Why is this set is closed in the subspace $([0,\infty)\times \mathbb{R}) \cup (\mathbb{R}\times \{0\})$ of $\mathbb{R}^2$ I am trying to show the set $\{(x,y)\in \mathbb{R}^2\ : xy=1, x>0\}$ is closed in the subspace $X := ([0,\infty)\times \mathbb{R}) \cup (\mathbb{R}\times \{0\})$ of $\mathbb{R}^2$.
To do this I need to show its complement equals $U \cap X$ for some open subset $U$ of $\mathbb{R}$. But the complicated structure is making it difficult.
If I'm not mistaken, the complement of $X$ is $\{(x,y)\in \mathbb{R}^2: x\leq 0$ or $y\leq 0$ or $xy\neq 0\}$

The map $f \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y) = xy$ is a continuous function and the set $\{ (x,y) : xy = 1 \} = f^{-1}(\{1\})$ is the pre-image of a closed set $\{1\}$ so it's closed by continuity in $\mathbb{R}^2$.
Call your set $A = \{ (x,y) : xy = 1, x > 0 \}$.

Let $X$ be a subspace of $Y$. A set $A$ is closed in $X$ if and only if there exists a closed $C$ in $Y$ s.t. $C \cap X = A$.

Pick $Y = \mathbb{R}^{2}$ and $C = f^{-1}(\{1\})$.
