Preimage of Closed Set Under a Continuous Map Isn't Closed?

Question

Problem: In the $xy$-plane, the graph of $x^{\log y} = y^{\log x}$ is

(A) Empty

(B) A single point

(C) A ray in the open first quadrant

(D) A closed curve

(E) The open first quadrant

Here is my reasoning. Clearly $(1,1)$ satisfies the equation, as do $x = e^a$ and $y=e^b$ for all $a,b \in \Bbb{R}$. Let $f(x,y) = x^{\ln y} - y^{\ln x}$ . Then clearly we are interested in describing the set $\{(x,y) ~|~ f(x,y) = 0\} = f^{-1}(\{0\})$, and since $f$ is continuous and $\{0\}$ a closed set, this preimage must be closed. This means that we can very easily rule out all choices but (D). Curiously, the answer is (E). By "a closed curve," I suspect they mean a closed path.

After typing this I now see why the answer is (E), but I cannot see the flaw in the above argument. Here is my reason for thinking the answer is (E). As I noted above, $(e^a,e^b) \in f^{-1}(\{0\})$ for every $a,b \in \Bbb{R}$. Since the $e^x$ is a homeomorphism from $\Bbb{R}$ to $(0,\infty)$, it isn't too difficult to see $(0,\infty) \times (0,\infty) \subseteq f^{-1}(\{0\})$. But $x$ and $y$ cannot be nonpositive, so we must have $(0,\infty) \times (0,\infty) = f^{-1}(\{0\})$.

Answer 1

The domain of $f$ is the open first quadrant, so that is the relevant space in which you're considering open and closed sets. The whole space is always closed in itself.

Answer 2

For all positive $x, y$ we have $$x^{\log y}=(e^{\log x})^{\log y}=e^{(\log x)(\log y)}=e^{(\log y)(\log x)}=(e^{\log y})^{\log x}=y^{\log x}.$$

And $\log x, \;\log y\;$ are not both defined when $x\leq 0$ or $y\leq 0.$

Therefore the answer is (E).

Let $D$ be the sub-space of $\mathbb R^2$ on which the function $f(x,y)=x^{\log y}-y^{\log x}$ exists. Then $f:D\to \mathbb R$ is continuous, so $f^{-1}\{0\}$ is a closed subset of the space $D$, but $f^{-1}\{0\}$ is not necessarily a closed subset of $\mathbb R^2.$ It also happens that $f^{-1}\{0\}=D.$