Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 4 '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 4 '13 at 13:32
up vote 4 down vote accepted

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 $B(p,\delta)$.
share|cite|improve this answer
Thanks very much for your answer, can you give me a hint in how you would show that they are attainable? – user53076 Jan 4 '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 4 '13 at 17:46

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 $U$.

share|cite|improve this answer
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – Stefan Hansen Jan 4 '13 at 13:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.