Let $T$= $ \{ (x, \sin(\frac{1}{x}))\in\mathbb R^n \mid x>0 \} \cup \{ (0,y)\in \mathbb R^2\mid y\in[-1,1] \} \subseteq \mathbb R^2$
I've looked through questions regarding the topologist's sine curve and have came across different definitions for various versions of it, so I'm a little confused.
Is this one, $T$, compact?
My guess is yes (because it is closed and bounded? Heine-Borel?), could anyone demonstrate how to prove this?
The set $\{ (x, \sin(\frac{1}{x}))\in\mathbb R^2 \mid x>0 \} \cup \{ (0,y)\in \mathbb R^2\mid y\in[-1,1] \} \subseteq \mathbb R^2$ is unbounded because the first of the two terms in the union is unbounded. If instead of $x>0$ you had said $0<x\le 1$, then the union of two sets that you wrote would be closed and bounded; hence compact.