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Show that an open ball in $\mathbb{R^n}$ is a connected set.

Attempt at a Proof: Let $r>0$ and $x_o\in\mathbb{R^n}$. Suppose $B_r(x_o)$ is not connected. Then, there exist $U,V$ open in $\mathbb{R^n}$ that disconnect $B_r(x_o)$. Without loss of generality, let $a\in B_r(x_o)$: $a\in U$. Since $U$ is open, for some $r_1>0$, $B_{r_1}(x_o)\subseteq U$. Since $(U\cap B_r(x_o))\cap (V\cap B_r(x_o))=\emptyset$, $a\not\in V$. Thus, $\forall b\in V, d(a,b)>0$. But then for some $b'\in V: b'\in B_r(x_o)$ and some $r>o$, $d(a,b')>r$. Contradiction since both $a$ and $b'$ were in the ball of radius $r$.

Is this the general idea?

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What is $A$? And, your definitions of connectedness is with disjoint open sets, I assume. –  user21436 Feb 25 '12 at 22:39
    
sorry about the ambiguity, edited. my definition of connectedness deals with open sets disjoint in $B_r(x_o)$. –  Emir Feb 25 '12 at 22:56
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4 Answers 4

An easy way to see this : a convex set is by construction path-connected, since $\forall x, y \in C$ with $C$ convex, $\lambda x + (1-\lambda)y \in C$ by convexity (so that you can choose the line between $x$ and $y$ as a path). Therefore since the unit ball is convex (show it if you wish to), and since path-connected sets are also connected, you're done.

I'm not quite sure your proof is correct though. Why would $d(a,b') > r$? I don't see an explicit reason for this.

Hope that helps,

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we haven't done anything about convexity or path connectedness yet, thanks though. Anyway, my logic for the proof is, because $\forall b\in V, d(a,b)>0,$ for some $r\in\mathbb{R}$ and some $b'$ $d(a,b')>r$. –  Emir Feb 25 '12 at 22:55
    
Then something's wrong, because you fixed $r$ when you began your proof, thus you cannot change it afterwards. –  Patrick Da Silva Feb 25 '12 at 23:22
    
you're right. Is there still a way to show that $d(a,b)>r$ for some $b'\in B_r(x_o): b'\in V$, perhaps by taking $b=\sup V$? Or am I barking up the wrong tree with this proof attempt? –  Emir Feb 25 '12 at 23:30
    
I've given some thought about it but right now this is the only way I see fit. What have you seen in your class? Perhaps that might give me an idea of where I could start for you. –  Patrick Da Silva Feb 25 '12 at 23:33
    
we've covered compactness, connectedness and separability and all of the related notions, we just haven't seen anything dealing with path-connectedness and convexity (though if they are essential to the proof, I would have to ask the prof to go over them) –  Emir Feb 25 '12 at 23:44
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The last line, "Contradiction since both a and b′ were in the ball of radius r" is NOT a contradiction. It is only a contradiction if one of the points is the center of the ball, otherwise the farthest two points can be is 2r.

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Hint: An open ball in $\mathbb R^n$ is convex, as can be shown with the triangle inequality. And convex set is path-connected, thus connected.

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As mentioned, the simplest way to see this is via path connectedness. However, I think induction works too:

We already know that open balls in $\mathbb{R}$ (i.e. open intervals) are connected. Suppose that open balls in $\mathbb{R}^{n-1}$ are connected. Consider an open ball $B(x,r)$ in $\mathbb{R}^n$, and view it as $$B(x,r)=\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}\cup\bigcup_{\varepsilon\in(x_n-r,x_n+r)}\{y\in B(x,r):y_n=\varepsilon\}.$$ Now $\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}$ is an interval (and is then connected) and each $\{y\in B(x,r):y_n=\varepsilon\}$ is homeomorphic to an open ball in $\mathbb{R}^{n-1}$ (so that each such set is then connected). Furthermore $\{(x_1,\ldots,x_{n-1},y):|x_n-y|<r\}\cap\{y\in B(x,r):y_n=\varepsilon\}\neq\emptyset$ for each $\varepsilon\in(x_n-r,x_n+r).$ There is a theorem (likely in your textbook) that now allows you to conclude that $B(x,r)$ is connected.

In particular the theorem in question states that $X\cup\bigcup_\alpha Y_\alpha$ is connected if $X$ is connected and each $Y_\alpha$ is connected, and furthermore $X\cap Y_\alpha\neq\emptyset$ for each $\alpha.$ This is a straightforward corollary of the more basic theorem that $\bigcup_\alpha Y_\alpha$ is connected if each $Y_\alpha$ is connected and $Y_\alpha\cap Y_\beta\neq\emptyset$ for each $\alpha,\beta$.

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