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I'm trying to prove that if you remove a finite number of points from $\Bbb R^2$, you get a connected set. Let $X\subset R^2$ be a finite set. Every open set in the induced topology of $\Bbb R^2 - X$ is of the form $O-X$, where $O$ is open in $\Bbb R^2$. We must show that if $O-X$ is also closed in $\Bbb R^2 - X$, then it's either the whole space, or the empty set. So assume it's closed (in the induced topology).

I was hoping to show that therefore $O$ would have to be closed in $\Bbb R^2$, and since $\Bbb R^2$ is connected, therefore $O$ is either the plane or the empty set. So if $x\not\in O$, then if $x\not\in X$, we can wrap $x$ in a neighborhood that doesn't touch $O$, since $O-X$ is closed in the induced topology. However, if $x\in X$, we can't use that argument, and this is where I get stuck.

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why dont you try to prove that $\mathbb R^2-X$ is path connected istead –  omar Jan 25 at 8:31

3 Answers 3

up vote 3 down vote accepted

A slightly slick way that ignores the direct topological proof is to prove the stronger condition that the set is path-connected. Fix two points $a, b \in \mathbb{R}^2 - X$; choose an $\alpha \in \mathbb{R}$ such that the line through $a$ with slope $\alpha$ does not encounter $X$. This is possible since $X$ is finite, but there are infinitely many choices for $\alpha$.

Likewise, choose a $\beta \in \mathbb{R} - \{\alpha\}$ such that the line through $b$ with slope $\beta$ doesn't meet $X$. The two lines have different slopes, so intersect; now connect segments appropriately and you're done.

Proving that a path-connected space is connected is fairly easy, and is done e.g. here.

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Let's step back and look at the general strategy. Note that the analogous statement for the line $\mathbb R^1$ is blatantly false! Although $\mathbb R^1$ is connected, removing even a single point leaves a disconnected space. Therefore, a successful proof of the desired statement for $\mathbb R^2$ must somehow exploit the difference between $\mathbb R^2$ and $\mathbb R^1$. It will not be enough to juggle the definitions for connected spaces and subspace topologies; any such argument is doomed to failure.

What, then, is the difference between $\mathbb R^2$ and $\mathbb R^1$? Well, the most obvious difference is the dimension: $2$ versus $1$. In a $2$-dimensional space, you can have $1$-dimensional families of lines. Playing with families of lines, you can find your way to the answers by T. Bongers and copper.hat.

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Choose two points $p_1,p_2 \in C=\mathbb{R}^n \setminus X$. Let $v$ be any non-zero vector such that $v \bot (p_1-p_2)$. Such a vector exists for all $n > 1$.

Let $t \in \mathbb{R}$, and define $\gamma_t:[0,1] \to \mathbb{R}^n$ by $\gamma_t( \lambda ) = \lambda p_2+(1-\lambda) p_1 + t\lambda (1-\lambda)v$. Note that if $t \ne s$, then $\gamma_t((0,1)) \cap \gamma_s((0,1)) = \emptyset$.

Let $T = \{ t | \gamma_t((0,1)) \cap X \ne \emptyset \}$. Since $X$ is finite, we see that $T$ is finite, hence if $t \notin T$, then $ \gamma_t([0,1]) \subset C$ and hence $C$ is path connected.

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