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I'm trying to prove that if $U \in \mathbb{R}^2$ is open and path connected, then for a point $ p \in U$ we have $U \smallsetminus \{p\}$ still path connected.

Start by taking $x,y \in U$. As path connected there exists continuous $\gamma : [0,1] \to U$ such that $\gamma (0) = x$ and $\gamma(1) = y$.

Take $p \in U$.

If $p$ is not on the the path $\gamma$. Then $x$ and $y$ are still path connected.

If $p$ is lying on the path $\gamma$ then as $U$ is open, there exists a $\delta >0$ such that the ball $B(p,\delta) \subset U$. The path $\gamma$ crosses the boundary of this ball. Now one can create a new path which traverses round the edge of the ball. Hence $x$ and $y$ are still path connected.

Is this proof valid? Can anyone think of a simpler way of doing it?

user53076
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    The heart of your argument, "The path $\gamma$ crosses the boundary of this ball. Now one can create a new path which traverses round the edge of the ball.", is rather unconvincing. How exactly do we create this new path? Can we be sure our method will work if, for example, $\gamma$ crosses the boundary infinitely many times? – Chris Eagle Jan 04 '13 at 13:13
  • And what if $x$ and $y$ are in fact both inside the ball? Then the path need not cross the ball's boundary at all. – Chris Eagle Jan 04 '13 at 13:32

2 Answers2

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Your idea is good, but the proof is not rigourous enough.

  1. You should choose $\delta>0$ such that $\delta < \min\{\|x-p\|, \|y-p\|\}$ and $B(p,c\delta)\subset U$ for some $c>1$. The first condition is needed because you want to keep $x$ and $y$ outside the open disc. The second condition is needed, otherwise the boundary of $B(p,\delta)$ may not lie inside $U$.
  2. As Chris Eagle mentioned in his comment, the original path $\gamma$ may cross the closed disc $\bar{B}(p,\delta)$ multiple times. So you should consider the first entry point $t_1 = \inf\{t: \|\gamma(t)-p\|\le\delta\}$ and the last exit point $t_2 = \sup\{t: \|\gamma(t)-p\|\le\delta\}$. You should show that these infinum and supremum are attainable minimum and maximum. Then you can remove the part of $\gamma$ on $[t_1,t_2]$ and replace it by a circular arc on the boundary of $B(p,\delta)$.
user1551
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  • Thanks very much for your answer, can you give me a hint in how you would show that they are attainable? – user53076 Jan 04 '13 at 17:11
  • Show that $S={t:|\gamma(t)-p|\le\delta}$ is bounded, closed and nonempty. Hence it has a maximum and a minimum. – user1551 Jan 04 '13 at 17:46
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This is a comment: what if one of or both $x, y$ are inside the open nbd you have picked?

added How about this: Pick a point $q$ different from $p$ and consider the set of points in $U-\lbrace p \rbrace$ that can be path connected to $q$. This set and its complement are both open. If the complement is non-empty then it will provide a separation for $U$.

Apoha
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