Homework Help: Difference between closed set and open set is a closed set

1. Aug 18, 2016

JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'd like to check with you guys if I tackled that problem correctly. I might have a few theoretical questions along the way :)

Prove that the difference $A \setminus B$ of a closed set $A \subset \mathbb{R}^2$ and an open set $B \subset \mathbb{R}^2$ is a closed set.

2. Relevant equations

First I'd like to define open/closed sets in $\mathbb{R}^2$:
- a set $M_1 \subset \mathbb{R}^2$ is called open, if none of its boundary points is included in the set;
- a set $M_2 \subset \mathbb{R}^2$ is called closed, if it contains all of its boundary points.

I will use also the following theorems:
1. If $X$ is a topological space and $U$ is a subset of $X$, then the set $U$ is called closed when its complement $X \setminus U$ is an open set.
2. The intersection of two closed sets is a closed set.

3. The attempt at a solution

My first question concerns the topological space. I have read numerous articles (mostly Wikipedia I admit, but not only) about what a topological space is, and I still don't get what it really is. Neither did I understand any of the few examples I saw... I used the theorem nevertheless, assuming $\mathbb{R}^2$ is a topological space, and if not I theorised it may be a subset of $\mathbb{R}^2$ containing both $A$ and $B$.

Applying the 1st theorem, I can develop as follows:
Let $X = \mathbb{R}^2$ be a topological space (...or not).
$B$ is an open set $\implies$ its complement $B^c = X \setminus B$ is closed
$\implies\ A \setminus B = A \cap B^c$ is closed, because of theorem 2 ($A$ and $X \setminus B$ are closed sets).

Is that a valid proof? Do the general definitions of open/closed sets hold in $\mathbb{R}^n$?

Julien.

2. Aug 18, 2016

Orodruin

Staff Emeritus
This is not a theorem, it is the definition of a closed set.

2 is a theorem which follows trivially from the definition of a topological space (in particular what the open sets need to satisfy).

It is difficult to tell you without just repeating the definition. What particular part of the definition do you have trouble with?

Note that R^2 by itself does not uniquely define a topological space - there are several different possible topologies on R^2. However, the standard topology is often assumed where sets are open if, for each point in the set, there is a small ball around the point that is entirely in the set.

Yes, but you really should understand the basic definitions of a topological space before going through it.

As said, there are many possible topologies on R^n. This theorem would hold regardless of the topology you use.

3. Aug 18, 2016

Krylov

@JulienB, I hope you do not find it presumptuous of me, but I think it would help if you would get a good book, say on the foundations of analysis in $\mathbb{R}^n$. (Various possibilities are discussed in the textbook forum.) From what I read from you here and there, it seems you know a little something about a lot of different topics that fall within this subject, but you miss a coherent view. Reading bits and pieces here and there is probably not going to remedy that.

Studying such a book calmly and carefully, you would be introduced to the usual Euclidean topology on $\mathbb{R}^n$ before meeting the general definition of a topological space. You would also learn about the behaviour of functions on $\mathbb{R}^n$, including their continuity and differentiability properties and how these depend on the underlying topology. This way you will understand how things are built on top of each other and how everything comes together in the end.

Of course, just to be clear, others and I are willing to help.

4. Aug 18, 2016

JulienB

Hi @Orodruin and first, thanks a lot for your answer. I understand the proof I wrote except, indeed, for the concept of topological space. If you don't mind, I will quote the definition of a topological space (from Wolfram.com) and describe what I don't understand. I hope I don't abuse your time, I appreciate your help very much,

A topological space is a set $X$ together with a collection of open subsets $T$ that satisfies the four conditions:
1. The empty set $\emptyset$ is in $T$.
2. $X$ is in $T$.
3. The intersection of a finite number of sets in $T$ is also in $T$.
4. The union of an arbitrary number of sets in $T$ is also in $T$.

Okay, uh...I guess my first question would be: is a topological space some sort of duplet? Like: in order to define a topological space, do we need to define a set $X$ and a collection of sets $T$? Or does the set so to say "self-defines" what the collection of sets $T$ is?

For example, in my proof I have the set $\mathbb{R}^2$. In order to make it a topological space (according to what I just wrote, if that is correct), I would have to define a collection of sets $T$ that contains $\mathbb{R}^2$ and possibly other subsets of $\mathbb{R}^2$ (I guess that would not be necessary for my proof, or?). Then if I run again through the definition:
1. $\emptyset \subseteq \mathbb{R}^2 \implies$ condition fulfilled.
2. $\mathbb{R}^2$ is in $T \implies$ condition fulfilled.
3. $\mathbb{R}^2 \cap$ any other potential set in $T \in \mathbb{R}^2 \in T \implies$ condition fulfilled.
4. $\mathbb{R}^2 \cup$ any other potential set in $T \in \mathbb{R}^2 \in T \implies$ condition fulfilled.

Sorry the two last conditions were a bit confused. I hope you get the point, and that you can correct my misunderstandings.

Thank you very much, I appreciate it.

Julien.

5. Aug 18, 2016

JulienB

Hi @Krylov and thanks for your answer. I don't find it presumptuous at all, and I appreciate your comment.

It's true that I mostly base my knowledge on the alas not so detailed script of my teacher. I have a German book as well that I find pretty good, but I unfortunately stand under the pressure of upcoming exams like many students, though I'd love to spend more time on each topic. :( Posting on this forum has already helped me a great deal in the past, especially in clearing up concepts that I usually read in German (I'm French so that's a challenge!).

I will in any case think about it and see if I can invest more time in reading math theory. Do you have a specific book in mind? Thank you for your suggestion.

Julien.

6. Aug 18, 2016

Krylov

If you already like the book in German that you have and you learn from it, then there is no reason to change a winning strategy. The exam pressure factor I understand completely. I will think a little bit more about it (you can also look at the textbook forum) and perhaps get back to you if something useful comes up in my head. This I then do by sending you a message, because otherwise the present thread will become derailed.

7. Aug 18, 2016

Orodruin

Staff Emeritus
You need a base set X, which is the set you want to define a topology on, and a collection T of subsets of X. The sets in T must satisfy the mentioned conditions and we then define these to be open sets. The collection T just defines what sort of sets you call open and generally there are many possibilities for any given set.
No, the intersection should not be a subset of R^2 - it should be an open set (ie, in T). This is a stronger requirement. Similar for the union. The union of any open sets must be in T, ie, be an open set.

8. Aug 19, 2016

JulienB

@Orodruin Thanks for your answer. I do one more recap, it's slowly becoming clearer with your explanations. So in the case of my proof, the base set on which I want to define a topology is $\mathbb{R}^2$ (of which $A$ and $B$ must be subsets), and $T$ would then be any collection of open sets in $\mathbb{R}^2$ that satisfies the four conditions. What I found confusing (and still do I guess) is that the topological space is written as $X$ and not as, say, $(X,T)$ in the definition of a closed set above. Maybe it is because of the many possibilities of $T$ for a given set $X$? Like "we don't mention it, because any collection of sets $T$ satisfying the four conditions (and there are a lot) would make the proof valid"?

Julien.

9. Aug 19, 2016

Orodruin

Staff Emeritus
This is really an abuse of notation, without specifying what open sets are, X is not a topological space. Often, as in the case of $\mathbb R^2$, a standard topology is often assumed.

Yes, the proof is valid for any topological space. It does not matter what the space is (it does not need to be $\mathbb R^2$, it can be the set {1,2,3,20}, a functional space, or the empty set, or any other set as long as you define a topology).

10. Aug 19, 2016

JulienB

@Orodruin Nice, thanks! I think that pretty much cleared my question, thx a lot!

Julien.