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Possible Duplicate:
Arcwise connected part of $\mathbb R^2$

As the topic,if $A\subset\mathbb{R^2}$ is countable, does $\mathbb{R^2}\setminus A$ path connected??? I know the answer is it is path connected but not sure how to prove it.

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marked as duplicate by Brian M. Scott, Did, MJD, JSchlather, EuYu Nov 19 '12 at 8:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
@BrianM.Scott Gosh, I just duplicated your answer... –  Did Nov 19 '12 at 8:14
    
@did: Well, it is the natural thing to do, I think. –  Brian M. Scott Nov 19 '12 at 8:15

1 Answer 1

Here is a marginal variation:

Choose $x_0,x_1 \in A^c$. Let $d \neq 0$ be orthogonal to $x_1-x_0$. Define the collection of paths $\gamma_\alpha(t) = x_0 + t(x_1-x_0) + \alpha t(1-t) d$. Since $\{\gamma_\alpha\}_{\alpha \in [0,1]}$ is uncountable, there exists $\alpha_0 \in [0,1]$ such that $\gamma_{\alpha_0}[0,1] \cap A = \emptyset$. Hence $A^c$ is path connected.

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