So I was given this question and have been thinking about it for a while but I can't seem to figure out where to go with it exactly. From other answers online people refer to using the fact that $[0,1]$ is convex but I am not familiar with this idea. The other method I have seen is using the fact that intervals in $\mathbb R$, like $[0,1]$, are path connected and then constructing continuous functions to show a given ball in $\mathbb R^n$ is connected as well. If anyone could give a more detailed explanation of this method that would be really appreciated.
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By "any ball", do you mean "for all $n$, the Euclidean balls", or do you mean "for all $n$, for all norms, the balls with respect to the norm"? – Oct 29 '19 at 07:49
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Oct 29 '19 at 07:50
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You should give an example of this "other method", so that someone knows how much more detailed they need to be. – Calvin Khor Oct 29 '19 at 08:40
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@JoséCarlosSantos Thank you for the links I will be giving them a read soon. – Mohammed Oct 29 '19 at 21:40
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@CalvinKhor I am going off of the explanation I saw in this post https://math.stackexchange.com/questions/1284893/how-can-i-prove-that-any-ball-in-mathbbrn-is-connected when I was writing the question. – Mohammed Oct 29 '19 at 21:41
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@Gae.S. the question is exactly what I have as the title so I believe it is for all n, the Euclidean balls. – Mohammed Oct 29 '19 at 21:44
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If $x$ and $y$ are two points in a ball then $\gamma (t)=tx+(1-t)y$ defines a path from $x$ to $y$ lying in the ball.
In the case of an open ball $B(x_0,r)$ we have $\|x-x_0\| <r$ and $\|y-x_0\| <r$ implies $$\|tx+(1-t)y-x_0\| =\|tx+(1-t)y-(tx_0+(1-t)x_0)\|$$ $$\leq t \|x-x_0\| +(1-t)\|y-x_0\|<tr+(1-t)r=r$$. The proof is similar for a closed ball.
Kavi Rama Murthy
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I think it's possible to be confused by this. As it stands right now, isn't this a straight line? – David Oct 29 '19 at 08:10
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@David If you take two points in a ball in $\mathbb R^{n}$ the line segment connecting them is inside the ball. – Kavi Rama Murthy Oct 29 '19 at 08:15
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@KaboMurphy can you show explicitly why the segment is always contained in the ball? I can't seem to get the inequality to be < r – Mohammed Oct 30 '19 at 22:53
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Let $\|\cdot\|$ be any norm on a real vector spave $V$. Then for all $v_0\in V$ and $r>0$, the ball (with respect to this norm) $B_r(v_0)=\{\,v\in V:\|v-v_0\|<r\,\}$ is path connected. Indeed, if $v_1,v_2\in B_r(v_0)$, consider $\gamma\colon[0,1]\to V$ given by $$\gamma(t):=\begin{cases}(1-2t)(v_1-v_0)+v_0&0\le t\le \frac12\\(2t-1)(v_2-v_0)+v_0&\frac12\le t\le 1\end{cases} $$ an verify by short computations that a) $\|\gamma(t)-v_0\|<r$ for all $t\in(0,1]$, and b) $\|\gamma(t_2)-\gamma(t_1)\|<2r|t_2-t_1|$. This makes $\gamma$ a continuous map into $B_r(v_0)$, i.e., a path from $v_1$ to $v_2$.
Hagen von Eitzen
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