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I am trying to solve the following problem

Let $S^1 \subset \Bbb R^2$ and $W \subset \Bbb R^2 $ be the curve given by $r=\frac{\theta}{1+\theta}, \theta \ge 0$ in polar coordinates. Let $Z=S^1 \cup W$ be the subspace of $\Bbb R^2$ in the usual Topology.

(a) Sketch $Z$, and show that $Z$ is connected

(b) Is Z path- connected? Prove your answer.

My trial : Actually, I was even stuck in drawing the curve. So, I just utilized graphic tool to draw this curve as follows.

enter image description here

And, I realized that the relation could be satisfied as follows : $W \subset W\cup S^1 \subseteq \overline W$. Thus, it suffices to show that $W$ is connected and and all limit points of $W$ contain $S^1$. the former is trivial and the latter can be shown by constructing sequence $\lim_{n\to \infty} r(\theta + 2\pi n)=1$ for all $0\le \theta \le \pi$, which means every points in $S^1$ can be approximated by some sequence of $W$

When it comes to (b), I thought it has similar characteristics with the topologist's sine curve. So I tried to show there is a contradiction if we suppose a path from the orgin to $(1,0)$, But I failed to do it. So hear is my questions.

First, How could I draw the curve $W$ without a graphic tool

Second, could you give me a few hint to solve the problem (b)

Thanks!

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Clearly $S^1$ and $W$ themselves are connected. Hence, if $Z=U \sqcup V$ with $U$ and $V$ open neither $U$ or $V$ could intersect both $S^1$ and $W$. Thus we are forced to conclude that WLOG, $U=S^1$ and $V=W$, but this is a contradiction as $S^1$ is not open. So $Z$ is connected.

If $p: [0,1] \rightarrow Z$ is a path between the origin and, say, $(0,1)$, define $s=\sup\{t: p(t)\in W\}$. Then $p(s)\in S^1$ by continuity since $S^1$ is closed. By continuity there exists a $\delta>0$ such that $p((s-\delta, s+\delta))\subseteq B(p(s); 1)$. Then it is easy to see that $p((s-\delta, s+\delta))$ is a disconnected set, which is a contradiction as $p$ is continuous and supposed to map the connected set $(s-\delta, s+\delta)$ to a connected set.

Jonathan Hole
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