This must be easy but I'm stuck on it. This is exercise 43 page 64 of Armstrong Basic Topology (1983 Springer Verlag version). I'm not trying to get somebody to do my homework, I took this class in 1989, I'm just reviewing it. Thank you in advance for anything that helps, a hint or the full answer.

Convert $\{0\}\cup\{1/n|n\in\mathbb N\}$ into a subspace of the plane that is path-connected but not locally path connected.

"$X$ is locally path-connected" means for each $x\in X$ and open set $U$ containing $x$, $\exists$ an open set $V$ s.t. $x\in V\subseteq U$ and $V$ is locally path connected.

  • $\begingroup$ Are you trying to construct a form of the topologist's sine curve? $\endgroup$ – Michael Burr Mar 19 '15 at 13:06
  • $\begingroup$ Well, the topologist's sine curve was an example in this section. But I don't see how it helps here. $\endgroup$ – Gregory Grant Mar 19 '15 at 13:07
  • $\begingroup$ It's confusing that the question says "subset of the plane" and not "subset of $\mathbb R$. If it works in $\mathbb R^2$ then wouldn't it work in $\mathbb R$ with the subspace topology? $\endgroup$ – Gregory Grant Mar 19 '15 at 13:08
  • $\begingroup$ I don't think it is possible for a countable set to be path-connected. $\endgroup$ – Thomas Andrews Mar 19 '15 at 13:09
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    $\begingroup$ I'm interpreting this question to read: Find a set $X$ such that $X$ is path connected, but not locally path connected such that $X\cap\mathbb{R}$ is the set above. $\endgroup$ – Michael Burr Mar 19 '15 at 13:11

you try to prove that comb space (http://en.wikipedia.org/wiki/Comb_space) is such a space

  • $\begingroup$ Thank you, I see how the comb space is connected but not locally path connected. But I don't see yet how to use that to do the same for $\{0\}\cup\{1/n\mid n\in\mathbb N\}$. $\endgroup$ – Gregory Grant Mar 19 '15 at 13:14
  • $\begingroup$ Oh, maybe I misunderstood the question. Maybe he doesn't want the set $\{0\}\cup\{1/n\mid n\in\mathbb N\}$ itself to be given such a topology, he wanted us to come up with the whole comb space by ourselves (that's what he meant by "of the plane"), in which case the comb space is exactly the answer. It's probably not possible to endow just that countable set with a topology with this property. No wonder I was stuck! $\endgroup$ – Gregory Grant Mar 19 '15 at 13:16

Consider $U = (\{0\} \times [-2, 2]) \cup ([0, 1] \times \{-2\}) \cup (\{1\} \times [-2, 0]$

and $V$, the graph of $y = \sin(\frac{\pi}{x})$ for $0 < x < 1$.

Then $U \cup V$ is a set containing your original set that's connected but not locally pc at $(0,0)$.

Perhaps this is what's meant by "convert", although that's certainly a badly phrased exercise.

  • $\begingroup$ Thank you, your suggestion is akin to the comb space, and I think your answer is what the Armstrong was looking for. I got confused by "of the plane" since $\{0\}\cup\{1/n\mid n\in\mathbb N\}$ is just a subset of $\mathbb R$. But I think you're right, this is what he was looking for. And thank you for corroborating my contention that this is a poorly phrased question (from an otherwise excellent book!). $\endgroup$ – Gregory Grant Mar 19 '15 at 13:23
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    $\begingroup$ Yeah, I like Armstrong's book -- generally excellent. Best of luck catching up with that 1989 course! $\endgroup$ – John Hughes Mar 19 '15 at 13:26

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