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I have a mathematics problem, but I have no idea. please help me...

The problem is "Give an example of a function $f(x)$ defined on an interval I whose graph is connected, but is is not continuous at some point(s) of $I$"

In my idea, a solution is topologist's sin curve.

\begin{gather*} f\colon[0, 1] \to [0, 1] \\\\ f(x) = \begin{cases} \sin(1/x) &(x\neq 0)\\ 0 &(x=0). \end{cases} \end{gather*}

Is this function connected on $\mathbb{R}^2$ space but not continuous at $x=0$ ?

Please help me...

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  • $\begingroup$ Your example is good. In fact, it is not terribly hard to prove that almost all functions from an interval to an interval are discontinuous at some point. It's also true, but harder to prove, that almost all of them are discontinuous at every point. $\endgroup$ – dfeuer Sep 29 '13 at 17:14
  • $\begingroup$ @dfeuer Thank you for your comment and edit my question!! $\endgroup$ – Edmond.Cho Sep 30 '13 at 12:12
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This is a good example. The graph of the topologist's sine curve, which includes the point $(0,0)$ as you have indicated, is indeed connected. However, it is not continuous. To see this, try and produce a sequence of points $x_n$ converging to $0$ for which $\sin(1/x_n) = 1$.

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  • $\begingroup$ Thank you!!! It is very helpful!!! $\endgroup$ – Edmond.Cho Sep 30 '13 at 12:05
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Yes, your function works perfectly. The graph of this curve is connected for the following reason: any open set that contains $(0,0)$ is going to contain another part of the curve.

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  • $\begingroup$ Thank you for your comment!!! $\endgroup$ – Edmond.Cho Sep 30 '13 at 12:10

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