1

We have $$F:R^{2}\rightarrow R , F\left( x,y\right) =x^{3}+2x^{2}y+y^{3}$$ and for $D\subset R^{2}$ we know $$0\in F\left( D\right),4\in F\left( D\right) \text {but } 2\not\in F\left( D\right)$$ How to show that $D$ is not a connected set?

user190080
  • 3,701
FMath
  • 941
  • Anything tried so far? You sure mean $F:\mathbb R^2\mapsto \mathbb R$, don't you? – user190080 Apr 27 '16 at 15:04
  • 2
    $f$ is obviously continuous, so if we restrict its domain to $D$ but we can find $a,b\in D$ so that $f(a)=0,f(b)=4$, then if $D$ was connected we could find a path $\Gamma$ in $D$ connecting $a,b$ and $f$ would be continuous along the path. – almagest Apr 27 '16 at 15:05
  • 2
    @almagest, a set can be connected without being path connected. – Barry Cipra Apr 27 '16 at 15:11
  • I did some editing on your question, if you feel something went wrong feel free to revert back to your original question or let me know – user190080 Apr 27 '16 at 15:16
  • Is there any particular reason you want to use that F? The first thing that came to my mind was to use a locally constant function: http://math.stackexchange.com/questions/44850/locally-constant-functions-on-connected-spaces-are-constant. Maybe there is a way to reduce the problem to this? – Chill2Macht Apr 27 '16 at 15:19
  • @William All you know about $D$ is implicitly given by the image of $F$, so you sort of need to use this particularly $F$ to gain information about $D$ – user190080 Apr 27 '16 at 15:22
  • Good point, thanks – Chill2Macht Apr 27 '16 at 17:35

1 Answers1

4

There is a general theorem that the continuous image of a connected set is connected. The polynomial function $F$ is certainly continuous, while the stipulated conditions $0,4\in F(D)$ but $2\not\in F(D)$ imply the image $F(D)$ is not connected: That is, $F(D)$ is covered by the union of disjoint open intervals $(-\infty,2)\cup(2,\infty)$ but not by either open interval by itself. Thus $D$ cannot be connected.

Community
  • 1
Barry Cipra
  • 79,832