A question says find a connected subset of $\mathbb{R}^2$ which is not path connected, the answers say that the graph of $y = \sin(1/x), \, \, x \in (0,1) \, \, \text{with} \, \, \{0\}\times [-1,1]$ is connected but not path connected but it doesn't provide any proof. Can someone tell me how to prove this? Thanks!
Find a connected subset of $\mathbb{R}^2$ which is not path connected
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0You're right, t.b. The argument is the same. I also cast$a$vote to reopen. Why don't you post your answer when the question is reopened, @rschwieb? Probably won't take too long. – 2012-05-29
1 Answers
I'm sure you can see that $C_1=\{0\}\times[0,1]$ and $C_2=\{(x,\sin(1/x)\mid x\in(0,\infty) \}$ are connected sets, and that you can also see that $C_1$ contains many points of closure of $C_2$. I think this makes a nice Lemma: If $C_1$ and $C_2$ are connected sets, and $\overline{C_2}\cap C_1\neq \emptyset$, then $C_1\cup C_2$ is connected.
To show it isn't path connected, think about what will happen if a continuous function $g$ from $[0,1]$ begins at the point $(0,0)$, stays within our graph, and goes to some point $(a,\sin(1/a))$ for some $a>0$.
To add another hint: you will be able to trap $g$ within the vertical interval $[-1/2,1/2]$ close to $(0,0)$, but on the other hand, $f$ breaks out of that interval no matter how closely you approach $(0,0)$
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0@BrianM.Scott Ah thank you! A lot of little features like that fly under my radar... I was looking for something like the drop-down menu you get when editing the OP. – 2012-05-29