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Prove that if $A$ is both open and closed then $A = \mathbb{R}$ also as one suggested let $A \neq \emptyset$

You may use what ever definition of open and closed you would like, just avoid going into metric spaces, haven't covered that topic yet. My question is well essentially how to prove this statement but considering I like to do things myself, I was wondering if anyone had any particular suggestions to help me solve this.

attempted proof: Let $a\in A$ then since $A$ is open there exists an open interval $N(a,\epsilon)$ such that $$a\in N(a,\epsilon)\subset A$$ Then $a$ must be an interior point of $A$, but since $A$ is also closed then $a$ must also be an accumulation or limit point of $A$ as well.

I am going to stop here because I am not sure if this is the right approach here, any suggestions would be greatly appreciated.

Wolfy
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    You need to state that $A\ne\emptyset$. – Tim Raczkowski Apr 24 '15 at 14:17
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    It is not correct to say that every point in a closed set is an accumulation point of the set. "Closed" means that every accumulation point is a member, not that every member is an accumulation point. ${}\qquad{}$ – Michael Hardy Apr 24 '15 at 14:21
  • @MichaelHardy But isn't every member the limit of the constant sequence? Therefore also every member is an accumulation point. – Jolien Apr 24 '15 at 14:37
  • @Jolien : That does not follow. It is true that every member is the limit of a constant sequence. That doesn't make it an accumulation point. A point $a$ is an accumulation point of a set $A$ precisely if every open interval $(a-\varepsilon,a+\varepsilon)$, no matter how small, contains some point of $A$ other than $a$ itself. Many closed sets contain points that are not accumulation points of the set, but are isolated points of the set. – Michael Hardy Apr 24 '15 at 14:41
  • That is why the concept of a "perfect set" exists and is separate from the concept of a closed set. A perfect set is a set that is equal to its set of accumulation points. – Michael Hardy Apr 24 '15 at 14:41
  • @MichaelHardy Ah, in that case you're right. Thank you. – Jolien Apr 24 '15 at 14:45

2 Answers2

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Suppose a set $A$ of real numbers is both open and closed, and that it is neither $\varnothing$ nor $\mathbb R$. Since $A\ne\varnothing$, there exists some $a\in A$. Since $A\ne\mathbb R$, there exists some $b\not\in A$.

Case 1: $a<b$.

Let $c=\sup (A\cap [a,b])$. This exists since $A\cap[a,b]$ is non-empty (since it contains $a$) and bounded above (since $b$ is an upper bound). Either $c\in A$ or $c\not\in A$. If $c\in A$ then $c\ne b\not\in A$ so $a\le c<b$. Since $A$ is open, some open interval centered at $c\in A$ is a subset of $A$. But that open interval contains numbers between $c$ and $b$, contradicting the fact that $c$ is an upper bound of $A\cap [a,b]$. Thus we rule out the possibility that $c\in A$ and conclude $c\not\in A$. Since no number smaller than $c$ can be an upper bound of $A\cap[a,b]$, every interval $(c-\varepsilon,c)$ contains some member of $A$. That means $c$ is an accumulation point of $A$. Since $A$ is closed, that means $c\in A$, and now we have a contradiction.

Case 2: $b<a$.

In this case, let $c=\inf(A\cap [b,a])$ and then proceed as above (but with $(c,c+\varepsilon)$ where $(c-\varepsilon,c)$ appeared above).

In either case, the assumption that $A$ is both open and closed and equal to neither $\varnothing$ nor $\mathbb R$ leads to a contradiction.

Summary: The idea is to find a point $c$ at the boundary between $A$ and $\mathbb R\setminus A$ and show that since $A$ is both open and closed it must both include and exclude that point.

Michael Hardy
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  • In reference to your summary, I don' t see how showing that a point on the boundary of $A$ and $\mathbb{R} \setminus A$ is both included and excluded in $A \implies A = \mathbb{R}$. – Zduff Sep 15 '18 at 22:41
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Use that X is path connected implies that it's connected and that a space is connected iff the only sets that are both open and close in X are X itself and the empty set

Belgi
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