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?
Connected subset of the plane
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0Related questio$n$: http://math.stackexcha$n$ge.com/questions/34103/for-a-function-from-mathbbr-to-itself-whose-graph-is-connected-in-mathbbr – 2011-10-03
1 Answers
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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0The 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. – 2011-10-04