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Let A and B be two finite subsets of $\mathbb R$. Describe a necessary and sufficient condition for the spaces $\mathbb R\setminus A$ and $\mathbb R\setminus B$ to be homeomorphic.

I think $|A|=|B|$. Am I right?

t.b.
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poton
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3 Answers3

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I’m going to assume that you know that all open intervals in $\Bbb R$, including the open rays of the form $(x,\to)$ and $(\leftarrow,x)$, are homeomorphic to one another; if you don’t, you should try to prove it. In what follows I’ll use the term open interval to include the open rays.

If $A$ is finite, then $\Bbb R\setminus A$ is the union of $|A|+1$ open intervals, so it’s homeomorphic to the disjoint union of $|A|+1$ copies of $(0,1)$. Clearly, then, $\Bbb R\setminus A$ is homeomorphic to $\Bbb R\setminus B$ whenever $|A|=|B|$: both are homeomorphic to the disjoint union of $|A|+1$ copies of $(0,1)$.

Suppose that $|A|=m$ and $|B|=n$, where $m<n$. To show that $\Bbb R\setminus A$ and $\Bbb R\setminus B$ are not homeomorphic, you need only show that the disjoint union of $m$ copies of $(0,1)$ is not homeomorphic to the disjoint union of $n$ copies of $(0,1)$. This is true because homeomorphisms, and indeed continuous maps in general, preserve connectedness. Suppose that $h:\Bbb R\setminus A\to\Bbb R\setminus A$ is continuous. Each of the $m$ copies of $(0,1)$ making up $\Bbb R\setminus A$ is a connected subset of $\Bbb R\setminus A$, so $h$ must send it to a connected subset of $\Bbb R\setminus B$. Every connected subset of $\Bbb R\setminus B$ lies entirely within some one of the $n$ copies of $(0,1)$ making up $\Bbb R\setminus B$, so the image of $\Bbb R\setminus A$ under $h$ can fill up at most $m$ of the $n$ copies of $(0,1)$ making up $\Bbb R\setminus B$. Thus, $h$ cannot possibly be a surjection, and $\Bbb R\setminus A$ cannot be homeomorphic to $\Bbb R\setminus B$.

Brian M. Scott
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  • Hello sir I am having a problem how did you get this result that $\mathbb R\setminus A$ is homeomorphic to... $(0,1)$. – Learnmore Jan 12 '15 at 05:01
  • It will be very nice of you to help me .thanks @Brian M.Scott sir – Learnmore Jan 12 '15 at 05:01
  • @learnmore: Do you mean the second paragraph? Let $A={a_1,\ldots,a_n}$, where $a_1<\ldots<a_m$. Then $X\setminus A$ is the union of the $m+1$ open intervals $(\leftarrow,a_1),(a_1,a_2),\ldots,(a_{m-1},a_m)$, and $(a_m,\to)$. If $B$ also has $m$ elements, say $b_1<\ldots<b_m$, $\Bbb R\setminus B$ is a similar union of $m+1$ open intervals. $(\leftarrow,a_1)$ is homeomorphic to $(\leftarrow,b_1)$, $(a_1,a_2)$ to $(b_1,b_2)$, and so on, so $\Bbb R\setminus A$ is homeomorphic to $\Bbb R\setminus B$. – Brian M. Scott Jan 12 '15 at 16:43
  • yes how did you get the homeomorphism relation sir @Brian M Scott – Learnmore Jan 13 '15 at 02:06
  • @learnmore: Do you know that $(a,b)$ and $(c,d)$ are homeomorphic whenever $a<b$ and $c<d$? – Brian M. Scott Jan 13 '15 at 02:13
  • yes I know that sir @Brian M. Scott. – Learnmore Jan 13 '15 at 02:21
  • @learnmore: That's all you need: $(\leftarrow,a_1)$ is homeomorphic to $(\leftarrow,b_1)$ by some homeomorphism $h_0$, $(a_1,a_2)$ is homeomorphic to $b_1,b_2)$ by some homeomorphism $h_1$, and so on, until we get to $(a_n,\to)$ homeomorphic to $(b_n,\to)$ by some homeomorphism $h_n$. Now just combine the homeomorphisms $h_0,\ldots,h_n$, and you have the desired homeomorphism between $\Bbb R\setminus A$ and &\Bbb R\setminus B$. – Brian M. Scott Jan 13 '15 at 02:33
  • lets give an example suppose we have $A={0}$ $B={0,1}$ then $R\setminus A=(-\infty,0)\cup (0,\infty)$ and $R\setminus B=(-\infty,0)\cup (1,\infty)\cup (0,1)$ right ?? – Learnmore Jan 13 '15 at 03:02
  • By using your point I can say $R\setminus A=(0,1)\cup (0,1)$ and $R\setminus B=(0,1)\cup (0,1)\cup (0,1)$ – Learnmore Jan 13 '15 at 03:04
  • Can u now please say what is meant by copies of $(0,1)$ – Learnmore Jan 13 '15 at 03:05
  • Is my reasoning correct @Brian M. Scott – Learnmore Jan 13 '15 at 03:06
  • @learnmore: I’m sorry: I completely forgot about these comments. Your examples of $\Bbb R\setminus A$ and $\Bbb R\setminus B$ are correct. You can’t say that $\Bbb R\setminus A$ is equal to $(0,1)\cup(0,1)$ (which is just $(0,1)$), or even to the union of two disjoint copies of $(0,1)$, but it is homeomorphic to the union of two disjoint copies of $(0,1)$. The simplest way to interpret copies of $(0,1)$ in this context is to say that the union of $n$ copies of $(0,1)$ is the space ${1,2,\ldots,n}\times(0,1)$, where ${1,\ldots,n}$ has the discrete topology. – Brian M. Scott Jan 14 '15 at 20:16
  • thank you sir for your help – Learnmore Jan 15 '15 at 06:16
  • @learnmore: You’re welcome. Again, I apologize for forgetting to answer promptly. – Brian M. Scott Jan 15 '15 at 14:07
  • Please dont apologize; you are a professor and you are engaged in so many works.I am pleased that you have shown so much respect towards my questions.love you and have a good day @Brian M. Scott – Learnmore Jan 15 '15 at 14:16
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Try proving by induction on $n$ that if $A$ is a finite subset of $\mathbb{R}$ with $|A| = n$, then $\mathbb{R} \setminus A$ is homeomorphic to a disjoint union of $n+1$ copies of $\mathbb{R}$. This suffices.

Pete L. Clark
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Assume $A=\{a_1, \ldots, a_n\}$ with $a_1<\ldots<a_n$ and $B=\{b_1, \ldots, b_m\}$ with $b_1<\ldots<b_m$. If $n=m$ we define a homeomorphism $f:A\to B$ (or vice versa) in the obvious manner: $$\phi(x)=\begin{cases}x+b_1-a_1 & x<a_1\\\frac{b_{k+1}-b_k}{a_{k+1}-a_k}(x-a_k)+b_k & a_k<x<a_{k+1}, 1\le k<n\\x-b_n-a_n& x>a_n\end{cases}$$

If $n\ne m$, then $\mathbb R\setminus A$ and $\mathbb R\setminus B$ cannot be homeomorphic because we can recover $n$ from $A$ as a topological invariant: It is one less than the number of connected components, i.e. the maximal number of disjoint (relatively) open sets that can cover the whole space. To see this you need to know that each interval is connected, which is not hard.

Hagen von Eitzen
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