Give an example of a set that is closed but not compact nor bounded. Prove your answer.

Question

Let $X = (0,\infty)$ with the usual topology in $\mathbb{R}$ and the the usual metric.

Consider $A \subset X$ where $A = [1, \infty)$.

Then $A$ is closed as $A' = (0,1) \subset X$.

My attempt is as follows. Please critique my proofs for the unboundedness of A. Do they both work?

A is not bounded attempt 1

By contradiction

Since $A = (0,\infty)$, clearly $R \in A$. Now, take $y_1, y_2 \in A$ and $y_1<y_2$ and $y_2 > R$. Then $d(x,y_1) < R < d(x,y_2)$. $\rightarrow\leftarrow$

A is not bounded attempt 2

By Contradiction

Assume Sup$A$ is the LUB of $A$. THen $\forall a \in A$, Sup$A \geq a$. Since $A = [1,\infty)$, then $\forall a \in A, a+1 \in A$. So Sup$A \geq$ Sup$A+1$. $\rightarrow\leftarrow$

A is not compact

Let $\{G_{\alpha} \}_{\alpha}$ be an open cover for $A$ where $G_{\alpha_n}=\{[1,\infty):n \in \mathbb{N}\}$

Then $A \subset \bigcup_{n=1}^\infty G_{\alpha_n}$

For $A$ to be compact, then $\exists n: A \subset G_{\alpha_1} \cup G_{\alpha_2} \cup \cdots \cup G_{\alpha_n}$

However, since $A = [1,\infty)$, then $\nexists n: [1,\infty) \subset [1,n)$

$\therefore A$ is not compact

Answer 1

In $\mathbb R^n$, by Heine-Borel theorem a set is compact iff it is closed and bounded.So no need for such a long proof.Consider $x-$axis in $\mathbb R^2$ which is closed but not bounded hence not compact

Answer 2

The positive integers work fine... They're closed because they contain no limit points (suppose one exists, derive obvious contradiction, etc) and they're unbounded by the archimedian property.