Assume $X$ is a connected open subset of $\mathbb{R^2}$ show that $X$\ $\{c\}$, $c\in X$, is connected. I have thought two ways of proving this but none of which looked convincing to me.

Method 1: Assume for a contradiction that $X$\ $\{c\}$ is disconnected then $X$\ $\{c\}$ $= A \cup B$ for some $A,B$ open sets and $A\cap B=\emptyset$. Hence $X=A\cup B\cup\{c\}$. Furthermore, since $X$ is open, there exists $\epsilon>0$ such that $B(c,\epsilon)$\ $\{c\}$$\subset A\cup B$. Then I am not too sure how to continue this argument. I thought about using the fact that $c\notin A\cup B$ and the fact that $A\cap B=\emptyset$.

Method 2 (sketch): This way involves showing $X$\ $\{c\}$ is path connected. Take $x,y\in X$\ $\{c\}$. If $x,y,c$ are not collinear then it is trivial. Suppose $x,y,c$ are collinear then since X is open, there exists $z$ such that $x,y,z$ are not collinear. Then it is easy to find a path from $x$ to $z$ and another path from $z$ to $y$, compose two together we have find a path for any two points in $X$\ $\{c\}$. Lastly since path connectedness implies connectedness and hence it is shown?

  • $\begingroup$ despite the (just accepted) answer to the contrary, your method 2 is incomplete, you seem to implicitly assume that $X$ is also convex (that is, every line segment with endpoints in $X$ is a subset of $X$). Your method 1 (or any method) would not work in $\Bbb R^1$ (since the result does not hold), it would only work in $\Bbb R^n,n\ge2$, so whichever method you use should take something specific for $\Bbb R^2$. Perhaps take a small circle $S\subset X$ centered at $c$ and construct paths in $X$ from points to/from this circle, avoiding $c$. Perhaps there is something easier. $\endgroup$ – Mirko Oct 27 at 21:07
  • $\begingroup$ For a non-convex open connected set think of $X=$ the plane with the negative part of the $x$-axis (including the origin) removed. Now also remove from $X$ the point $c=(1,0)$ and find a path in $X\setminus\{c\}$ from $x=(-1,1)$ to $y=(-1,-1)$. Then $c,x,y$ are not collinear, yet the line segment from $x$ to $y$ is not a path in $X$, since it intersects the negative $x$-axis, that is it does not stay in $X$. $\endgroup$ – Mirko Oct 27 at 21:15

Your method (2) might work if you provide more details, to make sure it works without the extra (implicit) assumption that $X$ is convex. The additional detail required may make that approach a bit long.

Here is how method (1) could be completed.

Using that $X$ is open take a small positive $\varepsilon$ such that the ball $E=\{p:dist(c,p)<\varepsilon\}$ is contained in $X$.

Note that the deleted ball (or disk, since we work in the plane) $D=\{p:0<dist(c,p)<\varepsilon\}$ is (obviously) path-connected. You could use your proof (method 2), using line segments, since $E$ is convex. Any two points in $D$ could be connected by either a line segment missing $c$, or by a path consisting of two such line segments.

Now take any two disjoint open sets $A$ and $B$ such that $X\setminus\{c\}=A\cup B$. Assume without loss of generality that $A\cap D\neq\varnothing$. Since $D$ is connected, we must have $D\subseteq A$. Let $C=A\cup\{c\}$. Then $C$ is open, disjoint from $B$, and $X=C\cup B$. Since $X$ is connected, $B$ must be empty. Therefore one cannot represent $X\setminus\{c\}$ as the union of two nonempty disjoint open sets $A$ and $B$, showing that $X\setminus\{c\}$ is connected.


Your Method 2 works, so the answer is yes.

In Method 1 you falsely conclude that $B(c, \epsilon) \subset A \cup B$, when in fact it only follows that $B(c, \epsilon) \subset X = A \cup B \cup \{c\}$. I doubt that there is an easy way to avoid constructing a path, but maybe someone else has an idea.

What you could do in the specific case of $X = \mathbb R^2$ (or, similarly, a ball) is the following. The set $\mathbb R^2 \setminus \{c\}$ deformation retracts onto $S^1$, so has the same homotopy type. In particular, $\mathbb R^2 \setminus \{c\}$ is path-connected since $S^1$ is.

Edit: In general, you might want to do something like this. Let $B \subset X$ be a ball around $c$. Then $B \setminus \{c\}$ is path-connected (there your argument works). Since $X$ was path-connected (you use here again that $X$ is open), so is $X \setminus \{c\}$: Any path going through $c$ can be replaced by using connectedness in $B$.

  • $\begingroup$ Method (2) does not work, could you provide more details why you think it does? It only works when $X$ is convex, but not if $X$ is the plane with the negative part of the $x$-axis (including the origin) removed. Now also remove from $X$ the point $c=(1,0)$ and find a path in $X\setminus\{c\}$ from $x=(-1,1)$ to $y=(-1,-1)$. Then $c,x,y$ are not collinear, yet the line segment from $x$ to $y$ is not a path in $X$, since it intersects the negative $x$-axis, that is it does not stay in $X$. $\endgroup$ – Mirko Oct 27 at 21:19
  • $\begingroup$ See my edit; it requires a small trick. $\endgroup$ – Levi Oct 27 at 21:26
  • $\begingroup$ ...replaced by ?what?? using connectedness in $B$ $\endgroup$ – Mirko Oct 27 at 21:29
  • $\begingroup$ Suppose you have a path between two points going through $c$. Then this path does this by first doing something, then getting into $B$ and reaching a point $a \neq c$, then maybe passing through $c$ a few times, then reaching a point $b \neq c$, and then leaving $B$ without going over $c$ again. Now replace whatever the path did between $a$ and $b$ by some path in $B \setminus \{c\}$ between $a$ and $b$. The path obtained in this way is a path in $X \setminus \{c\}$ between your original two points. $\endgroup$ – Levi Oct 27 at 21:33
  • $\begingroup$ I know, but describing this precisely is not easy, and the "passing through $c$ a few times" part could actually be very messy. (It could pass through $c$ infinitely many times.) The "getting into $B$" and "leaving $B$" part may also require a more careful description, you may need to talk about the circle that goes with $B$ and how/when the path crosses it. $\endgroup$ – Mirko Oct 27 at 21:57

Let $\ c\in X\subseteq\mathbb R^2,\ $ where $\ X\ $ is open in $\ \mathbb R^2.\ $ (The case of $\ x\in\mathbb R^2\setminus X\ $ is trivially true).

Let $\ G\ H\ $ be disjoint open subsets of $\ R^2\ $ such that $\ G\cup H=X\setminus\{c\}.\ $ Consider an open ball

$$ B\ := \mathscr B(c, r)\ \subseteq X\qquad\qquad (\mbox{where}\quad r>0) $$

Then each circle $\ \mathscr C(c, t),\ $ where $\ 0<t<r,\ $ is contained in one of $\ G\ H,\ $ (and is disjoint with the other one). By Dedekind axiom of real numbers we see that all these circles are contained in the same set $\ G\ $ or $\ H;\ $ say, in $\ G.$

Then $\ G\cup\{c\},\ $ i.e.

$$ G\cup\{c\}\ =\ G\cup B\ \subseteq\ X $$

is open; furthermore

$$ (G\cup\{c\})\ \cup\ H\,\ =\,\ X $$


$$ (G\cup\{c\})\ \cap\ H\,\ =\,\ \emptyset $$

This means that $\ X=G\cup\{c\},\ $ i.e. $\ X\setminus\{c\}=G\ $ cannot be decomposed into two non-empty open disjoint subsets $\ G\ H.\ $ This means that $\ X\setminus\{c\}\ $ is connected.



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