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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!

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What have you tried? Can you prove the given set is connected? Can you prove it is not path-connected? –  Chris Eagle May 29 '12 at 18:58
    
@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. –  Jyrki Lahtonen May 29 '12 at 19:10
    
To 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. –  Jyrki Lahtonen May 29 '12 at 19:25
    
Oops. I won't need more than one of those tiny squares. The compactness step was unnecessary. Shouldn't post when feverish :-( –  Jyrki Lahtonen May 29 '12 at 19:27
    
@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. –  t.b. May 29 '12 at 19:28
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1 Answer

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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@BrianM.Scott is there a way to view edit history of an answer? I'm curious what my typo was. –  rschwieb May 29 '12 at 20:53
    
Click on the time stamp next to ‘edited’. But I can tell you: I changed sin to \sin twice. –  Brian M. Scott May 29 '12 at 20:57
    
@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. –  rschwieb May 29 '12 at 21:09
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