Given the following definition of topology, I am confused about the concept of "open sets".

2.2 Topological Space. We use some of the properties of open sets in the case of metric spaces in order to define what is meant in general by a class of open sets and by a topology.

Definition 2.2. Let $X$ be a nonempty set. A topology $\mathcal{T}$ for $X$ is a collection of subsets of $X$ such that $\emptyset,X\in\mathcal{T}$, and $\mathcal{T}$ is closed under arbitrary unions and finite intersections.
    We say $(X,\mathcal{T})$ is a topological space. Members of $\mathcal{T}$ are called open sets.
    If $x\in X$ then a neighbourhood of $x$ is an open set containing $x$.

It seems to me that the definition of an open subset is that subset $A$ of a metric space $X$ is called open if for every point $x \in A$ there exists $r>0$ such that $B_r(x)\subseteq A$. What is the difference of being open in a metric space and being open in a topological space?

Thanks so much.

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    $\begingroup$ Well, by definition an open set is a member of the topology. That's all. When you have a metric space, you can use the metric to define some "distinguished" sets (the definition you remarked) and then check that they indeed form a topology. Hence they deserve the name of open sets. $\endgroup$ – User3773 Feb 20 '17 at 10:46
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    $\begingroup$ Often a course in topology starts with metric spaces. Then a definition of "open set" arises in that context. It can be shown that the collection of these sets has the properties mentioned in def 2.2. and for such a collection a name allready exists: a topology. That opens the door to reach out for a more general definition of open set wich is not linked only with metric spaces, but with topology. Nothing goes wrong here. A metric space induces a topology and the sets that belong to it are exactly the sets that were defined to be open at first hand. $\endgroup$ – drhab Feb 20 '17 at 10:49
  • $\begingroup$ @ButterMath May I ask you what's your textbook? $\endgroup$ – Eric Feb 20 '17 at 13:16
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    $\begingroup$ Please give a citation for the quoted text. $\endgroup$ – David Richerby Feb 20 '17 at 19:51
  • $\begingroup$ When I study general topology, the definitions are given axiomatically as followed: A topological space is defined to be a nonempty set $X$ equipped with a topology. A topology is a subset of $P(X)$ closed under finite intersection and arbitrary union. A set is said to be open if it is in the topology. $\endgroup$ – Alex Vong Feb 21 '17 at 14:50

In an abstract topological space, "open set" has no definition!

You simply decide (as part of making your topological space) which sets you want to call open -- those are the sets you put into $\mathcal T$. Whatever you decide to call open will be called open, as long as your decision meets the condition "$\emptyset, X\in\mathcal T$ and $\mathcal T$ is closed under arbitrary unions and finite intersections".

A metric space becomes a topological space by deciding that the "open sets" in this particular topological space are going to be exactly the ones that are open according to the metric-space definition.

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    $\begingroup$ This is not correct, an abstract topological space has (by definition) an assigned topology, and thus specified open sets. $\endgroup$ – diracdeltafunk Feb 21 '17 at 5:11
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    $\begingroup$ I found the sentence "In an abstract topological space, "open set" has no definition!" confusing. As said by @diracdeltafunk, and the rest of your answer, "open set" has a definition in topological spaces. $\endgroup$ – Maxim Feb 21 '17 at 10:28
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    $\begingroup$ @Maxim I think what Henning meant was that there is no "intrinsic" or "natural" way to decide which sets are open or not (unlike in, say, $R^n$ where the OP has a clear grasp of what's "naturally" open). Though, of course, there are other topologies on $R^n$ beyond the "natural" one. This is all basic/obvious stuff, of course, but I'm writing in the hope that it might somehow elucidate things for OP or other learners. $\endgroup$ – Felix Goldberg Feb 21 '17 at 10:42

In the topological space $( X, \mathcal T)$ a subset of $X$ is simply called open iff it is an element of $\mathcal T$. Nothing else. This terminology is justified by the fact that the set of open sets (by the definition that you know) in a metric space is a topology and that some of the theorems about open sets, continuity and so on carry over from metric spaces to topological spaces. There will surely be some examples shortly after the definition in your book.

  • $\begingroup$ Sorry for the late reply, but just to make sure, the converse is not true i.e. if $S$ is a member of the topological space $(X, \mathcal T)$, $S$ doesn't have to be open in the metric space $(X, d)$, where an open set is defined using the distance definition? Thank you! $\endgroup$ – harlem Jul 19 '18 at 13:11

Given a metric space $X$, we get a topological space $(X, \mathcal{T})$ in the following way: we declare a set $U$ to be open if for all $x \in U$, there exists $r > 0$ such that $B_r(x) \subset U$. (It is an exercise to check that this meets the conditions of Definition 2.2: namely, that the empty set and the whole space is open, and that it is closed under arbitrary unions and finite intersections.)

However, two different metric spaces $X$ may give rise to the same topological space. For example, we could take $\mathbb{R}$ with the usual metric, or with the metric defined by $d(x, y) = 2|x-y|$; these are different metric spaces, but give rise to the same topological space.

Finally, there exist topological spaces that do not come from a metric space at all. The easiest example (although this is not a very important example) is the space $X = \{x, y\}$ with the topology where $X$ itself and the empty set are open, but neither $\{x\}$ nor $\{y\}$ are open. Try proving to yourself that you can't put a metric on a set of two points which gives rise to this topology.

  • $\begingroup$ (In this answer I have indulged in a standard terminological shortcut: we say a set "is open" to mean it belongs to $\mathcal{T}$). $\endgroup$ – hunter Feb 20 '17 at 10:46

Though the already offered answers are correct, I'd like to try to give a view of why topologies are defined this way.

The idea of an open set is that for each point in the set, the set should also contain all other points that are sufficiently close to it. Or stated in another way, you should not be able to approach a point within an open set without entering the set. So openness is an expression of what it means to be close. In a metric space, we can define closeness by means of distance. But in a more general setting, this is not possible. So instead we define closeness by simply listing what sets we want to be open. This "list" is just a set containing all the open sets in the space, called the topology.

The topology has to capture the basic properties of what closeness ought to be. It turns out that there are two properties that incapsulate the idea (I can expand on why "closeness" should satisfy these, but in the interest of brevity will not). These are that the union of any collection of open sets ought to be open, and that the intersection of any two open sets should also be open. For technical reasons, the entire space and empty set (both of which obviously must contain every point sufficiently close to any point inside them) are also considered open. So these four conditions are what is required to be a topology.

As an example of a useful non-metric topology, consider limits from above. For example: $$\lim_{x \to 0+} f(x)$$

The idea of a limit expresses an idea of what is close. So these limits should be expressible topologically. The limit says that how $f$ behaves above $0$ is important, but how $f$ behaves below $0$ is not. As far as the limit is concerned, what is below $0$ is not close. Only what is above. That is, the intervals $[0, \epsilon)$ define what is close to $0$.

So define $\scr T_1 = \{[a,b)\mid a < b \le \infty\} \cup \{\emptyset\}$. If $[a,b), [c, d) \in \scr T_1$, and WLOG, $a \le c \le d$, then $$[a,b)\cap[c,d) = \begin{cases} \emptyset & b\le c\\ [c, b) & b\le d\\ [c, d) & d < b\end{cases}$$ Hence the intersection of any two sets in $\scr T_1$ is also in $\scr T_1$. But unions are not. So we define $${\scr T} = \left\{\bigcup_{B \in \scr{A}} B \mid \scr{A} \subseteq \scr{T}_1\right\}$$

$\scr T$ is then a topology. Denote the standard topology on $\Bbb R$ by $\scr S$. Then for a function $f: \Bbb R \to \Bbb R$, the limit from above at a point $a$ is the limit at $a$ when the topology on the domain of $f$ is taken to be $\scr T$, but on the codomain the topology is still $\scr S$. This general topological definition of limit is:

  • $\lim_\limits{x \to a+} f(x) = L$ if for every $S \in \scr S$ with $L \in S$, there is an $A \in\scr T$ with $a \in A$ and $f(A \setminus \{a\}) \subseteq S$.
  • $\begingroup$ Re: scripted/unscripted $B$: I have changed it but please check that the result is what you wanted. $\endgroup$ – David Feb 21 '17 at 5:03
  • $\begingroup$ @David - Thank you. I didn't realize that putting \scr at the start of the markup would affect the entire markup like that. $\endgroup$ – Paul Sinclair Feb 21 '17 at 5:06
  • $\begingroup$ I haven't used \scr but I believe it's the same as \bb etc - it affects the whole group and not just a single argument. Compare AB\scr{C}DE $AB\scr{C}DE$ and AB{\scr C}DE $AB{\scr C}DE$. $\endgroup$ – David Feb 21 '17 at 5:28
  • $\begingroup$ Your phrasing "for each point in the set, the set should also contain all other points that are sufficiently close to it." is quite related to the intuitive notion stated in my answer that is somewhat like saying "it is clear-cut that the members are inside", which may be a concise way of "expanding on the idea of closeness". What do you think? $\endgroup$ – user21820 Feb 21 '17 at 11:40
  • $\begingroup$ @user21820 - While the two concepts may be relatable, I think "closeness" is more the idea that topology was invented to capture. However, your post does give an explanation on the ideas that I left out for brevity. However, it sounds like your RE sets are necessarily countable, and thus are a digression here. $\endgroup$ – Paul Sinclair Feb 22 '17 at 0:26

There are two definitions of open set here. An abstract topological space is just a set $X$ along with a designated set $\cal T$ that satisfy the axioms in your post; elements of $\cal T$ are called open sets. One common method of defining such a $\cal T$ is through a base $\beta$: a set of subsets of $X$ such that (i) $\bigcup_{B\in \beta} B = X$ and (ii) For any $B, B'\in \beta$ and any point $x\in B\cap B'$, there exists some $B''\subset B\cap B'$ in $\beta$ containing $x$. The unions of (arbitrary) elements of $\beta$ then form a topology $\cal{T}(\beta)$ on $X$. (The only nontrivial part of verifying that $\cal{T}(\beta)$ is a topology is its closure under finite interesections, and that follows from induction on condition (ii) above.)

For a metric space $(X, d)$, a set $A\subset X$ is often defined to be open if any $x\in U$ has an open ball $U_x = B_{\epsilon}(x)\subset A$ for some $\epsilon > 0$. In particular, $A = \bigcup_{x\in A} U_x$. Unraveling the definition above shows that these open sets are precisely the topology $\cal{T}(\beta)$ constructed above for $\beta$ the set of open balls $B_\epsilon(x)$ for $x\in X$ and $\epsilon > 0$. Thus the two definitions of open set--- the specific one for metric spaces, and the more general one for arbitrary topological spaces--- coincide, provided we give $X$ the topology ${\cal T}(\beta)$ above.


Here is an interesting intuitive definition for open set that actually gives rise to the topological axioms!

An open set is a collection of objects for which we can easily verify membership of any given member.

Note that it may not be easy to verify non-membership of non-members!

Now let us take a look at the axioms:

The empty collection is an open set.

Vacuously true since it has no members.

Any finite intersection of open sets is an open set.

True because you just verify membership in each open set one at a time.

Any union of open sets is an open set.

True because for each member in the union you just have to verify its membership in the correct open set of the union.

Notice how the restriction to "finite intersection" is crucial here because we may not be able to easily verify membership in infinitely many open sets! Here we are using the intuitive assumption that it is easy to do finitely many easy tasks.

Now of course one can ask whether there is a concrete representation of such intuitive notion of open sets. There is! In a metric space, suppose that you interpret that membership of $x$ in $S$ is easily verified iff there is some positive error margin $ε$ such that moving $x$ by a distance of at most $ε$ cannot make it no longer a member of $S$. Then indeed you can easily verify that metric-open sets have easily verifiable membership, and you can see why infinite intersections of metric-open sets may not be open (since those sets may require smaller and smaller error margins for that member).

Incidentally, there is a related notion of semi-decidable sets (namely the sets such that there is a program that accepts the input iff it is a member) or equivalently recursively enumerable (RE) sets (namely the sets such that a program can enumerate all their members by appending them one by one to the end of the designated output list). RE sets are slightly related to open classes in linguistics, in that the enumeration idea corresponds to the notion of a collection whose members are not fixed but get added over time but never removed.

We have the following analogous facts for RE sets:

The empty set is an RE set.

It is enumerated by a program that rejects everything.

Any finite intersection of RE sets is an RE set.

Take any RE sets $S_{1..n}$, and let $P_{1..n}$ be the corresponding programs. Then $\bigcap_{k \in \{1..n\}} S_k$ is enumerated by a program that runs all the programs $P_{1..n}$ in parallel and appends an object to the output list whenever it has been appended to all of their output lists. (Again note how the finiteness restriction for intersections naturally arises from the fact that we can only run finitely many programs in parallel!)

Any RE union of RE sets is an RE set.

(Note that an RE set is represented by a program, so by "RE union of RE sets" we mean "union of RE sets whose programs form an RE set".) Take any set $S$ of RE sets such that there is a program $P$ that enumerates all the programs corresponding to the RE sets in $S$. Then $\bigcup S$ is enumerated by a program that runs $P$ and in parallel runs every program output by $P$ (which we shall call a child program) and appends an object to the output list whenever it is appended for the first time to the output list for some child program.

  • $\begingroup$ See also this post for another related way of interpreting "easily verifiable membership" that lies somewhere between the abstract intuitive notion I described here and the concrete notion in metric spaces. $\endgroup$ – user21820 Feb 21 '17 at 11:42
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    $\begingroup$ Excellent, this should be the accepted answer IMO. $\endgroup$ – leftaroundabout Feb 21 '17 at 12:07
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    $\begingroup$ It also translates easily to: a closed set is a collection of objects for which we can easily verify non-membership. $\endgroup$ – leftaroundabout Feb 21 '17 at 12:11
  • $\begingroup$ @leftaroundabout: I'm biased and also think my answer is at least on par with the rest, but I answered a day later than the top-voted answer and as usual observe the disparity in accruing votes. =P By the way I was considering including your second comment in my remarks, but thought it was obvious enough from the definition of closed sets in topology. =) $\endgroup$ – user21820 Feb 21 '17 at 12:29
  • $\begingroup$ @Winter96: Thank you very much for spotting the accidental error and fixing it! =) $\endgroup$ – user21820 May 9 '18 at 13:57

I somewhat agree with Carsten S that it doesn't necessarily mean anything much – openness is just an abstract, axiomatic concept. Different topological spaces may have very different-looking open sets; for any intuitive explanation you could probably come up with a counterexample space where the analogy fits rather badly.

Nevertheless, openness isn't completely arbitrary. The conditions on $\mathcal{T}$ are crucial. For instance, you cannot have a topology that does not permit you to cover the space with open sets. Also you cannot have two open sets that overlap on anything but an open set.

Here's my stab at an intuitive definition: an open set is a subset that does not contain any sharp edges. If you can completely divide a space $X$ into disjoint nonempty subsets $A$ and $B$, then either $X$ must already be “naturally partitioned” into these regions (any embedding of $X$ into, say, $\mathbb{R}^n$ would then contain a gap between $A$ and $B$), or one of the two subsets must contain a sharp edge which shows you that someone has been making cuts in the space.
Now, that just kicks the stone further down the road: what does “sharp edge” mean? If you want to get rigorous, there's probably no better way to define it than to refer back to open sets: a sharp edge is a subset $R \subset X$ such that any cover with open sets will be strictly bigger than $R$. In other words, the edge is too thin to completely fill any open set that actually contains it.
Trouble is, you'd now have to also specify what's meant by “contains a sharp edge”. This is not simply a subset relation, because certainly most open sets can be further cut apart and then one of the pieces would contain such a edge.

You don't really get around the general, abstract definition. You need concrete examples to really give it an intuitive meaning, but it's always good to keep apart what's specific to the example and what's true in general.

  • $\begingroup$ Hmm ... in my intuition, I think of the edge of, say, an open disc as being distinctly rougher than the edge of a closed disc (which to my mind's eye is smooth and polished). $\endgroup$ – Henning Makholm Feb 20 '17 at 13:24
  • $\begingroup$ You're right. s/rough/sharp/g. $\endgroup$ – leftaroundabout Feb 20 '17 at 13:31
  • $\begingroup$ @HenningMakholm: And there's also the convention of using squares for closed sets and ovals for open sets... =P $\endgroup$ – user21820 Feb 21 '17 at 11:34

Short answer: It's a generalization of the concept of an open interval on the real line to arbitrary spaces.

You'll see statements that the concept of 'open set' has no special meaning. That's only true in the most abstract sense. Really, it's just a convention.

Point-set topology is usually developed from the concept of 'open set' where a topology is defined by defining what its open sets are, and the topology's properties develop from that. (A metric space, of course, defines its open sets from the metric.)

However, the Kuratowski closure theorems show that we can just as easily start by defining the closed sets. It doesn't gain us anything but it's interesting to know as we go along with convention and proceed by defining the open sets first.

Topology developed as a way to understand problems of real analysis, and of course we always try relate any abstract concept of topology back to the Euclidean topology $E^n$ of $\mathbb{R}^n$ (sometimes called the 'usual topology' to emphasize the abstraction).

What are the open sets in $E$? Collections of open intervals. Everything else proceeds from that.

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    $\begingroup$ It would be difficult to claim that modern topology, even modern point-set topology, is still an adjunct to real analysis or should be informed by the example of $\mathbb{R}^n$. $\endgroup$ – anomaly Feb 21 '17 at 5:42
  • $\begingroup$ @anomaly Does the scope of 'modern topology' really matter when discussing why open sets are called 'open sets'? The 'usual topology' $E^n$ is still a frequently-used example of a topology, and properties of more exotic topologies are still compared to it. $\endgroup$ – Spencer Feb 21 '17 at 18:03

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