Is $\{ (r, \theta) \in \mathbb R^2 \mid 0 \le r \le 1 + \cos(\theta) \land 0 \le \theta \le 2 \pi\}$ compact?

Question

Is $S =\{ (r, \theta) \in \mathbb R^2 \mid 0 \le r \le 1 + \cos(\theta) \land 0 \le \theta \le 2 \pi\}$ compact on $\mathbb R^2$ with the standard topology?

This set is, of course, bounded. It also seems to be closed, because its complement will not contain its boundary, and so will be open.
This problem seems to be very easy, however, this $r$ and $\theta$ suggest the polar coordinates, and I can't draw this set there. Is it enough to draw it in the Cartesian system and check if it's compact?
Or, in other words, is compactness an invariant of the transformation
$$T: \text{Cartesian coordinates} \to \text{Polar coordinates}$$

Answer 1

To show that $S$ is closed ,take a convergent sequence $((r_n, \theta_n))$ in $S$ with limit $(r_0, \theta_0)$.

Then we have $r_n \to r_0$ and $\theta_n \to \theta_0$. From

$0 \le r_n \le 1 + \cos(\theta_n)$ for all $n$ we get, with $ n \to \infty$, $0 \le r_0 \le 1 + \cos(\theta_0)$ .

Hence $(r_0, \theta_0) \in S$.

Answer 2

It's much easier to note that it is closed because it is the inverse image of a closed set with respect to a continuous function.