Showing the closedness of a set in two ways

Question

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:

1. 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?
2. 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?
3. Is my first attempt correct and if yes, could it be shortened?
4. Any hints to complete my second attempt?

Thank you in advance.

Answer 1

1. I can't read minds, but I am quite sure that whoever created that exercise had that topology in mind.
2. 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.
3. Yes, it is correct.
4. 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.

Answer 2

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 $\mathbb{R}_{\geq 0} \times \mathbb{R}$.