Showing that a Set in $\mathbb{R}^2$ is not bounded.

Question

Using the following definition:

A set $S \subseteq \mathbb{R}^2$ is bounded iff $\exists \ M>0$ s.t $S \subseteq B(\textbf{0},M)$ where $B(\textbf{0},M)$ is a ball of radius M around the point $(0,0) \in \mathbb{R}^2$.

Take the following set $S = \{(x,y) \in \mathbb{R}^2:1<x+y, xy<1\}$

I wish to prove that this set is unbounded.

I have found a sequence of points ${x_n} \in S$ such that $$\lim_{n\to \infty} ||x_n|| = \infty$$ Namely, $x_n = (n, \frac{1}{2n})$ where $n \in \mathbb{Z}^{+}$.

How do I argue, using this information that the set $S$ is then not bounded?

I was thinking of using proof by contradiction:

Suppose $S$ is bounded. Hence, $\exists M > 0 $ s.t $S \in B(\textbf{0},M)$. This implies that $\forall \ x_0 \in S$ we have $||x_0|| < M$.

Take the following sequence of points in $\mathbb{R}^2$, $x_n = (n,\frac{1}{2n})$ where $n \in \mathbb{Z}^{+}$. Now, $n + \frac{1}{2n} > 1$ since $n \geq 1$. We also have $(n)(\frac{1}{2n}) = \frac 12 < 1$. Hence, $x_n \in S$.

Now, $$\lim_{n \to \infty} ||x_n|| = \lim_{n \to \infty} \sqrt{n^2 + \frac{1}{4n^2}} = \infty$$

I am now tempted to say "this is a contradiction. Hence, $S$ is not bounded. I am not comfortable with saying this, and feel like some elaboration may be needed. How should I continue?

Does this work?

Hence, for sufficiently large $n$, we have $||x_n|| > M$. But $x_n \in S$ and $x_n \notin B(\textbf{0},M)$. This is a contradiction. Hence, $S$ is not bounded.

Answer 1

A little bit simpler is the following argument: if $x>1$, then $(x,0) \in S$ and $||(x,0)||=x$.Hence, to each $M>0$, there is $X_M \in S$ with $||X_M||>M.$

You can take $X_M=(M+1,0).$

Answer 2

Your proof works ! You can add the following:

we have $||x_n||>n$ for all $n \in \mathbb N$, hence, if $n \ge M$, then $||x_n||>M$, a contradiction.