# Prove that $\mathbb{R}^{n}-A$ with the standard topology is connected where $n \geq 2$ and $A \subset \mathbb{R}^{n}$ is countable.

I've been stuck on this proof for quite a while. While I realize it is much easier to show using arcwise connectedness or pathwise connectedness, I would like to complete the proof without resorting to more powerful results. I do know that $\mathbb{R}$ is connected.

• As a special case, consider $n=2$ and $A=\emptyset$. So you are asking how to prove that $\mathbb{R}^2$ is connected without resorting to path connectedness. I don't know how to do that without first reproducing the entire proof of connectivity of $\mathbb{R}$. Jul 31, 2014 at 0:53
• But connectivity of $\mathbb{R}$ differs, by at most one line of proof, from the theorem that path connectivity implies connectivity. Jul 31, 2014 at 0:56
• From what I understood of the question, they don't want a proof that $\Bbb R^n-A$ is path-connected. They want an alternative proof that $\Bbb R^n-A$ is connected with more elementary methods.
– user123641
Jul 31, 2014 at 0:58
• Let me ask, what connected spaces are you allowing us to assume? For instance, are you allowing us to assume that $\mathbb{R}$ is connected? Jul 31, 2014 at 1:05
• On second thought, contradiction isn't working as I had hoped. However, I do know that if $A,B$ are connected spaces then $A \times B$ is connected. So letting $A=B=\mathbb{R}$ and knowing $\mathbb{R}$ is connected then $\mathbb{R} \times \mathbb{R}$ is connected. Jul 31, 2014 at 3:00

Since you know that $\mathbb{R}$ is connected, let's use that.

The underlying idea will come from the following (pretty horrible) proof that $\mathbb{R}^2$ is connected. Just to set up some notation, given any $D \subseteq \mathbb{R}^2$ and $a,b \in \mathbb{R}$ I'll denote $$D_{\langle a,- \rangle} := \{ y \in \mathbb{R} : \langle a,y \rangle \in D \}; \qquad D_{\langle -,b \rangle} := \{ x \in \mathbb{R} : \langle x,b \rangle \in D \}.$$

Suppose that $U , V \subseteq \mathbb{R}^2$ are disjoint nonempty open sets whose union is $\mathbb{R}^2$. Picking $\langle a,b \rangle \in U$, note that if $V_{\langle a,- \rangle} = \varnothing$ then $U_{\langle a,- \rangle} = \mathbb{R}$ and so both $U_{\langle -,d \rangle}$ and $V_{\langle -,d \rangle}$ are nonempty for any $\langle c,d \rangle \in V$. We may then without loss of generality assume that $a \in \mathbb{R}$ has been picked so that $U_{\langle a,- \rangle}$ and $V_{\langle a,- \rangle}$ are both nonempty.

But note that $U_{\langle a , - \rangle} , V_{\langle a , - \rangle}$ are disjoint nonempty open subsets of $\mathbb{R}$ whose union is $\mathbb{R}$, which is impossible!

Now we'll modify the above to show that $X := \mathbb{R}^2 \setminus A$ is connected for countable $A \subseteq \mathbb{R}^2$. The trick will be to avoid the set $A$ in order to arrive at the same contradictory conclusion as above.

Suppose that $U, V$ are open subsets of $\mathbb{R}^2$ such that

1. $U \cap X \neq \varnothing \neq V \cap X$;
2. $X \subseteq U \cup V$.
3. $(U \cap V ) \cap X = \varnothing$.

Claim. If $W \subseteq \mathbb{R}^2$ is open and nonempty, then there is a $\langle a,b \rangle \in W$ such that $A_{\langle a,- \rangle} = \varnothing = A_{\langle -,b \rangle}$.

proof. Picking any $\langle a^\prime,b^\prime \rangle \in W$, since $W$ is open there is an $\varepsilon > 0$ such that $\langle a,b \rangle \in W$ for all $a \in (a^\prime - \varepsilon , a^\prime + \varepsilon)$ and $b \in (b^\prime - \varepsilon , b^\prime + \varepsilon)$. Since $A$ is countable there are only countably many $a \in A$ such that $A_{\langle a,- \rangle} \neq \varnothing$ and countably many $b \in \mathbb{R}$ such that $A_{\langle -,b \rangle} \neq \varnothing$. Thus there must be $a \in (a^\prime - \varepsilon , a^\prime + \varepsilon)$ such that $A_{\langle a,- \rangle} = \varnothing$ and a $b \in (b^\prime - \varepsilon , b^\prime + \varepsilon)$ such that $A_{\langle -,b \rangle} = \varnothing$. $\dashv$

Picking $\langle a,b \rangle \in U$ as in the claim, note that if $V_{\langle a,- \rangle} = \varnothing$ then $U_{\langle a,- \rangle} = \mathbb{R}$. Picking any $\langle c,d \rangle \in V$ as in the claim, it follows that $U_{\langle -,d \rangle}$ and $V_{\langle -,d \rangle}$ are both nonempty. So without loss of generality there is an $a \in \mathbb{R}$ such that $A_{\langle a,- \rangle} = \varnothing$ and both $U_{\langle a,- \rangle}$, $V_{\langle a,- \rangle}$ are nonempty.

But now we're in the same situation as above: $U_{\langle a,- \rangle}$ and $V_{\langle a,- \rangle}$ are disjoint nonempty open subsets of $\mathbb{R}$ whose union is $\mathbb{R}$, which is impossible!

The basic idea can be extended (it won't be pretty, though) to show that $\mathbb{R}^n \setminus A$ is connected for all $n \geq 2$ and countable $A$.

• +1. And to get back to the point of my comments to @user166967, assuming that $\mathbb{R}$ is connected is very close to assuming "Theorem: Path connectivity implies connectivity". In fact, some of the details in this answer are basically repeating, over and over again, details of the proof of that Theorem. Jul 31, 2014 at 17:40
• I'm not following the proof as of yet, I'll have to mull it over for a while. Thank you for contributing this though! It looks like there is some good stuff here. Jul 31, 2014 at 19:52

Given two points $x$ and $y$ we find a path from $x$ to $y$ avoiding $A$. Consider perpendicular bisector of the segment form $x$ to $y$. For each point $z$ on this line take the straight line from $x$ to $z$ then the straight line from $z$ to $y$. As $z$ varied these paths are all disjoint. Since there are uncountablely many $z$ and $A$ is countable one of these paths will not contain any point for $A$.

• The OP specifically asked to avoid pathwise connectedness. Jul 31, 2014 at 0:57