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.

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    $\begingroup$ 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 $\mathbb{R}$. $\endgroup$
    – Lee Mosher
    Jul 31, 2014 at 0:53
  • $\begingroup$ But connectivity of $\mathbb{R}$ differs, by at most one line of proof, from the theorem that path connectivity implies connectivity. $\endgroup$
    – Lee Mosher
    Jul 31, 2014 at 0:56
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    $\begingroup$ 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. $\endgroup$
    – user123641
    Jul 31, 2014 at 0:58
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    $\begingroup$ 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? $\endgroup$
    – Lee Mosher
    Jul 31, 2014 at 1:05
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    $\begingroup$ 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. $\endgroup$
    – graydad
    Jul 31, 2014 at 3:00

2 Answers 2


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 $A$.

  • $\begingroup$ +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. $\endgroup$
    – Lee Mosher
    Jul 31, 2014 at 17:40
  • $\begingroup$ 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. $\endgroup$
    – graydad
    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 $A$.

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    $\begingroup$ The OP specifically asked to avoid pathwise connectedness. $\endgroup$ Jul 31, 2014 at 0:57

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