Showing the closedness of a set in two ways 
Let $S:=\{(x,y)\in\mathbb{R}^2: x>0 \,\text{ and }\, y\geq x^{-1}\}$. I want to show that $S$ is closed.

My first attempt is based on a sequence argument: Let $(x_n,y_n)_{n\in\mathbb{N}}$ be a sequence in $S$ that converges to $(x,y)\in\mathbb{R}^2$. We then have $x_n>0$ and $y_n\geq x_n^{-1}$ for all $n\in\mathbb{N}$. Assume that $x_n\rightarrow 0$ for $n\rightarrow\infty$, i.e. there exists $n_0\in\mathbb{N}$ such that $x_n<\varepsilon$ for all $n\geq n_0$. Let $N\in\mathbb{N}$. Then set $\varepsilon:=\frac{1}{N}$ to obtain $y_n \geq x_n^{-1} > \frac{1}{\varepsilon} = N$ for all $n\geq n_0$. Therefore, the sequence $(y_n)_{n\in\mathbb{N}}$ would be unbounded, a contradiction to the assumption that it converges. So we can assume $x_n\rightarrow x$ for $n\rightarrow\infty$, where $x\neq 0$. We can even infer $x >0$ by taking the limit on both sides of $x_n>0$ for all $n\in\mathbb{N}$. Then $x^{-1}>0$ and $x_n^{-1}\rightarrow x^{-1}$ for $n\rightarrow\infty$. Taking the limit on both sides of $y_n\geq x_n^{-1}$ for all $n\in\mathbb{N}$ gives $y\geq x^{-1}$. All in all, we have shown $(x,y)\in S$ and $S$ is closed.
My second attempt uses the topological definition of continuity: Let
\begin{equation}
f: \mathbb{R}_{>0}\times \mathbb{R}\rightarrow\mathbb{R},\, (x,y)\mapsto y-x^{-1}. 
\end{equation}
We then have $S = f^{-1}([0,\infty))$. As $[0,\infty)$ is closed in $\mathbb{R}$ and $f$ is continuous, we conclude that $S$ is closed in $\mathbb{R}_{>0}\times\mathbb{R}$. I am now stuck a little bit because $\mathbb{R}_{>0}\times\mathbb{R}$ is open in $\mathbb{R}^2$ and I would need a closed subspace to conclude that $S$ is also closed in $\mathbb{R}^2$.
My questions:

*

*I am not really sure what the topology on $\mathbb{R}_{>0}\times\mathbb{R}$ is. Is it a subspace topology of the product topology?

*If yes, I would also appreciate ideas how to show formally that $f$ is continous with respect to this topology. Does the sequence criterion also work with subspace topologies?

*Is my first attempt correct and if yes, could it be shortened?

*Any hints to complete my second attempt?

Thank you in advance.
 A: *

*I can't read minds, but I am quite sure that whoever created that exercise had that topology in mind.

*The function $f$ is continuous because $f=f_1-f_2$, with $f_1(x,y)=y$ and $f_2(x,y)=\frac1x$, both of which are continuous. I suppose that this is clear for $f_1$. concerning $f_2$, it is equal to $\iota\circ f_3$, with $f_3(x,y)=x$ ($(x,y)\in\Bbb R_{>0}\times\Bbb R$) and $\iota(x)=\frac1x$ ($x\in\Bbb R_{>0}$), both of which are continuous. And the sequence criterion works on any metric space.

*Yes, it is correct.

*Consider the function $g(x,y)=xy-1$ ($(x,y)\in\Bbb R^2$) instead. Then$$g(x,y)\geqslant0\iff xy\geqslant1$$and so $S=g^{-1}\bigl([0,\infty)\bigr)\cap\{(x,y)\in\Bbb R^2\mid x\geqslant 0\}$, which is the intersection of two closed sets.

A: Not directly answering the questioning, but the shortest proof I could find was this:
Note that $\{(x, y) \in \mathbb{R}^2 \mid x > 0, y \geq x^{-1}\} = \{(x, y) \in \mathbb{R}^2 \mid x \geq 0, x \cdot y \geq 1\}$.
The latter is clearly the intersection of $S_1 := \{(x, y) \in \mathbb{R}^2 \mid x \geq 0\}$ and $S_2 := \{(x, y) \in \mathbb{R}^2 \mid x \cdot y \geq 1\}$.
Now $S_1 = p_1^{-1}([0, \infty))$ where $p_1 : \mathbb{R}^2 \to \mathbb{R}$ is the first projection. Since $p_1$ is continuous and $[0, \infty)$ is closed, $S_1$ is closed.
And $S_2 = \cdot^{-1}([1, \infty))$ where $\cdot : \mathbb{R}^2 \to \mathbb{R}$ is the multiplication function. Since multiplication is continuous and $[1, \infty)$ is closed, $S_2$ is also closed.
Therefore, $S_1 \cap S_2$ is closed.
Edit: actually, this is related to OP's second proof attempt. The relevant closed subspace is $\mathbb{R}_{\geq 0} \times \mathbb{R}$.
