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Let $k \in \mathbb{R}$ and: $f(x) = \sin \frac{1}{x}$ , for $x\neq 0$ with $f(0) = k$.For what value(s) of $k$, the graph of $f$ is NOT a connected subset of the plane?

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Do you have any thoughts about the likely answer? For example, what do you think happens if $k=17$? If $k=-10583421$? Any difference if $-1\leq k\leq 1$? –  Arturo Magidin Oct 3 '11 at 19:10
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Is this a homework problem? –  Stefan Geschke Oct 3 '11 at 19:36
    
Related question: math.stackexchange.com/questions/34103/… –  Chris Eagle Oct 3 '11 at 20:38

1 Answer 1

Notice that the graph of $f|_{(0,\infty)}$ is connected as the image of a connected set $(0,\infty)$ through the continuous function $\mathrm{id} \times f$. Let's call this graph $G_1$.

If $k \in [-1,1]$, then $(0, k) = (0,f(0)) \in \mathrm{cl}(G_1)$. Therefore, in this case, $G_1 \cup \{(0,f(0))\}$ is connected because it is "between" a connected set and its closure.

The same goes for $G_2$, the closure of $f|_{(-\infty,0)}$. Since $G_1$ and $G_2$ have one point in common, their union is connected. Now, notice that their union is exactly the graph of $f$.

For the case where $k > 1$, it is easy to see that the graph is disconnected. For example, take the sets $\mathbb{R} \times (\frac{1+k}{2},\infty)$ and $\mathbb{R} \times (-\infty,\frac{1+k}{2})$... they separate $(0,k)$ from the rest of the graph. The same is true for $k < 1$.

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The key point is of course that if $C$ is connected and if $C\subseteq D\subseteq\mathrm{cl}(C)$, then $D$ is connected as well. You might wish to add a proof of that. –  Did Oct 4 '11 at 7:34

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