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Let $R \subset \mathbb{R}^2$ denote the unit square $R = [0,1] \times [0,1]$. If $F \subset R$ is finite, is $R \backslash F$ connected?

Listing
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Jimmy Xiao
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1 Answers1

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You can even prove that $R\setminus F$ is connected if $F$ is countable. In fact, it’s path-connected.

HINT: Given two points $p$ and $q$ in $R\setminus F$, draw their perpendicular bisector; note that this has uncountably many points within $R\setminus F$. Show that there must be a point $r$ on this bisector such that the segments $\overline{pr}$ and $\overline{rq}$ lie entirely within $R\setminus F$.

Brian M. Scott
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  • You can even prove that the statement holds if the respective projections of $F$ on the axis are a null-set as shown by joriki: http://math.stackexchange.com/questions/77791/separation-of-two-points-with-null-sets – Listing Nov 26 '13 at 20:38
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    @Listing: True, but you have to work a little harder. The nice thing about the countable case is that it’s no more work at all. – Brian M. Scott Nov 26 '13 at 20:42
  • This might seem naive, but what if F bisected R. Then wouldn't R \ F be disconnected? – Jimmy Xiao Nov 26 '13 at 20:45
  • @Ralph: It takes a whole line segment to cut $R$ in two; a line segment is infinite, so a finite set can’t do it. In fact a line segment is uncountable, so a countable set can’t do it. – Brian M. Scott Nov 26 '13 at 20:47