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I'm trying to better understand connectedness vs. path connectedness for the Topologist's Whirlpool.

EDIT: I'm specifically trying to understand the point I mention below. If I have a neighborhood $N$ around $p(y)$, how do I show that it's open? Can I just assume that's it open for the purposes of the problem?

If I write the whirlpool set as $W = A \cup S^1$, then I understand why $W$ would be the closure of $A \in \Bbb R^2$. If $W = \overline A$, then I know the closure of a connected space is connected.

I was told that showing $W$ is not path connected means showing that all paths starting in $S^1$ are contained in $S^1$.

I would start by saying $p: [0,1] \rightarrow W$ is a path where $p(0) \in S^1$. Then I have to prove $p^{-1}(S^1)$ is both open and closed in $[0,1]$.

It seems like closed is simple:

(1) I am aware via a number of methods that $S^1$ is closed in $\Bbb R^2$. For example, say the set $\{1\}$ is closed and the map $f:\Bbb R^2⟶\Bbb R$ is continuous.

(2) Then the circle $\{(x,y)\in \Bbb R^2:x^2+y^2 = 1\} = f^{−1}(\{1\})$ is closed in $\Bbb R^2$. If the unit circle is closed in $\Bbb R^2$, it's closed in any subspace of $\Bbb R^2$ that contains it. Such as $W$. If $p$ is continuous, then $p^{-1}(S^1)$ is closed in $[0,1]$.

Proving open seems tougher:

I know the basis for the topology on $[0,1]$ can be obtained by taking the intersection of $[0,1]$ and open intervals on $\Bbb R$.

(1) I would start with $y \in p^{-1}(S^1)$ and and have to find a basis element that contains $y$ and is in $p^{-1}(S^1)$.

(2) I would use polar coordinates. $p(y)$ must have the form $(1, \theta_0)$. Say $N \subset \Bbb R^2$ is some neighborhood of $y$ such that $N = \lbrace(r,\theta) \mid (.9 < r < 1.1), (\theta_0 - 0.1 < \theta < \theta_0 + 0.1) \rbrace$.

If I could prove that $N$ is open in $\Bbb R^2$, I could use $N' = N \cap W$ and prove the rest for openness. I can't figure out how though.

Any suggestions? Any thoughts on what I have?

EDIT: By "the rest", I mean that I could use $N$ being open to show $N'$ is open and contains $y$. Then I would use some basis element $B$ with $y$ in it such that $B \subset p^{-1}(N')$. Then I could prove $B \subset p^{-1}(S^1)$ and conclude openness.

Edward Peterson
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  • I would start with noting that $W$ is not locally connected at any point of $S^1$. And then notice that any continuous map $[0,1]\to X$ is a quotient map and thus its image has to be locally connected. – freakish Apr 19 '22 at 17:52
  • This would be an alternate start for the entire proof? I'm familiar with the idea of local connectedness, but probably would have to prove that $W$ is not locally connected at points of $S^1$ before I could take it as given. – Edward Peterson Apr 19 '22 at 18:08
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    Yeah, I guess it is an alternative approach. I don't know how to proceed with yours. Of course, yeah, showing that $W$ is not locally connected will require some calculations. Basically you can show that any small enough ball in $W$ around a point on $S^1$ is homeomorphic to infinite sequence of "convegent" lines. – freakish Apr 19 '22 at 18:35
  • Good to know that the other post exists. It doesn't really answer the question I asked, though. I was looking for clarification about finding out how $N$ would be open in the context of my answer. – Edward Peterson Apr 19 '22 at 20:24

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