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The motivation for this question comes from a proof I just saw demonstrating that a circle in the plane and $[a,b]$ are not homeomorphic. The argument consists of removing a point from the circle, and also a point from $[a,b]$ which i not an endpoint. Upon removal of this point from both we observe that the circle remains connected while the interval is disconnected.

From my understanding, $\mathbb{R}$ is connected and therefore has no proper clopen subsets. But if we can take $[a,b] \in \mathbb{R}$ and disconnect it by removing a point which is neither $a$ or $b$, then this subset of $\mathbb{R}$ has a proper clopen subset, which means that $\mathbb{R}$ has a proper clopen subset. I know that this reasoning is somehow wrong, but I'm not sure where my misunderstanding lies and I would appreciate any help.

Also, what would be the proper clopen subset of $[a,b]$ minus a single point that makes it disconnected?

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    $\begingroup$ the phrase "which means that ℝ has a proper clopen subset ..." is the problem. $\endgroup$
    – zhw.
    Commented Oct 4, 2015 at 2:46
  • $\begingroup$ So you're saying that proper clopen subsets of a subset are not necessarily a proper clopen subset of the whole metric space? $\endgroup$
    – AnotherPerson
    Commented Oct 4, 2015 at 2:48
  • $\begingroup$ @SirJective: Quite right. $(-\infty, 0)$ is closed in $(-\infty, 0) \cup (0,\infty)$, but obviously not in the whole space. $\endgroup$
    – user98602
    Commented Oct 4, 2015 at 2:48
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    $\begingroup$ Also: cute username. $\endgroup$
    – user98602
    Commented Oct 4, 2015 at 2:48

2 Answers 2

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Note that a set $U \subset [a, b]$ is open in $[a, b]$ if $U = [a, b] \cap V$ for some open set $V$ in $\mathbb R$. The term "open in $[a, b]$" is not the same as "open in $\mathbb R$". Indeed, for each $c\in (a, b)$, $[a, c)$ is open in $[a, b]$ but not open in $\mathbb R$. Thus a clopen set in $[a, b]$ might not be a clopen set in $\mathbb R$

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  • $\begingroup$ So if we are considering $[a,b]$ the metric space, then I can see that it is closed since removing the point from the interval removes that point from the whole space. But for the openness of the interval, I'm having trouble understanding that I can make a neighborhood of radius $r>0$ at the endpoints so that it would be simultaneously open. $\endgroup$
    – AnotherPerson
    Commented Oct 4, 2015 at 2:54
  • $\begingroup$ The term "open" and "closed" are relative terms. For example, $[a, b]$ is open $[a, b]$, but not in $\mathbb R$: @SirJective $\endgroup$
    – user99914
    Commented Oct 4, 2015 at 2:58
  • $\begingroup$ So I was given the definition of openness as follows: for any $p \in [a,b]$ there exists an $r>0$ such that $d(p,q)<r \implies q \in [a,b]$ Since the whole metric space consists of just the interval, are we just able to say that since nothing exists outside of it we can create neighborhoods at the endpoints without contradicting the definition? $\endgroup$
    – AnotherPerson
    Commented Oct 4, 2015 at 3:03
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    $\begingroup$ @SirJective Yes. Perfect. You are considering only the set $[a, b]$. You don't care about anything not in $[a, b]$. $\endgroup$
    – user99914
    Commented Oct 4, 2015 at 3:06
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The space $[a,c)\cup(c,b]$ does have two proper clopen subsets. But the fact that $[a,c)$ is closed in $[a,c)\cup(c,b]$ does not mean $[a,c)$ is closed in $\mathbb R$. Open subsets of $[a,c)\cup(c,b]$ are interections of $[a,c)\cup(c,b]$ with open subsets of $\mathbb R$. But they are not open subsets of $\mathbb R$.

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  • $\begingroup$ $\Bbb R^{\color{Red}2}$?? $\endgroup$
    – Akiva Weinberger
    Commented Oct 4, 2015 at 4:19

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