# in topology a set is open IFF?

So far, my understanding of an open set is limited. I understand that a set is closed if it's complement is open. That's clear and concise, but I can't find anything so clear to define open.

I found one definition here that a set is closed iff it includes all the limit points.

So I'm confused as to which of open/closed is a stand-alone definition, and which is the dependent definition. I've found a collection of answers that imply both. However, that's a paradox.

Can open be shown on a set of numbers, like { 1, 2, ... 10 }. I understand that {empty} and the whole set are closed sets, but what else would be a limit?

Adding another question: Assuming that we choose which sets are open, and they must represent a valid topology O o

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You might find some information in the answer I gave here: math.stackexchange.com/questions/15007/… – Asaf Karagila May 31 '11 at 20:47

In abstract topology, if that is what you are talking about, open sets are usually not defined but declared to be open.

That is, to define a topology you choose a collection of sets having certain properties, and you decide that these will be the sets that you will call "open". You could actually do the same thing with closed sets and end up with an equivalent definition of a topology.

If you are talking about metric topology, then the open sets are the sets which are unions of open balls.

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thank lor there's someone who makes sense. Ok, so union of open balls includes open balls themselves, as an open ball is the union of itself and itself. – Lee Louviere May 31 '11 at 20:48
It that much is true, then why are most people stating that the set X over which a topology is chosen T, means that X and {empty} are closed. How can this be decided if I get to choose what is an open set? Is it because X and {empty} are both automatically open, and therefore closed, and therefore clopen? – Lee Louviere May 31 '11 at 20:50
I didn't explicitly state the conditions that collection of open sets must satisfy in order to be called a topology, but one of them is that the empty set and $X$ have to be in it (so yes, they are both automatically open). Since they are each others' complement, so both complements of an open set, they are also both closed. – Vhailor May 31 '11 at 20:56
@Xaade: In order to be a topology, you can't just pick any sets to be open, you have to follow a few rules. First of all, any union of open sets must be open. Secondly, any intersection of a finite number of open sets must be open. And finally, the empty set and $X$ itself must be open. That last one implies that $X$ and the empty set are also closed, since they are complements of each other. – MartianInvader May 31 '11 at 21:00

It is true that a set is closed if and only if it contains all of its limit points. However, you can't define what it means to be a limit point (or what it means for a sequence to converge) unless you have a notion of an open set.

A topological space is a set and the additional information of which subsets are considered open sets, subject to certain axioms. In general, any set has many different topologies. There are a few general ones worth mentioning.

Let $X$ be a set, possibly with additional structure (which we shall specify below). The following are all natural topologies on $X$

The trivial topology The only open sets are $X$ and $\emptyset$.

The discrete topology Every subset of $X$ is open, and hence closed.

The finite complement topology The closed sets in $X$ (other than $X$ itself) are the finite subsets of $X$.

The order topology If $X$ is a linearly ordered set, we can declare all open intervals of the form $(a,b)=\{x\in X \mid a< x < b\}$ to be open. Then, all unions of these are also open. Edit: we also want to include intervals of the form $(a,\infty)=\{x\in X \mid a< x \}$ and $(\infty,b)= \{x\in X \mid x < b\}$. Otherwise, the endpoints would give us the wrong topology, e.g., on $[0,1]$.

The metric topology If $X$ has a metric, we declare all open balls $B_r(p)=\{x\in X \mid d(x,p)<r\}$ to be open. Just as with the order topology, we must take all unions of these balls to get a topology.

The first three examples are somewhat strange if you haven't seen them before, and the first two tend to be more for the sake of counterexamples than anything else.

The last two will give you all the examples of topological spaces you saw before you saw the definition of a topology.

For example, $\mathbb{R}$ is both a linearly ordered set and a metric space. Therefore, there are at least two natural topologies we could think to put on $\mathbb{R}$. In fact, they coincide.

More generally, $\mathbb{R}^n$ can be made into a metric space in many different ways. If we define the $\ell_p$ norm by $||(x_1, \ldots x_n)||_p=\left(\sum |x_i|^p \right)^{1/p}$, this norm generates a metric. Each of these metrics generates a topology. However, all these topologies are the same. In fact, all metrics on $\mathbb{R}^n$ coming from linear norms generates the same topology. This does not happen for infinite dimensional vector spaces.

In summary, there are lots of good choices for a topology on a space, and you must fix one before you can talk about things like closed sets, convergence, or most other things considered to be topological.

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thanks, this is very clear water in the mud out there. What would be considered an open ball in 1D space, a linear set in $\mathbb{R}$ ? (x, y) – Lee Louviere May 31 '11 at 21:17
@Xande Yes, given an interval $(x,y)$, we can also write it as the open ball of radius $(y-x)/2$ centered at $(x+y)/2$. Similearly, every open ball in $\mathbb{R}$ is an open interval. Of course, $\mathbb{R}$ is special, as most linearly ordered spaces don't have a notion of distance, and most spaces that have a notion of distance aren't linearly ordered. That's the difficulty when you're used to dealing with nice spaces. There is so much good going on that there are many different ways to generalize. – Aaron May 31 '11 at 21:43
Little correction: the order topology should also include all open segments (it's the minimal topology that contains all segments) like $(-\infty,a)$ and $(b,+\infty)$, otherwise the minimal and maximal elements (if they exist) do not have non-trivial neighbourhoods. – Henno Brandsma Jun 1 '11 at 12:56