# Curve Selection Conjecture.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Hello MHB,

I have the following conjecture which I cannot seem to settle either way:

Let $f:[0,1]\to\mathbb R^2$ be a differentiable function such that $f(0)=(0,0)$.
Then there exists a continuous function $g:[0,1]\to\mathbb R^2$ such that:
1) $g(0)=(0,0)$
2) $g([0,1])\cap f([0,1])=\{(0,0)\}$.
3) $g$ is not a constant function. (Forgot to add this earlier.)

Basically what I am trying to prove is that if we have a differentiable curve in $\mathbb R^2$ whic passes through origin then we can find a continuous curve in $\mathbb R^2$ which intersects the gives curve only at origin.

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#### HallsofIvy

##### Well-known member
MHB Math Helper
What's wrong with defining, say, g(x)= f(x)+ (x2, x2)?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
What's wrong with defining, say, g(x)= f(x)+ (x2, x2)?
It might so happen that $g(x_1)=f(x_1)+(x_1^2,x_1^2)=f(x_2)\neq (0,0)$.

#### Opalg

##### MHB Oldtimer
Staff member
[From my phone by Tapatalk, so this has to be short. ]

Suppose $f (t)=t^3 \sin (1/t) e^{i\pi/t}$, with $f (0) =0$. This is differentiable, and I don't see how a function $g$ can get away from the origin without crossing the graph of $f$.

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