# Bijection of a segment in $\mathbb{R}^n$ to an interval. Path connectedness.

So this question is probably trivial.

It seems to me that I have a problem understanding how $\mathbb{R}^n$ generally looks like, and proving things with it.

Context: I started from the fact that I want to prove that any open connected subset of $\mathbb{R}^n$ is path connected. During the proof I use that between any 2 points in an open ball in $\mathbb{R}^n$ you can draw a line segment included in the ball.

So from this I want to show that the ball is path connected. Fix x and y in B. If there is a bijection from such a ball (call it $B$) such that f(x) = a and f(y) = b.then exists $f\colon B \rightarrow[a,b]$ bijective and continuous, and then $f^{-1}\colon[a,b]\rightarrow B$ will be bijective (here, the fact that it is also continuous is not self implied) with $f(a) = x$ and $f(b) = y$. There exists a bijection from g\colon [0,1] \rightarrow [a,b] with g(0)= a, and g(1) = b, so $f^{-1}. g\colon [0,1] \rightarrow B$ such that $(f^{-1}. g)(0) = x$ and $(f^{-1}. g)(1) = y$.

The issue is that I have to find $f$ because $f^{-1}$ is not guranteed to be continuous.

So what came to mind is projecting the line segment from $x$ to $y$ one of of the axis of the space. This should be a segment, but it is not necessarily bijective.

So what I need is a continuous bijective function from an interval $[a,b]$ to two points connected by a segment in $\mathbb{R}^n$ (ie. any $2$ points in $\mathbb{R}^n$).

Thank you! (I dumped a lot of my train thought here, but I think that might for someone who wants to explain this to me)

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I don't think there will exist a continuous bijection from a ball in $\Bbb R^n$ to an interval in $\Bbb R$ in general. My suggestion to prove this: pick a point $x_0\in U$, where $U$ is your given open, connected set, and consider the set of points that can be connected by a path to $x_0$. If you can show this set is open and closed in $U$, you'll be done since $U$ is connected. – Clayton May 4 '13 at 16:49
That is exactly what I am doing. But in order to show that, I need to show that there exists a path between any 2 points in an open ball in $R^n$. I do not claim that there exists a bijection from a ball in $R^n$ to a closed interval in $R$, but from a segment in a ball in $R^n$ to a closed interval (a closed segment) in $R$. – elaRosca May 4 '13 at 16:53
From my question:"So what I need is a continuous bijective function from an interval [a,b] to two points connected by a segment in Rn (ie. any 2 points in R^n)." – elaRosca May 4 '13 at 16:54
Also from your question: "If there is a bijection from such a ball (call it B) such that $f(x) = a$ and $f(y) = b$, then exists $f:B−>[a,b]$ bijective and continuous," – Clayton May 4 '13 at 16:58

Without loss of generality you can assume that your ball is the one described by $x_1^2+\ldots+x_n^2<1$, otherwise use a translation (to bring the center of the ball to the origin) and a homothety to make the ball of the appropriate size. Your two points, $a,b$ will satisfy then $|a|<1,|b|<1$.
Consider the following path $C:[0,1]\to\mathbb{R}^n,\quad C(t)=ta+(1-t)b$, what we need to prove to conclude that $B$ is path connected is to show that $|C(t)|<1$ for all $t$, but this is very simple: $|C(t)|=|ta+(1-t)b|\leq t|a|+(1-t)|b|<t+(1-t)=1$
The first inequality is due to the fact that $\mathbb{R}^n$ is metric vector space and that $t>0, t-1>0$.