Exercise 18.12

Proof of $(a)$. Since $F$ is symmetric in interchanging $x\leftrightarrow y$, we only have to prove $\forall <!--\; FUNCTION\; APPLICATION\; -->y_0\in Y$, $h(x)=F(x\times y_0)$ is continuous as a function $ℝ\to ℝ$. For $y_0=0$, this is trivially true for the image of $h$ is $(0,0)$ with preimage $ℝ$. Now suppose $y_0\ne 0$; then we have $h(x)=x{y}_0∕(x^2+y_0^2)$. This is continuous since $x{y}_0$ and $x^2+y_0^2$ are both continuous, and so their quotient is also continuous (since also $x^2+y_0^2\ne 0$), using the $𝜖$-$\delta$ definition of continuity (see Theorem 21.5). □

Proof of $(b)$. Since $F(x\times x)$ for $x\ne 0$ equals $x^2∕(x^2+x^2)=x^2∕2x^2=1∕2$, we have

Proof of $(c)$. We claim $F(x\times y)$ is not continuous along the line $L=\{x=y\}$ at $(0,0)$, i.e., $F{\text{|}}_L$ is not continuous at $(0,0)$; this suffices by Theorem $18.2(d)$. Note that the line $L$ in the subspace topology is homeomorphic to $ℝ$, where the homeomorphism is given by either of the coordinate projection maps. Now the preimage of the closed set $\{1∕2\}\subseteq ℝ$ is $L\setminus \{(0,0)\}$, which is not closed since $ℝ\setminus \{0\}$ is not closed, hence $F{\text{|}}_L$ is not continuous, and neither is $F$. □

Proof. It is easy to see that $F$ is continuous in $x$. For any real $y_0$ we generally have that

so long as one of $x$ and $y_0$ are nonzero. If $y_0=0$ then $x=0$ implies that $x\times y_0=0\times 0$ so that $h(x)=F(0\times 0)=0$ by definition. If $x\ne 0$ then we have $h(x)=0∕x^2=0$ again. Thus $h$ is the constant function $h(x)=0$ and so is continuous when $y_0=0$. If $y_0\ne 0$ then $y_0^2>0$ so that $x^2+y_0^2>0$ since also $x\ge 0$. Thus the denominator is never zero the $h(x)$ is given by the expression above, which is continuous by elementary calculus. Hence $h$ is always continuous. The same arguments show that $F$ is continuous in $y$ as well. □

Proof. First consider the function $f:ℝ\to ℝ\times ℝ$ defined simply by $f(x)=x\times x$. This function is clearly continuous by Theorem 18.4 since it can be expressed as $f(x)=f_1(x)\times f_2(x)$ where the identical functions $f_1(x)=f_2(x)=x$ are obviously continuous. Then $g=F\circ f$, where $g$ is the function from part (b) since we have $g(x)=F(x\times x)=F(f(x))$ for any real $x$. Now, clearly, $g$ as calculated in part (b) has a discontinuity at $x=0$ so that it is not continuous. It then follows from Theorem 18.2 part (c) that either $F$ or $f$ is not continuous since $g=F\circ f$. As we know that the trivial function $f$ is continuous, it must then be that $F$ is not as desired. □