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The question is:

Show that if $A⊆\Bbb R^n$ is finite and $n≥2$, then the set $\Bbb R^n\backslash A$ is connected in $(\Bbb R^n,τ_d)$ where $τ_d$ is the topology induced by the Euclidean distance in $\Bbb R^n$.

My attempt:

Denote $A^C=\Bbb R^n\backslash A$ and suppose $A=\{a_1,…,a_n\}$. Assume by contradiction that $A^C$ is not connected, then there exists $O_1,O_2∈τ_{A^C }$ st. $A^C=O_1⋃O_2,O_1⋂O_2=∅$. Here $\tau_{A^C}$ is a subspace topology.

Intuitively, part of $A$, denoted by $A_1$ is "enclosed" in $O_1$ and the rest, denoted by $A_2$ are "enclosed" in $O_2$. Then $G_1=O_1\bigcup A_1$ is an open set in $\tau_d$, and $G_2=O_2\bigcup A_2$ is also an open set in $\tau_d$, and $\Bbb R^n=G_1\bigcup G_2$. However, $G_1, G_2$are disjoint, contradicting with the fact that $\Bbb R^n$ is connected.

However I got stuck when trying to implement this idea. Can anyone provide some help? Thank you!

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    $\begingroup$ Try showing the stronger fact that it is path-connected. You can do that by considering a straight line path, then "correcting" the straight line whenever it passes through one of the "bad" points. $\endgroup$
    – Ian
    Commented Sep 6, 2015 at 23:41
  • $\begingroup$ Very related (the same method works for any $n \ge 2$). $\endgroup$ Commented Sep 7, 2015 at 9:00

3 Answers 3

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Let's show that $\mathbb{R}^n \setminus A$ is path-connected. Take any two points $x$ and $y$. One would like to exhibit a path from $x$ to $y$. It suffices to show that there is at least one point $p$ so that $x$ and $y$ can both be connected to $p$ via a straight line that misses all the points in $A$.

For each point $a \in A$, let $L_{a}$ be the line passing through $x$ and $a$, and let $N_{a}$ be the line passing through $y$ and $a$. This describes finitely many lines. $\mathbb{R}^n$ is not a finite union of lines, because $\mathbb{R}^n$ intersects the sphere of radius $1$ at infinitely many points (the whole sphere), but a line can only intersect a sphere at $2$ points, and hence finitely many lines can only intersect a sphere at finitely many points. Thus there is some point $p$ that is not contained in the union of these lines. Call this point $p$. Can you see why this $p$ works?

[This proof generalizes nicely to the situation when $A$ is countably infinite]

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  • $\begingroup$ So, if $p\notin\bigcup\limits_{a\in A} L_a\cup N_a,$ this means there is no $a\in A$ such that $\boldsymbol{x, p, a}$ or $\boldsymbol{y,p,a}$ are collinear, right? So, if we assumed there is $a\in A$ such that a line through $\boldsymbol x$ and $\boldsymbol p$ or $\boldsymbol y$ and $\boldsymbol p$ is also passing through $a,$ we would've arrived at a contradiction because that would imply $p\in\bigcup\limits_{a\in A} L_a\cup N_a$? $\endgroup$
    – PinkyWay
    Commented Sep 24, 2021 at 11:28
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The pathwise-connected approach will work.Here is an"internal" proof : Suppose $ R \backslash A =O_1 \cup O_2$ where $O_1,O_2$ are disjoint open subsets of $R^n \backslash A$.......(1) : We show that every $ p \in R^n$ has a neighborhood $E(p)$ such that $ E(p)\backslash A \subset O_j$ for some $j \in \{1.2\}$....... (1-1) : For $p \in A$ there exists $r>0$ such that$ D= A \cap\{b : d(b,p)\le r\}$={p} ( where $d$ is the metric). Let $C=\{b : d(b,p)=r \}$. We have $ C=(C \cap O_1) \cup (C \cap O_2)$ and $ C$ is a connected subspace so $C\subset O_j$ for some $j \in \{1,2\}$. Now for each $q \in C$ let $S(q)$ be the half-open line segment from $q$ to $p$, including the point $q$ but not the point $p$. We have $S(q)=(S(q)\cap O_1) \cup (S(q)\cap O_2)$, and $S(q)$ is a connected subspace which intersects $O_j$ (because $ q \in S(q)\cap O_j$) so $S(p) \subset O_j$. So let $E(p)=\{b :d(b,p) <r\}$........(1-2) : For $p \in R^n\backslash A$ there exists $r>0$ such that $E=\{b : d(b.p)<r\}$ is disjoint from $A $, so we have $ E=(E\cap O_1)\cup (E\cap O_2)$. And $E$ is a connected subspace so $ E=E\cap O_1$ or $ E=E\cap O_2$. So let $ E(p)=E.$.......(2) : Finally for each $j\in \{1,2\}$ let $ G_j=\cup \{E(p) : E(p)\backslash A \subset O_j\}$. The sets $G_1,G_2$ are disjoint open subsets of $R^n$ and $G_1\cup G_2=R^n$ so one of them is empty.So one of $O_1,O_2$ is empty because $O_1 \subset G_1$ and $O_2 \subset G_2$.Observe that this proof is valid verbatim when we replace $R^n$ with any connected open subspace $S$ of $R^n$ and for any closed discrete subspace $A$ of $S$.

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You can use the fact that if a region is path-connected, then it is connected.

Since $A$ is finite, for any pair of points $x,y\in\Bbb{R}-A$, there is a polyline $L$ connecting $x$ and $y$ and $L\cap A=\varnothing$, i.e. we can always find a set of piecewise straight lines that has more lines than number of points in $A$ to connect $x$ and $y$ bypassing all the points of $A$. So $\Bbb{R}-A$ is path-connected, and thus connected.

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    $\begingroup$ 1/ You forgot an exponent on the $\mathbb{R}$. 2/ The whole point of the proof is to show that "it is always possible to find a curve that gets by the finite number of points in $A$", something you haven't proved. It may be intuitively obvious, but it still requires a proof. $\endgroup$ Commented Sep 7, 2015 at 8:59
  • $\begingroup$ @NajibIdrissi Isn't the "between 2 points there must be a line" enough for this? I mean, if I want to connect a and b, assume that the proper line goes through the restricted area, then add another point c, outside, and connect a to b, and b to c, and so on. Is it enough? thx in advance $\endgroup$ Commented Sep 7, 2015 at 11:19
  • $\begingroup$ @CIsForCookies It's possible that the points a and c are still aligned with one of the points you removed... $\endgroup$ Commented Sep 7, 2015 at 11:20
  • $\begingroup$ @NajibIdrissi so I can continue the process as long as I need in order to get set n lines between n+1 points such that every line is contained in the space. I think $\endgroup$ Commented Sep 7, 2015 at 11:23
  • $\begingroup$ More formally, for any $c$ in a hyperplane separating two given points $a,b$ in $\mathbb R^n-A$ there is the polygonal path $acb$ and these paths are pairwise disjoint (apart from $a,b$) for different choices of $c$. As long as we can make more than $|A|$ choices, the space is path connected. So this argument even works for countably infinite $A$ because the hyperplane has continuum-many points (this is where we use $n\ge 2$) $\endgroup$ Commented Sep 7, 2015 at 20:51

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