B What is a clopen set?

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1. Jun 9, 2016

ProfuselyQuarky

I feel like I ask too many questions here, so I'm sorry about that. But, anyway, what is a clopen set? I was watching something the other day that a friend sent me, and, out of the blue, the guy starts referring to a set as being "clopen" with no explanation. I tried to break all the definitions I know down into little bits, but I still find it confusing. Apparently a clopen set is both open and closed. How?? Letting $A\subseteq \mathbb{R}$, I know that a set A is open if every point of A is an interior point of A . A is closed, however, if and only if $\mathbb{R}\setminus A$.

From Wiki:
I know that a complement of a set is all the things outside of the set, but I just don't understand. It's not the (singular) set itself that's both open and closed, right? All I get out from reading is that you can have a closed set with an open complement and an open set with an open complement. I can't see how that makes a set clopen. It is still either open or closed.

When it says "which leaves the possibility of an open set....." are they talking about a completely different set or something related to the initial closed set...?

Obviously I am wrong and lost, so clarification would be greatly appreciated. I am going in circles.

2. Jun 9, 2016

Staff: Mentor

From the wiki article they have this example:

3. Jun 9, 2016

ProfuselyQuarky

Well, I already read that. It doesn't really help though Another example that was given was:
I feel like this is something I should understand easily, but I'm not. I have to leave now, but will come back shortly, hopefully with different reasoning.

4. Jun 9, 2016

The Bill

Recall that in the definition of a topological space (X,O), where O is the set of all sets defined to be open, both X and $\emptyset$ are defined to be open. That is, both X and $\emptyset$ are placed in O "by hand" in the very definition of a topological space.

Now, what is the complement of X? What is the complement of $\emptyset$? Are these complements open?

5. Jun 10, 2016

ProfuselyQuarky

The complements are closed, right?

6. Jun 10, 2016

The Bill

In this specific case, X is closed because its complement $\emptyset$ is defined to be open, and $\emptyset$ is closed because its complement X is defined to be open. Therefore they're each both closed and open.

7. Jun 10, 2016

ProfuselyQuarky

Ah! I think I understand now. This statement is ever so concise, thank you for the simple explanation.

As a side note, regarding the definition saying that "A set is open if its complement is closed", this obviously makes sense. However, I have seen a few (very few) places that say that "A set is open iff its complement is closed". This is an error, correct? I haven't read anything implying that the statement is supposed to be biconditional and the places that have mentioned it only noted it briefly without elaborating. If the biconditional version is correct, could you direct me to a proof for it?

8. Jun 10, 2016

PeroK

In a definition "if" is often used instead of more precisely "iff". In this case, a set is closed iff its complement is open; which means a set is open iff its complement is closed.

9. Jun 10, 2016

micromass

It is a mistake to think that open and closed are two extremes. I understand that an open door and a closed door are as far apart as can be. But this situation is not true in mathematics anymore. I think the terminology open/closed is a very bad one, but it has unfortunately stuck.

There are many types of closedness in mathematics. But they all share the same characteristic. A set is closed if whatever you "do", you cannot go out of the set. Much like a prison. In topology, this translates to: if you are in the set and if you take infinitely many steps in the set, your position at infinity will still be in the set. More formally, if you have a sequence $(x_n)_n$ that converges to $x$, then if $x_n$ is in the set for each $n$, then so is $x$. (purists will no doubt tell me that I have defined sequentially closed and that closedness is something else. They're right, but for $\mathbb{R}^n$ they coincide).

As for openness, it is very different. A set is open if wherever you are in the set, you can find some comfortable area all around you that is still in the set. In $\mathbb{R}$ this corresponds to saying that around every point $x$, there is some interval $(a,b)$ that is a subset of the set and such that $x\in (a,b)$. You can rephrase this with sequences: if you take infinitely many steps and if you end up in the set, then you must have been in the set after finitely many steps. More formally: if you have a sequence $(x_n)_n$ that converges to $x$ and such that $x$ is in the set, then there is some $N$ such that $x_n$ is in the set for all $n\geq N$.

As you can see, closed and open are notions that are not at all extremes of eachother, or inverses of eachother. You can easily see $\mathbb{R}$ is open using these two explanations.

10. Jun 10, 2016

The Bill

I think using the words "all around you" is misleading here. I know you're trying to say that for every point in an open set one can find an open neighborhood set which contains the point and is in turn contained in the open set. However, since we're talking about topological spaces, an open neighborhood containing a point won't necessarily be intuitively "all around" that point.

Take the point 1 and the set [1,2) as a neighborhood of it in the Sorgenfrey line topology, for example. The set [1,2) isn't intuitively "all around" 1, but it is a neighborhood of 1 because the right half open intervals are defined as the basis of the Sorgenfrey line's topology, and therefore open.

Other than that quibble, I like your description of the false dichotomy of open and closed.

11. Jun 10, 2016

micromass

You're aware the OP is in high school, right? Talking about exotic topologies won't really do her much good.

12. Jun 10, 2016

The Bill

In post #3, ProfuselyQuarky mentioned topological spaces. That led me to believe we're talking about more than just the standard open ball topology on Euclidean space.

Also, more people than ProfuselyQuarky might read this thread in the future. I want to do what I can to preempt what confusion I can predict. I found this site by doing web searches related to questions I had.

13. Jun 11, 2016

TeethWhitener

This might have already been cleared up for you, but an example never hurts (right?). It turns out that in $\mathbb{R}$, the only clopen sets are $\emptyset$ and $\mathbb{R}$ itself. (This actually leads to another topological property known as connectedness.) However, as an example we can consider the set $(0,1)\cup(2,3)$ with the standard topology as a topological space. In this space, the complement of the open set $(0,1)$ is $(2,3)$, which is itself also an open set. But since it's the complement of an open set, that means it's closed, so it's both closed and open: clopen.

14. Jun 12, 2016

ProfuselyQuarky

Thank yoouuuuu. This clears up a lot. I can really see how a set can be both open and closed now Regarding openness, "uncomfortable" area be where $x\notin(a,b)$, right?
Thank you, too :) Looking at an example was helpful. I just looked to see what connectedness was and I've some questions about that, as well.

By the way (as a side question), the terms open and closed are really unfortunate (as micromass pointed out) and confusing. If you all could rename them, what would you choose? Obviously it has to stay "open" and "closed", but I like to come up with my own names/descriptions to scratch on the margins of my own notes just to help sometimes with tricky definitions and terminology.

15. Jun 12, 2016

ProfuselyQuarky

And thanks for clearing this up, too, PeroK!

16. Jun 12, 2016

micromass

Depends. For example $(0,+\infty)$ is open. A comfortable area for $1$ would be $(0,2)$. Everything to the left of $0$ woldn't be in the set and wouldn't be comfortable.

17. Jun 12, 2016

ProfuselyQuarky

What about everything to the right of $2$? Same thing?

I like the name "fat set" and since I won't be delving deeper into exotic spaces any time soon, thinking of open sets as fat would be nice.

18. Jun 12, 2016

micromass

No, because they are in the set. So the set $(0,6)$ is "comfortable".

19. Jun 12, 2016

ProfuselyQuarky

It's just that you said that a comfortable area for $1$ would be the interval $(0,2)$ since $1\in(0,2)$. How is everything to the right of $2$ within the set?

20. Jun 12, 2016

micromass

We're talking about the set $(0,\infty)$ being open. This means that for every element in the set there must be a comfortable area (= an open interval). $(0,6)$ is an open interval inside the set $(0,\infty)$ so it's a comfortable area.

21. Jun 12, 2016

ProfuselyQuarky

Oooh, that's what you meant. Thanks, micromass.