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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general-topology
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0What have you tried? Can you prove the given set is connected? Can you prove it is not path-connected? – 2012-05-29
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0@t.b. the Topologist's sine curve in the other thread does not include the line segment. I'm afraid I don't remember, whether the line segment is needed to make this set connected. But that's the formulation of the question I recall once having solved:-) Anyway, I'm **not** voting to close because of this difference in the two questions. – 2012-05-29
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0To prove that this is connected it is enough to show that any open set containing the vertical line segment will intersect with the other path-connected component. So let $U$ be such a set. For each point $P(y)=(0,y), -1\le y\le1$ there is a $\delta(y)>0$ such the open square centered at $P(y)$ with side length $\delta(y)$ is contained in $U$. The line segment is compact, so it can be covered by finitely many such squares. Let $a$ be the smallest $\delta(y)$ appearing in that finite cover. Here $a>0$ and $(-a,a)\times [-1,1]$ is contained in $U$. The claim follows. – 2012-05-29
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0Oops. I won't need more than one of those tiny squares. The compactness step was unnecessary. Shouldn't post when feverish :-( – 2012-05-29
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0@Jyrki: For connectedness (as opposed to compactness) it doesn't matter whether you include a segment on the $y$-axis or only the origin, the argument is the same. I cast a vote to re-open and thanks for looking more closely than I did. – 2012-05-29
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0Ah apologies for the duplicate, didn't realise it was called the Topologist's sine curve – 2012-05-29
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0Dang, I was almost done with my answer too. – 2012-05-29
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0Post it in the comments? – 2012-05-29
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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