Definition 1

Let $X$ be a set of points. A collection of subsets $U = \left\{U_{\alpha }\right\}$ forms a topology on $X$ if

  1. Any arbitrary union of the $U_{\alpha }$ is another set in the collection $U$.

  2. The intersection of any finite number of sets $U_{\alpha }$ in the collection $U$ is another set in $U$.

  3. Both the empty set $\emptyset$ and the whole space $X$ must be in $U$.

The pair $(X, U)$ is called a topological space.

The sets $U_{\alpha }$ in the collection $U$ are called open sets.

My Question

  1. Is a finite isolated points set an open set? At present I think it is a closed set.

  2. Does Definition 1 suit for a finite isolated points set? It seems it just call $U_{\alpha }$ open sets without defining?

  • $\begingroup$ What do you mean by "isolated points?" What is a closed set, as you are using the term here? (In topology, being "closed" is not the same as "not open.") $\endgroup$ – Thomas Andrews Jun 6 '13 at 14:20
  • $\begingroup$ @Thomas Andrews A set C is closed if its complement X-C is open. Some sets are both open and closed. isolated points means such like {1,2,3} not an interval [0,1] or (0,1) $\endgroup$ – HyperGroups Jun 6 '13 at 14:23
  • 1
    $\begingroup$ But an interval is already in a space where you have a notion of "topology." The above definition is much more general - it applies to "spaces" that look nothing like the real line. For example. Work from the definition above, don't work from your intuition from the real line or any other space. For example, there are topologies where $X$ is a finite set, so there are necessarily finite open subsets. There are also topologies where not all finite sets are closed. $\endgroup$ – Thomas Andrews Jun 6 '13 at 14:27
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    $\begingroup$ If every singleton is an open set, you get discrete space. It is perhaps worth pointing out that knowing that a set is closed does not mean that it is not open; a subset of a topological space can be both closed and open at the same time. $\endgroup$ – Martin Sleziak Jun 6 '13 at 14:35
  • $\begingroup$ Just curious, does your book/teacher really use a plain $U$ for the topology, or does it use something like $\mathcal U$? Because a plain $U$ is bound to get a bit confusing. $\endgroup$ – Thomas Andrews Jun 6 '13 at 15:22

To answer your first question: Finite sets can be open, closed, both or neither depending on your topology. I'll give some examples of topological spaces where these possibilities occur in the form $(X,U)$. For $X$, I will always choose $\mathbb{R}$, but my $U$ will vary:

  • $(\mathbb{R},\{\varnothing,\{0\},\mathbb{R}\})$ - In this topological space $\{0\}$ is open but not closed.
  • $\mathbb{R}$ with its usual (metric) topology - In this space any finite set is closed, but not open.
  • $(\mathbb{R},2^\mathbb{R})$, the so called discrete topology on $\mathbb{R}$ - In this topology every set (finite or infinite) is open and closed.
  • $(\mathbb{R},\{\varnothing,\mathbb{R}\})$, the so called indiscrete topology on $\mathbb{R}$ - In this topology the only open sets as well as the only closed sets are $\varnothing$ and $\mathbb{R}$, hence all finite non-empty sets are neither open nor closed.

It is a good exercise to check that all the above are actually topological spaces by your definition.

To answer your second question: I think you are confused about what is being defined. Definition 1 is a definition of topological spaces, not of open sets.

  • $\begingroup$ Not sure that OP will know what you mean by $(\mathbb{R},\{\varnothing,\{0\},\mathbb{R}\})$. Could be confusing. $\endgroup$ – Thomas Andrews Jun 6 '13 at 14:42
  • $\begingroup$ @ThomasAndrews It definitely looks a bit odd, but I think it shouldn't be too hard for him/her to walk through the axioms and see what happens. $\endgroup$ – Abel Jun 6 '13 at 14:44
  • $\begingroup$ Yes, I mean, I don't think he'll get that you are writing $X=\mathbb R$ and $\mathcal U=\{\varnothing,\{0\},\mathbb{R}\}$. He did not state that a topological space was an ordered pair, so it is not clear that this is what you mean. $\endgroup$ – Thomas Andrews Jun 6 '13 at 14:46
  • $\begingroup$ But he did state that a topological space is a pair $(X,U)$ explicitly and literally. I'll elaborate a little bit anyway. Couldn't hurt, I guess. $\endgroup$ – Abel Jun 6 '13 at 14:48
  • $\begingroup$ @MJD You are partially right, in my opinion. I don't think any set (other than $\varnothing$ perhaps) is be defined to be open. What you can do is define a topology and call its elements its open sets. This is just nomenclature though, not definition. $\endgroup$ – Abel Jun 6 '13 at 15:01

You hit upon the right answer in your second question - for general topologies, a set is open if it belongs to the collection $U$, so indeed open is just a name for sets in your collection.

This means that any set $S\subset X$ can be open in some topology, providing you can find some collection $U$ containing $S$ that satisfies the three axioms you listed. As taking $U$ to be all subsets of $X$ works, we can conclude that any subset of $S$ is open in some topology on $X$.

The overall conclusion is that you cannot conclude that a particular set is open or not without considering the chosen topology. In particular, a finite set of points could be open, closed, neither or both depending on the chosen topology.

It's also worth noting that a point can only be called isolated once you have a notion of topology. I could put the topology on the set $S=\{x,y\}$ such that only $\varnothing$ and $S$ are open, and then neither point is isolated.

  • $\begingroup$ Hi, at the end of the answer, I'm not sure about what is not isolated. {x},{y} or {Phi} S={x,y}, neither is isolated? $\endgroup$ – HyperGroups Jun 8 '13 at 0:42
  • $\begingroup$ A point $x$ in a topological space $X$ is said to be isolated in $X$ if the set $\{x\}$ is open, i.e. one of the $U_\alpha$s is just $\{x\}$. In the example I give, no one point set is open, so there are no isolated points. I could define a new topology on $S$ in which $\{x\}$, $\{y\}$ or both were also open, and then there would be isolated points. $\endgroup$ – mdp Jun 9 '13 at 0:08

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