I'm self-studying general topology and getting incredibly confused with the notions of open and closed sets.

In the following, I'm considering the space X = $\Re$, the set A = {2}, the collection $\tau$ of subsets of $\Re$: $\tau$ = {∅, $\Re$, {2}}. Since $\tau$ satisfies the axioms of a topology, it is a topology. And since A belongs to $\tau$, it is an open set.

However I'm also aware of the proof that finite sets are closed in $\Re$, using the argument that the complement of the singleton set is open. Where did I go wrong with my argument?

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    $\begingroup$ Many, many different topologies can be defined on the set of real numbers. In this case, they're defining the topology explicitly, saying "the open sets are precisely the empty set, $\mathbb{R}$, and $\{2\}$". These are the only open sets, and the closed sets are the complements of the open sets. Namely, the closed sets are $\emptyset$, $\mathbb{R}$, and $\mathbb{R} \setminus \{2\}$. No other sets other than these listed can be called open or closed. $\endgroup$
    – Kaj Hansen
    Jan 25, 2018 at 21:00
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    $\begingroup$ If that is the topology then there are just three open sets. Is $A$ one of them? $\endgroup$ Jan 25, 2018 at 21:01
  • $\begingroup$ The title asks a Question that depends on which topology of $\mathbb R$ is involved. Change the topology and you change which sets are open. One definition of "a topology" is the collection of all the open sets. $\endgroup$
    – hardmath
    Jan 25, 2018 at 22:28

2 Answers 2


Note that the question in your title is incomplete. This is because we can't discuss "open sets of $\mathbb{R}$" without mentioning the topology --- A more pedantic (which may be useful at this stage of your learning) way of asking this is "Are finite sets in $(\mathbb{R},\{\emptyset,\mathbb{R},\{2\}\}$ open?". This is because to talk about open sets, we need a topological space, which is both the set ($\mathbb{R}$ here) and the topology on that set ($\{\emptyset, \mathbb{R},\{2\}\}$ here).

When the topology chosen is "standard", this second part can be omitted --- A question such as "What are the compact sets in $\mathbb{R}$?" would implicitly assume that the "standard topology" on $\mathbb{R}$ is being used, meaning the metric topology given by the euclidean norm. If you aren't considering this topology, it's imperative to mention it. I know you do in the body of the question, but as it appears you're new to this field, I just wanted to reinforce that both parts are equally important.

Now, for your actual question:

Open sets are by definition any element of $\tau$. Consider the finite set $\{1\}$ --- is this in $\tau$? What about $\{2\}$? If every finite set is in $\tau$, then every finite set is open. In this way, $\tau$ tells you precisely what sets are open tautologically. Especially in a case like this (where $\tau$ is written down as an enumeration of its elements) it should be rather easy to check. This will change later when you instead talk about topologies generated by some set of elements (So a certain subset of the topology is written down, and then you essentially say "add in everything else to make the topology satisfy the closure properties required to be a topology"). The relevant terms here are basis and sub-basis, but I expect you'll learn these in the next few weeks.

  • $\begingroup$ Thank you for your comment. I have another question if you don't mind. If X is space and A is a non-empty proper open subset that is both open and closed, then is the set B = X - A both closed and open simply because the definition of a closed set is that its complement is open? I understand it if I just follow the definition, but intuitively it doesn't make much sense, since it doesn't mention the openness/closeness of X and I don't see how the closed sets fit into that definition of a topology. $\endgroup$
    – ensbana
    Jan 25, 2018 at 21:39
  • $\begingroup$ @user3173159 I'd recommend you post this as a new question (as this is "better StackExchange etiquette"). I'll check your post history in a bit and answer it there. $\endgroup$
    – Mark
    Jan 25, 2018 at 21:43
  • $\begingroup$ I've just posted a new question. Thank you. $\endgroup$
    – ensbana
    Jan 25, 2018 at 21:56

A set is one of the simplest structures we can define in math. One may intuitively think of a set as a collection of objects (there are formal definitions that are used to define this more rigorously). With only this structure, we can do some things, but not much. If we have two sets, we may consider maps between them. We are often interested in structure preserving maps. At the level of sets, the only structure we really have is the cardinality of the set (i.e. the number of elements in it). So, the structure preserving map at the level of sets is one that preserves cardinality, or a bijection.

We can, however, add more structure to our sets. Suppose we have a set $X$. One of the next most basic structures we can add on top of $X$ is a topology. A topology is a collection $\mathcal{T}$ of subsets of $X$ that we are treating as "special" in some sense. Really, they are special just because we say that they are. We further require this collection of subsets to satisfy a few properties.

  • $\emptyset$ and $X$ are in $\mathcal{T}$
  • If $U, V$ are in $\mathcal{T}$, then so is $U \cup V$.
  • If a finite collection of sets $U_i$ are in $\mathcal{T}$, then so is $\bigcap_i U_i$

That's all that we need to define open sets. So a subset of $\mathbb{R}$ may be open if it is in some $\mathcal{T}$ , as long as the chosen $\mathcal{T}$ satisfies the above three properties. We are still interested in structure preserving maps, but now we have more structure. We may take our old sets and give them each a topology. Then, we may take our old bijection and add a further requirement that if we give the map one of our 'special' (open) sets in the domain, it gives us one of our special sets in the image. These structure preserving functions are called continuous functions, and we say that continuous functions map open sets to open sets.

You may be familiar with the notion of open sets from analysis. These sets are really only open in the sense that they are in a certain topology (one of many we could define for $\mathbb{R}$). This topology that gives us the familiar notion of open sets from real analysis is called the standard topology. Call an open interval a set of the form $$ (a, b) = \{ x \in \mathbb{R} \mid a < x < b \} $$

Then the standard topology $\mathcal{T}_{st}$ is defined as the collection of open intervals.


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