5

I think of an "open set" as being "roomy" or "spacious," in the sense that around every point, there is a little bit of room. This motivates the following definition.

Definition. An "open system" consists of an underlying set $X$ together with a collection of subsets of $X$ that are considered "open," such that for all $A \subseteq X$ it holds that if

  • for all $a \in A$ there exists open $B \subseteq A$ such that $a \in B$

then

  • $A$ is open.

Question. Is an "open system" just a topological space?

goblin GONE
  • 69,385
  • I assume that you mean $B\subsetneqq A$? Otherwise $A$ itself is such $B$, so every subset of $X$ is open. – Asaf Karagila Apr 02 '13 at 12:45
  • 2
    No, take $\mathbb R$ and consider a set to be open if it contains $0$ or $1$. The system of "open" sets then satisfies the condition you stated. – Stefan Hamcke Apr 02 '13 at 12:51

2 Answers2

6

In a topological space you require the open sets to be closed under unions (which trivially holds in your case), but also under finite intersection.

Your "open system", on the other hand, either can be any collection of subsets closed under unions ($A\subseteq A$ is open if $A$ is open, and $a\in A$ satisfies $a\in A$ as well); or if you require $B\subsetneqq A$, you disallow any isolated points because $\{x\}$ cannot be open, and by induction no finite set can be open (except the empty set, perhaps). In either case it doesn't seem to follow that "open systems" are closed under intersections.

Asaf Karagila
  • 405,794
  • I don't see why is it true that isolated points are not considered open in "open systems". could you point this out please ? @Asaf Karagila – FNH Jul 22 '14 at 11:46
  • 1
    "or if you require $B$ to be a proper subset of $A$, you disallow any isolated points". (I've parsed $\subsetneqq$ to English to make it clearer.) – Asaf Karagila Jul 22 '14 at 11:50
  • Oh yes , I got it now. I think if we considered $B$'s to be proper subset of $A$ then an annoying thing appears . Since if $A$ is infinite open set then there is another proper set $B$ of $A$ that is itself open. and this $B$ itself has another proprt set inside itself that is open and so on. So we get infinite series of open sets inside each other. There is no "base" open set if the word is convenient. Isn't that somehow weired ? @Asaf Karagila – FNH Jul 22 '14 at 12:07
  • 1
    Not necessarily, consider $\Bbb R$ as a metric space with the usual metric. In each non-empty open set there is a smaller one, but we still get a basis for the topology. Even a countable basis for the topology. – Asaf Karagila Jul 22 '14 at 12:19
  • You are right, When I used that word "base" I think I used another concept in my mind , I assumed that those bases open set are minimum open sets that is , there are no smaller open sets than them.Now , it is not weired! . Thank you for your time. @Asaf Karagila – FNH Jul 22 '14 at 12:25
5

Let $X$ be a set. Then a collection $\mathscr{C}$ of subsets of $X$ is an "open system" if and only if $\mathscr{C}$ is closed under unions.

One direction is easy. Suppose then that $\mathscr{C}$ is closed under unions. Let $A$ be a subset of $X$, and assume that for each $a \in A$, there exists a $B_a \in \mathscr{C}$ such that $a \in B_a \subseteq A$. Then $A = \cup_{a \in A} B_a$, so $A \in \mathscr{C}$ since $\mathscr{C}$ is closed under arbitrary unions.

Now it should be easy for you to find "open systems" that are not topological spaces, ie. collections of subsets that are closed under unions but not under finite intersections.

spin
  • 12,267
  • What if $B_a$ has some elements which $A$ doesn't have ? in this case taking the union of those $B_a$'s will not give us $A$ but a larger set , right ? – FNH Jul 22 '14 at 11:39
  • 1
    Yes. But we choose $ B_a $ so that it is contained in $ A $. – spin Jul 22 '14 at 12:28