ and $W\in U$. However, since the equivalence classes are all disjoint, it has to be that $W=V$ since $v\times 0\in V$ as well. Thus $V=W\in U$ so that $f^{-1}(B_d(x,\delta ))\subset U$ since $V$ was arbitrary. This suffices to show that $f^{-1}$ is continuous by Theorem 18.1 since $U$ was an arbitrary neighborhood of $U=f^{-1}(x)$. This completes the proof that $f$ is a homeomorphism so that $X^{\text{∗}}$ is homeomorphic to $ℝ$ as desired. □

Proof. Taking a similar approach to that in part (a), define $f:X^{\text{∗}}\to A$ by $f(U)=\Vert x_{}\times y_{}\Vert =\sqrt{x_{}^2+y_{}^2}$ if $x_{}\times y_{}$ is a point in the equivalence class $U$. Clearly we have $f(U)\ge 0$ so that $f(U)\in A$. It also follows that $f$ is well-defined since, if $x_0\times y_0$ and $x_1\times y_1$ are two points in the same equivalence class $U$, then $f(U)=\sqrt{x_0^2+y_0^2}=\sqrt{x_1^2+y_1^2}$ since $x_0^2+y_0^2=x_1^2+y_1^2$.

Now, if $U_0$ and $U_1$ are two distinct equivalence classes and $x_0\times y_0\in U_0$ and $x_1\times y_1\in U_1$ then it is not true that $x_0\times y_0\sim x_1\times y_1$ so that $f(U_0)=\sqrt{x_0^2+y_0^2}\ne \sqrt{x_1^2+y_1^2}=f(U_1)$ since $x_0^2+y_0^2\ne x_1^2+y_1^2$ and the square root function is injective on the nonnegative reals. This shows that $f$ is injective. Also, if $c$ is any element of $A$ then let $U$ be the equivalence class containing $c\times 0$, which exists since the classes form a partition on the plane. Then we have $f(U)=\sqrt{c^2+0^2}=\left|c\right|=c$ since $c\ge 0$, which shows that $f$ is surjective since $c$ was arbitrary. Hence $f$ is a bijection so that of course $f^{-1}$ is also a bijective function.

Next, consider the order topology basis of $ℝ$ and any corresponding basis element $B$ of the subspace $A$. Then clearly either $B=[0,b)$ (since then, for example, $B=A\cap (-1,b)$ and $(-1,b)$ is clearly a basis element of $ℝ$) or $B=(a,b)$ for some $0\le a<b$. In the former case clearly $f^{-1}(B)$ is the collection $U$ of equivalence classes $U$ such that $0\le f(U)<b$, the union of which is clearly the set of points $x_{}\times y_{}$ in the plane such that $\sqrt{x_{}^2+y_{}^2}<b$, which is an open filled circle of radius $b$ centered at the origin. This is obviously an open set of ${ℝ}^2$. In the later case clearly $f^{-1}(B)$ is the collection $U$ of equivalence classes $U$ such that $a<f(U)<b$, the union of which is the set of points in the plane $x_{}\times y_{}$ such that $a<\sqrt{x_{}^2+y_{}^2}<b$. As this is clearly an open annular ring centered at the origin, it is open in ${ℝ}^2$. Hence in either case the union of the collection $U$ is open in ${ℝ}^2$ so that $U=f^{-1}(B)$ is an open subset of $X^{\text{∗}}$. Since $B$ was an arbitrary basis element of $A$, this shows that $f$ is continuous.

Similar to what was done in part (a), we show that $f^{-1}$ is also continuous by utilizing the euclidean metric $d$ on ${ℝ}^2$ and the standard metric $d^{\prime }$ on $A$, i.e. $d^{\prime }$ is the standard metric $d$ on $ℝ$ restricted to $A\times A$, which is a metric for the subspace $A$ by the remarks at the beginning of §21. So consider any $x\in A$ so that $x\ge 0$, let $U$ be the equivalence class in $X^{\text{∗}}$ such that $U=f^{-1}(x)$ so that $f(U)=x$, and let $U$ be a neighborhood of $U=f^{-1}(x)$ in the quotient topology on $X^{\text{∗}}$. Then we have that $U$ is a collection of equivalence classes and that $x\times 0$ must be in $U$ since $\sqrt{x^2+0^2}=\left|x\right|=x=f(U)$. Clearly also the union $C={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{U^{\prime }\in U}{U}^{\prime }$ is then open in ${ℝ}^2$ since $U$ is open in the quotient space, and $x\times 0\in C$ since $x\times 0\in U$ and $U\in U$. Thus by Lemma 20.4.1 there is a $\delta >0$ where $B_d(x\times 0,\delta )\subset C$.

Now, consider the set $B_{d^{\prime }}(x,\delta )$, which is clearly a neighborhood of $x$ in $A$. Consider any $V\in f^{-1}(B_{d^{\prime }}(x,\delta ))$ and let $v=f(V)\in A$ so that it must be that $v\times 0\in V$ since $\sqrt{v^2+0^2}=\left|v\right|=v=f(V)$. Then $v=f(V)\in B_{d^{\prime }}(x,\delta )$ so that $d^{\prime }(v,x)=d(v,x)=\left|v-x\right|<\delta$. Then also

and hence $v\times 0\in B_d(x\times 0,\delta )\subset C={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{U^{\prime }\in U}{U}^{\prime }$. Therefore there is an equivalence class $W$ in $U$ such that $v\times 0\in W$ and $W\in U$. However, since the equivalence classes are all disjoint, it has to be that $W=V$ since $v\times 0\in V$ as well. Thus $V=W\in U$ so that $f^{-1}(B_{d^{\prime }}(x,\delta ))\subset U$ since $V$ was arbitrary. This suffices to show that $f^{-1}$ is continuous by Theorem 18.1 since $U$ was an arbitrary neighborhood of $U=f^{-1}(x)$. This completes the proof that $f$ is a homeomorphism so that $X^{\text{∗}}$ is homeomorphic to $A$ as desired. □