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I don't get why the boundary of a clopen set is empty. If you take $A = [0,1)$ in $\mathbb R$, then isn't the closure of this the smallest closed super set that contains $A$ which is $[0,1]$. Isn't the interior, $(0,1)$? So the boundary would be the closure minus the interior, aka, ${0}$ and ${1}$?

I'm confused. Thanks for all of your great and quick replies!! No wonder I was so confused, I was completely misunderstanding the definition of clopen. I've learned a lot from all of you!

Andy
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    The boundary of a set is its closure minus its interior. In the case of a clopen set, the two are equal. What you've written down is not a clopen set in $\mathbb{R}$. – Jared May 21 '13 at 00:25
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    Clopen means a set which is both closed and open, not "half closed, half open interval". The only non-empty clopen set in $\Bbb R$ with the standard topology is $\Bbb R$. – Asaf Karagila May 21 '13 at 00:25
  • I think my answer is the only one that addresses your actual concern about the boundary of $[0,1)$. Yes, you were confused about the meaning of clopen, but that does not resolve the real question. – Zev Chonoles May 21 '13 at 00:36
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    @Zev: Mathematically speaking, that's not correct. The question [appearing in the title] is vacuously answered by anyone who explains that $[0,1)$ is not clopen! ;-) – Asaf Karagila May 21 '13 at 00:47

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You are correct that the boundary of the set $A=[0,1)$, considered as a subset of $\mathbb{R}$, is the set $\{0,1\}$, by exactly the reasoning you described.

However, treating $A$ as a topological space in its own right (i.e., given the subspace topology, a.k.a. the relative topology), the boundary of $A$ is empty. The same is true of any topological space $X$: the boundary of $X$, as a subset of itself, is empty.

Why? Well, the closure of $X$ in $X$ is the smallest closed subset of $X$ that contains $X$. The only subset $X$ whatsoever that contains $X$ is $X$ itself; and $X$ is closed as a subset of $X$ (since it is the complement of the open set $\varnothing$), so the closure of $X$ as a subset of $X$ is just $X$ again. Similarly, because $X$ is already open as a subset of itself, the interior of $X$ is also $X$. Thus $$\partial X=\overline{X}\setminus\mathrm{int}(X)=X\setminus X=\varnothing.$$

In short, the issue is that "boundary" is a term that depends on two pieces of information: the set we are taking the boundary of, and the "ambient" space that we are working in. The same is true of "closure" and "interior"; there is no such thing as the closure or the interior of a set. It is always relative to some "ambient" set where everything takes place (which may be left implicit, if it is assumed that it will be clear from context).

Zev Chonoles
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  • Hmm interesting..as a side note, could you say that A is relatively closed and relatively open in A? Or is this unrelated? Thanks for you explanation. – Andy May 21 '13 at 00:38
  • That's exactly right. Given a topological space $X$, a subset $A\subseteq X$, and a subset $B\subseteq A$, we say that $B$ is relatively open in $A$ when it is an open subset of $A$ (considering $A$ as a topological space in its own right). Similarly with relatively closed. For any topological space $Y$ whatsoever, it is always true that $Y$ is open and closed as a subset of $Y$ (because, by the axioms for a topology, both $\varnothing$ and $Y$ are open). Thus, $A$ is always relatively open and relatively closed in $A$. – Zev Chonoles May 21 '13 at 00:41
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The half-open interval $[0,1)$ is not clopen. A set is clopen if it is both closed and open, and $[0,1)$ is neither closed nor open.

If a set is clopen, then it is open so it is equal to its own interior; and it is closed so it is equal to its closure. The difference between a set and itself is empty.

Asaf Karagila
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I think you're mistaking a clopen set, which is both open and closed, with an interval (set) that is half open, and half closed. The two are not the same. Indeed, in this case, the "half-open/half-closed" interval $[0, 1)$ is not clopen in $\mathbb R$: it is neither open nor closed.

Your reasoning about the boundary of $[0, 1)$ is spot-on: it is indeed the set of two elements: $\{0, 1\}$. Now that we know that $[0, 1)$ is not clopen, then, we have no contradiction. The boundary of a clopen set is indeed empty, as you know, but then, since $[0, 1)$ is not clopen, there's not conflict.

The link above (to the entry Wikipedia entry "clopen set") will help clarify what it means to be a clopen set, and you'll find some nice examples to help illustrate such sets.

amWhy
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$[0,1)$ is not clopen as a subset of $\mathbb{R}$.

A clopen set is by definition a set that is both open and closed. Hence, the closure as well as the interior would be equal to the set itself, leaving an empty boundary.

As you already remarked, the interior of $[0,1)$ is $(0,1)\neq[0,1)$, hence $[0,1)$ is not open. Furthermore, the closure of $[0,1)$ is $[0,1]\neq[0,1)$, hence $[0,1)$ is not closed either. Thus $[0,1)$ is certainly not clopen.

Abel
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