Let $S^1 \subset \mathbb{R}^2$ be the unit circle centered at the origin and let $W\subset \mathbb{R}^2$ be the curve given by $r=\frac{\theta}{1+\theta}$ with $\theta\geq 0$.
Let $Z = S^1 \cup W \subset \mathbb{R}^2$, then Is $Z$ path connected and connected?
Here is my trial:
I know for non-empty intersection sets of (path)connected sets, their union is (path)connected. Clearly, I see $S^1$, $W$ is path-connected.(so conneced) And since $\frac{\theta}{1+\theta} \neq 1$ for all $\theta$, they are disjoint, so their union $Z$ is not (path)connected.
After @KaviRamaMurthy's comment, I realized my trial was wrong.
How to prove $Z$ is path connected or connected?
After @jasnee's comment I draw the graph of $Z$ as follows
It seems the graph of $r=\frac{\theta}{1+\theta}$ touches $r=1$ as $\theta \rightarrow \infty$. But except that infinity, their graph disjoint. So I guess $Z$ is not path connected, since there are no path connecting points in $S^1$ and $W$.
Proof of connectedness: $\frac {2\pi n} {2\pi n+1} e^{2\pi in}=\frac {2\pi n} {2\pi n+1}$ is on the curve for any positive integer $n$. Suppose you write $Z$ as a union two non-empty disjoint open sets $U$ and $V$. Then $S^{1}$, being connected, would be contained in one of these open sets, say $V$. Now $1 \in S^{1} \subset V$ , $V$ is open and $\frac {2\pi n} {1+2\pi n} \to 1$. This proves that there are points on the curve $W$ which belong to $V$. Connectedness of the curve now implies that the entire curve is contained in $V$. But this contradicts the fact that $U$ is non-empty.