I can form a mental image of sets with structures like metrics or norms. But if I try to picture a topology/ topological space I fail every time. The information provided in Wikipedia confuses me quite a bit since the concept of topology is new to me.

So, is there a (preferably simple) explanation of a topology placed on a set? And how should someone picture such topologies/ topological spaces?

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    Don't try to explicitly picture it -- "picturing" comes with habit. You should first learn the definition and memorize it, then you should gather enough different examples of topological spaces, and the next time you see a topological statement, try to instantiate it with the examples of topological spaces you know. After some time, topological statements will become second nature, like metric spaces are second nature to you. – Alexandre C. Jun 21 '15 at 16:30

7 Answers7


Topological space come in very different flavors and therefore I don't think that there is one good mental picture to provide the general idea behind the concept of topological spaces.

Apart from the usual examples of a topological spaces like $\mathbb R^n$ with its metric, there are more exotic ones.

For example:

  • Let $X \subseteq \mathbb N$ be open iff $X \in \{\emptyset, \mathbb N \}$ or $X = \{1,2, \ldots, n \}$ for some $n \in \mathbb N$. This defines a topology.
  • Let $2^\mathbb N = \{ (a_n)_{n \in \mathbb N} \mid a_n \in \{0,1\} \text{ for all } n \in \mathbb N \}$ be the set of infinite 0 1 sequences. Then $$ d: 2^\mathbb N \times 2^\mathbb N \rightarrow \mathbb R, \ \left( (a_n)_n , (b_n)_n \right) \mapsto \frac {1}{ \min \{n \mid a_n \neq b_n \}} $$ defines a metric and thus a topology on $2^\mathbb N$. This topology is compact and zero-dimensional which gives it a very different flavor than the usual topology on $\mathbb R$.
  • For a fixed language $L$ we can define a compact topological space on the set of $\mathcal T$ of all $L$-Theories which gives rise to a nice proof of the famous compactness theorem (see here).
  • Another example is the Zariski-Topology on the spectrum of a fixed commutative ring, which is studied in classical algebraic geometry.
  • ...

It might be a good idea to play a little with some simple examples of topological spaces (including some "weird" ones) to get used to this definition.

Martin Sleziak
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Stefan Mesken
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If you start with a metric space $X$, then you define the open sets in $X$ to be those $U\subseteq X$ satisfying that for all $x\in U$ there exists $\varepsilon >0 $ such that $B_\varepsilon (x)\subseteq U$. The reason we are interested in open sets (and their complements, known as closed sets) is due to various elementary theorems of metric spaces which I assume you've seen (otherwise I will just say that the open sets allow to precisely capture the notion of continuity). It is quite easy to prove that the open sets in a metric space are closed under finite intersections and under arbitrary unions.

Now, by analysing proofs one discovers that some of the theorems we like can actually be stated in terms of open/closed sets only, and the proof can be obtained only by using the properties of open sets mentioned above. That means that if one is in a situation that you start with a metric space, note what the open sets are, and then suffer from amnesia whereby you totally forget what the metric was, you can still do quite a lot with the remaining 'space'. Of course, amnesia is not the reason for this, rather we realise that the metric information can be forgotten as long as one remembers the open sets, at least for some purposes (loosely speaking, those related to continuity). So at this level it's a game - there is a metric, but we pretend to only have the open sets it produces, and we see what we can still do with it.

Now, we change the setting. What if there really is no metric at all, only a bunch of subsets which are closed under finite intersections and arbitrary unions? Well, then this is a topology and the resulting thing is a topological space, which you can think of as the result of forgetting some metric that led to these open sets. This turns out to be extremely useful.

Now here is the nice thing. Any topological space is in fact coming from a metric, if one slightly generalises what metric means. The details of this is in Flagg's paper "Quantales and continuity spaces". The bottom line then is that any topology arises as the collection of open sets for some $V$-valued metric space. So you can really think of a topology as the result of temporary amnesia, forgetting the metric.

Ittay Weiss
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  • As I currently don't have access to Flagg's paper. Do you know of any free online source where I can look up the definition of a "V-valued space"? – Stefan Mesken Jun 21 '15 at 08:34
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    Section 2 of http://arxiv.org/abs/1408.3887 – Ittay Weiss Jun 21 '15 at 09:00
  • A norm measures the "length" of an element in a normed space. A metric measures the "distance" between two elements in a metric space. A topology measures ____________ in a topological space. How would you complete this sentence? – Arthur Jun 21 '15 at 09:32
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    @Arthur: "neighborhood". But the point of a topology (as opposed to a metric) is that it *doesn't* measure -- you can say that points are "close to" some other point, without having to be able to give a precise distance (as a metric would). – Christian Clason Jun 21 '15 at 10:00
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    @Arthur I might go with "connectivity" as far as intuition goes (e.g. we could regard, in $\mathbb R^2$ the topology to express the fact that any point is somehow "attached" to those in its neighborhoods - that is, it is attached above, below, left, and right of - so it's kind of a structure expressing the sense in which a piece fabric is held together, where no two points are adjacent, but each point is still held fixed by its neighborhoods). Unfortunately, "connected" has precise mathematical meaning which doesn't capture any of the above. – Milo Brandt Jun 21 '15 at 16:03

It is difficult to give a good visualization for topologies or topological vector spaces because they were introduced precisely for situations where the usual intuition from (non-)Euclidean vector spaces no longer carries. However, a useful way of thinking about topologies is that they are a direct mathematical formalization of the concept of "nearness" without recourse to any concept of "distance".

As you know, a topology on a set $S$ is a collection of subsets (closed under certain operations) that are called open. You can then define a neighborhood of a point $x\in S$ as a set $V\subset S$ which contains an open set containing $x$. (Intuitively, you have a bit of "wiggle room" in $V$ around $x$.) Why is this useful? Because this allows you to give a new mathematical formalization (and thus, generalization!) of the naive notion of continuous functions as "drawn as a single line": a function $f$ is continuous, if points in a neighborhood of $f(x)$ come from points in a neighborhood of $x$.1

Of course, if you have a metric on $S$, then the neighborhoods can be defined in terms of a distance from $x$ (such as the open balls $B_r(x):=\{y:d(x,y)< r\}$). Many topologies can be written in terms of such distances (they are called metrizable), and these are not bad models for picturing them. However, this is not true for all topologies: There are some where you can say that two points are neighbors (i.e, one lies in a neighborhood of the other) without being able to give a single number measuring the distance. A practically relevant example is the set of all continuous functions, where the topology is that of pointwise (as opposed to uniform) convergence.

1. In fact, the development of much of topology was driven by this question, and the theory might make more sense if you keep this in mind while learning it.

Christian Clason
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It is very common in mathematics that you start with an easily comprehensible, "real world-inspired" definition. But then you realize that some things cannot be proven if you insist e.g.on the triangle inequality, so you define a more abstract object without it.

I don't know how to picture a general topological space but wouldn't see that as a disadvantage, since such a picture might lead to wrong assumptions / wrong intuition. Proving abstract stuff generally means to stick to the letters of the definition. However I'd keep in mind that $(\mathbb R^n,\lVert\cdot\rVert_2)$ is a simple example of a topological space, and sometimes it can give you valuable counterexamples when you're lost in abstraction ;-)

Edit: Answering the question in the comments (which I might have misunderstood), that's how a $p$-norm induces a topology on $\mathbb R^n$:

For $p\ge 1$ and $x=(x_1,...,x_n)\in\mathbb R^n$, $\lVert x\rVert_p = \left(\sum_{k=1}^n\right)^{1/n}$ defines a norm. ($p=2$ for the Euclidean norm)

For $x,y\in\mathbb R^n$, the function defined by $d:(x,y)\mapsto\lVert x-y\rVert_p$ is a metric.

Then, setting the sets $B_r(x):=\{y\in\mathbb R^n\mid d(x,y) < r\}$ for any $x\in\mathbb R^n, r>0$ as open (and their unions and intersections) defines a topology on $\mathbb R^n$.

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  • How is R^n a topology? – Jonathan Hebert Jun 21 '15 at 08:07
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    For example with the topology induced by the Euclidean norm. – luckyrumo Jun 21 '15 at 08:08
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    @luckyrumo: I think what Jon may have been getting at is that when someone comes to you with a question at this level, it pays to split hairs a little bit with your words. I wouldn't mind you saying that $\Bbb R^n$ is a *topological space* (though strictly speaking it's not true), but to say it is a *topology* feels a bit misleading. – Eric Stucky Jun 21 '15 at 08:21
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    @Eric: Thank you, I updated the answer. – luckyrumo Jun 21 '15 at 08:25

The way I always thought of topology was as a way to describe what "nearness" and "farness" mean, i.e. to give a sense of "space" or "location" to a set. The issue here is that this only really has meaning for certain topologies. Examples where there is no obvious notion of convergence actually meaning getting close to anything would be the cofinite topology (i.e. $(X, \tau)$, where $\tau$ is the set of all subsets $S \subseteq X$ such that $X \setminus S$ is finite), or topologies like $\tau = \tau_{x} = \{ S \subseteq X : x \in S \}$. Our topology professor told us that unless your space is Hausdorff (i.e. for any two points $x, y \in X$, there exist open sets $U, V \in \tau$ such that $x \in U$, $y \in V$, and $U \cap V = \emptyset$), then you don't really have a topology that conveys these ideas of "space".

When you're not Hausdorff, there are still cool things that can be done with a topology (for example, you can show that the Tychonoff's theorem is equivalent to the axiom of choice). Alternatively, though this may be a stretch, you'll sometimes find that you can use other objects to induce topologies which are not themselves Hausdorff, but you'll find that the induced topology might convey other sorts of information.

One example is to consider a complete probability space $(X, \mathcal{F}, \mu)$, i.e. a probability space such that if $\mu(F) = 0$, and $E \subseteq F$, then $\mu(E) = 0$ (note that a Borel probability space is not generally complete). In this case, if we define $\tau = \{ F \subseteq X : \mu(F) = 1 \} \cup \{ \emptyset \}$, then we have a topology. Now, in general, this will not meet many of your separation axioms. For example, if we consider the probability space $([0, 1], \mathcal{L}, \lambda)$, where $\lambda$ is Lebesgue measure and $\mathcal{L}$ is the completion of the Borel $\sigma$-algebra, then you do not have separability. We can check that it's $\textrm{T}_{1}$, but it will not be $\textrm{T}_{2}$. To see this, assume for contradiction that $x, y \in [0, 1]$, and that $x \in U, y \in V, U \cap V = \emptyset, \mu(U) = \mu(V) = 1$. Then $\mu(U \cup V) = \mu(U) + \mu(V) = 2 > \mu([0, 1])$, a contradiction. That no topology generated this way is $\textrm{T}_{2}$ will follow in the same manner.

Second, let $\mathcal{U}$ be an ultrafilter, that is, a family of subsets of $X$ such that (i) every superset of an element of $\mathcal{U}$ is an element of $\mathcal{U}$, (ii) every finite intersection of elements of $\mathcal{U}$ is in $\mathcal{U}$, and (iii) for every subset $S \subseteq X$, either $S \in \mathcal{U}$, or $S^{\complement} \in \mathcal{U}$, but not both. It is trivial to see that $\tau = \mathcal{U} \cup \{ \emptyset \}$ is a topology (the closure of finite intersections is guaranteed by (ii), and the closure under arbitrary union is promised by (i)). In this case, we can show that such a topology is never Hausdorff, again because we can't have disjoint open sets. Suppose for contradiction that $x \in U \in \mathcal{U}, y \in V \in \tau, U \cap V = \emptyset$. Then since neither $U$ nor $V$ is empty, we have that both must be elements of $\mathcal{U}$. Thus $V \subseteq U^{\complement} \Rightarrow V^{\complement} \supseteq U \Rightarrow V^{\complement} \in \mathcal{U}$. But this means that both $V$ and $V^{\complement}$ are elements of $\mathcal{U}$, contradicting axiom (iii).

But what both topologies convey, i.e. the properties which define the open sets $\tau$, are one of those that are conveyed by, say, $\mathbb{R}$ with its standard metric, and that is a certain sense of bigness. A set of probability $1$ is big in the sense of measure, and an element of an ultrafilter is big in the sense of the ultrafilter. Similarly, an open set in $\mathbb{R}$ is "big" in the sense that it surrounds all its points, whereas a non-open set would not. Again, this might be a stretch, but it does suggest that even a non-Hausdorff space can be useful.

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For a topology, think of the three element set $$E= \{a,b,c\}.$$ A topology on the set will be the whole set E or the $\emptyset$, also you can construct many more.

Now think of this. Take the set $\Bbb{R}$ , now the set of all half open intervals on this set is the basis for the lower limit topology.

Also the subspace topology can be imagined like this:

Take some arbitrary set ,draw open balls lying in that set,these open balls are elements of the topology . Now take some arbitrary subset and see the sets which are lying inside the subset. These sets define the subspace topology.

Well the mathematical definition of the subspace topology on a set Y , where $Y\subseteq X$ is $$\tau_{Y}=\{Y\cap U|U\in \tau_{X}\}$$

I wish this helps.

  • the collection of all intervals in $\mathbb R$ is not a topology. You last paragraph is obscure. What do you mean by "take some arbitrary set, draw open balls lying in..." how can you draw open balls in an arbitrary set? what does that mean? The rest is just as unclear. – Ittay Weiss Jun 21 '15 at 13:30
  • Sorry, I mean the OP is trying to imagine topologies on a set right So that is why I want to be as frank as possible and make the OP think of an diagram@IttayWeiss –  Jun 21 '15 at 13:32
  • @IttayWeiss I have edited it . Is it fine now? –  Jun 21 '15 at 13:38
  • no, still not a topology and the last paragraph is still not good. – Ittay Weiss Jun 21 '15 at 14:15
  • @IttayWeiss Now I think I have got what I wanted to say... I just jumbled up basis and a topology –  Jun 21 '15 at 14:39

Largely, it depends on how the topology is given. If the topology is given as "two copies of $X$ with [some complicated rule for which pairs of the same point of $X$ are considered open and which aren't]", then the obvious picture is two copies of $X$ with whatever visualization glue you use to indicate open pairs and non-open pairs. To see that there are many different ways to give a topology/topological space and consequently different paths to visulaization, I'd recommend "Counterexamples in Topology" by Steen and Seebach.

A "picture" of a topology is the lattice of open sets with union and intersection given by join and meet. The axioms tell you that arbitrary unions are open, so the lattice goes "up" as far as you want (up to the entire space). Only finite intersections are guaranteed open, so it becomes interesting to determine how far "down" one can go -- what sorts of sets are the finite intersections of given open sets. This leads to finding bases (or subbases), a distinguished collection whose members are sufficient to generate the entire lattice. This does not give you any intuition of the space as a chunk of $\mathbb{R}^n$, but with practice, you eventually develop a sort of double vision: chains in the lattice correspond to nested sets in the space and one can move back and forth between these two points of view easily.

There are other ways of doing this, though I stand by my first statement -- it depends on how the space/topology is given what is the "best" way to visualize it.

Eric Towers
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