Is $\{(e^{-x}\cos(\theta), e^{-x}\sin(\theta)): x \geqslant 0, 0 \leqslant \theta \leqslant2\pi\}$ Compact?

Question

Is $\{(e^{-x}\cos(\theta), e^{-x}\sin(\theta)), x \in \mathbb{R}^{2}: x \geqslant 0, 0 \leqslant \theta \leqslant 2\pi\}$ compact?

Attempt:

I've been trying to devlop some sort of visual intuition of what this might look like but I'm not getting anywhere. So to illustrate that the set is compact, I can apply the Heine-Borel theorem. To me it appears as if it is bounded, but I am basing that off of how the individual components of $(e^{-x}\cos(\theta), e^{-x}\sin(\theta))$ behave when $x \geqslant 0$. As far as closed, I have no intuition of how to see if there exists a sequence for every limit point to show that the limit points are also in the set. As a matter of fact, I don't even know what the limit points are of the set.

Some help would be appreciated, more so along the lines of how to approach a situation like this.

Answer 1

No, it is not compact: the sequence$$\left(e^{-n}\cos(0),e^{-n}\sin(0)\right)_{n\in\mathbb N}\left(=\left(e^{-n},0\right)_{n\in\mathbb N}\right)$$is a sequence of elements of your set which converges to $(0,0)$, which is outside your set.

Answer 2

Note that your set is exactly the disc with the origin removed, i.e. setting $r=e^{-x}$, we see that $0<r\leq 1$ and your set is $$ \{(r\cos\theta, r\sin\theta: 0\leq \theta\leq 2\pi, \;0<r\leq 1 \} $$ the punctured unit disc in polar coordinates, which is bounded, but not closed.