The graph of $f$ is a connected subset of $\mathbb{R}^2$.

Asked Nov 14 '15 at 00:10

Active Nov 14 '15 at 00:46

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Let a function $f:[0,1]\to\Bbb R$ given by $f(x)=\sin(\frac lx)$ for $x\ne 0$ and $f(0) = 0$. Show that although $f$ is not continuous, the graph of $f$ is a connected subset of $\mathbb{R}^2$.

I was wondering on how to prove this.

I have read that the closure of a connected set is also connected, so if I can show the given set is a closure of some connected set then I am done.

Could you please help me out.

edited Nov 14 '15 at 00:46

Jesse P Francis

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asked Nov 14 '15 at 00:10

00012 suxn

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http://math.stackexchange.com/questions/317125/topologists-sine-curve-is-connected – John Douma Nov 14 '15 at 00:13
Did you mean $\sin \frac lx$ or $\sin\frac1x$? – Jesse P Francis Nov 14 '15 at 00:47
You might check as well http://www.math.uconn.edu/~kconrad/blurbs/topology/connnotpathconn.pdf – EA304GT Nov 14 '15 at 00:56

The graph of $f$ is a connected subset of $\mathbb{R}^2$.

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