# Proving the connectedness of this subset

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?

• 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. – Mirko Oct 27 at 21:07
• 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 $X$. – 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) $$D=\{p:0 is (obviously) path-connected. You could use your proof (method 2), using line segments, since $$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.

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 $$B$$.

• 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 $X$. – Mirko Oct 27 at 21:19
• See my edit; it requires a small trick. – Levi Oct 27 at 21:26
• ...replaced by ?what?? using connectedness in $B$ – Mirko Oct 27 at 21:29
• 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. – Levi Oct 27 at 21:33
• 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. – 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 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

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

while

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

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.

Great!