Struggling with arc-connectedness

[geoffrey159]

Homework Statement

Let $(E,N)$ a finite dimensional normed vector space. Let $A$ be an arc-connected subset of $E$, and $P$ be a non-empty subset of $A$ that is both open and close in $A$. Show that $P=A$

Homework Equations

The Attempt at a Solution

By contradiction, I assume $P\neq A$, which is $C_A(P) \neq \emptyset$.

I'd like to find a continuous function from $A$ to a subset of $S$ of $\mathbb{R}$ that is not a segment.
This would be a contradiction because the image of an arc-connected set by a continuous function is arc-connected, and the arc-connected sets of $\mathbb{R}$ are segments.

I tried with the caracteristic function of $P$, but I believe it doesn't work because for any open set $O$ of $\mathbb{R}$, $f^{-1}(O)\in\{\emptyset, P,C_A(P), A \}$, and for now, I can't convince myself that $C_A(P)$ is an open set of $A$.

Do you have any idea please ?

 

[Samy_A]

I tried with the caracteristic function of $P$, but I believe it doesn't work because for any open set $O$ of $\mathbb{R}$, $f^{-1}(O)\in\{\emptyset, P,C_A(P), A \}$, and for now, I can't convince myself that $C_A(P)$ is an open set of $A$.

I think you are almost there ...
Think of all the known properties of $P$.

 

[geoffrey159]

I think that thanks to you, I have that last point that bothered me :
Since $P$ is closed in $A$, there exists a closed set $C$ of $E$ such that $P = C \cap A$.
So $C_A(P) = C_E(P) \cap A = (C_E(C) \cup C_E(A))\cap A = C_E(C) \cap A$.
But $C_E(C)$ is an open set of $E$. Therefore $C_A(P)$ is open in $A$.

So now I am sure that the caracteristic function is continuous from $A\rightarrow \{0,1\}$ which is absurd because $\{0,1\}$ is not arc-connected.

Do you agree ?

 

[Samy_A]

I think that thanks to you, I have that last point that bothered me :
Since $P$ is closed in $A$, there exists a closed set $C$ of $E$ such that $P = C \cap A$.
So $C_A(P) = C_E(P) \cap A = (C_E(C) \cup C_E(A))\cap A = C_E(C) \cap A$.
But $C_E(C)$ is an open set of $E$. Therefore $C_A(P)$ is open in $A$.

So now I am sure that the caracteristic function is continuous from $A\rightarrow \{0,1\}$ which is absurd because $\{0,1\}$ is not arc-connected.

Do you agree ?

Yes, looks OK to me.

 

[geoffrey159]

Thanks!