I've been very confused with topology lately, and here's another one that I can't answer myself after spending hours staring at the definition.

So, here's the definition I am using

$X$ is a set. A topology on X is a set of subsets $\tau$ of $X$ with the following properties

  1. Whenever $(U_{i})_{i \in I}$ is a family(finite or not) of subsets of $X$ such that $U_i \in \tau$, $\forall i \in I$ then $\cup_{i \in I}U_i \in \tau$

  2. Whenever $U_1$, $U_2 \in \tau$ then $U_1 \cap U_2 \in \tau$.

  3. $\emptyset \in \tau$ and $X \in \tau$

So far so good. But Here's a remark in my notes that confuses me;

We call the elements of $\tau$ the open subsets of $X$. Thus "$U \in \tau$" and "$U$ is open in $X$" mean the same thing.

Firstly, "open subsets", in my understanding, is that "$U$ is an open subset of $X$, iff $\exists$ open ball $B_\epsilon(X)$ of radius $\epsilon >0, \forall x \in U \subseteq X$ such that $B_\epsilon(X) \subseteq X$" Casually speaking, for every point in this subset, I can find an open ball around it which is always in that subset.

But to define an open ball, we need a metric, by definition of a ball, yes? If the elements of $\tau$ are called the "open subsets" of $X$, then we clearly are talking about the existence of a ball, thus the existence of a metric. But just as in the definition, $X$ is a set, but no where does it say it is a "metric space" Is it always assumed that $X$ has some metric with it and we're building a "topology" over it? If so, then why do we have something called a "topology induced by a metric"? If we're always assuming metrics defined on $X$ to create a topology(because, without it, we cannot define "open" thus there would be no such thing as "open subsets"), doesn't that mean every possible topology must be induced by some metric?? I am so confused

And, for the remark, it's saying that being a member of $\tau$ is the same as being called an "open subset" and "being open in $X$". But, why? The definition does not mention anything about $U_i$ being open or closed or whatever,can I not take closed subsets of $X$ and still satisfy the $3$ conditions i.e. qualify them as a topology?

Please someone help me out here, every time I feel like I understood something, a new question comes up and I feel depressed...Every time I get something, it contradicts with my understanding...

  • $\begingroup$ I think the most important step for you is to temporarily unlearn all the stuff you have learnt about open sets from calculus. A topology is simply a rule of determining whether a subset of $X$ is "open". The rule you learnt from calculus is only one of the many different rules. In general topology we study a theory that's generally applicable to such "rules", so that if one day you think of a creative rule, you can have enough toys to play with. $\endgroup$ – Qidi Oct 22 '15 at 23:20
  • $\begingroup$ Switch the word "open" by any other word temporarily. When you realize the word is irrelevant, turn back and start using the word again. The word is confusing you more than anything. $\endgroup$ – Aloizio Macedo Oct 22 '15 at 23:54
  • $\begingroup$ Forget about the "$\exists$ open ball…" definition of open sets. That's talking about an open set in a metric space. Here, we're talking about open sets in a topological space. Completely different. (Well, technically, metric spaces can be thought of as a special case of topological spaces.) $\endgroup$ – Akiva Weinberger Oct 23 '15 at 0:43

This is something that also caused me a lot of troubles when I started topology. The key to understanding what is going on at first is not to think about what you learned in calculus about open sets. In topology, an open set is just an element of your topology $\tau$, and $\tau$ must satisfy the axioms you mentioned. Therefore, being open is something that is totally dependent on which topology you're considering and is not an intrisic property of subsets of $X$.

If $X$ is a metric space, you can define $\tau$ to be the set of subsets $U\subset X$ that satisfy the "ball condition". It is a good exercise to show that $\tau$ is indeed a topology. In that case, when you work with this topology, being open in the topological sense is the same as being open in the sense of calculus. But you could also consider other topologies on $X$ that don't have anything to do with the metric and you would get totally different open sets.

So you just have to think of the topology $\tau$ as some extra structure on $X$ telling you which subsets you can call open.

  • $\begingroup$ Hi, so, hang on, if I understood your first paragraph correctly....there are many different topologies we can find on some set $X$ and "some" might be "open" but some other may not...yes? Then it confuses me more, if it isn't intrinsic, then why does my remark say (or assert, rather) that we can call the elements of $\tau$ an open subset, as if they're "always" open?? I might be misunderstanding you, but you have said that "open"ness is not always in the nature of $\tau$ but in the last paragraph, you also said $\tau$ tells us which subsets of $X$ can be called open...it.. contradicts to me... $\endgroup$ – Melba1993 Oct 22 '15 at 22:52
  • $\begingroup$ Well it doesn't really mean anything to say that a topology is open. But what I wanted to say is that there are in general many different topologies on a given set $X$. And a set can be open in one of these topologies but not in the others. That's why if you want to be rigourous you should always specify what topology you are using. In particular for metric spaces, you can say something like "let $(X,d)$ be a metric space with the topology induced by the metric $d$". $\endgroup$ – Nitrogen Oct 22 '15 at 22:57
  • $\begingroup$ Or another example, sentences like "let $X$ be a topological space and $U\subset X$ be open" don't mean anything until you specify which topology you are considering. That's why the rigourous statement is "let $(X,\tau)$ be a topological space and $U \in \tau$ be open." $\endgroup$ – Nitrogen Oct 22 '15 at 23:01
  • $\begingroup$ Thanks for answering back...but yes, so some are open, some are not and I just can't get my head around how it doesn't mean anything to say it's open...is there no reason behind it? Is it something I just should disregard...? $\endgroup$ – Melba1993 Oct 22 '15 at 23:02
  • $\begingroup$ Apparently you are just starting to learn topology, so it is totally normal to be confused, everybody is when starting to learn this subject. But you will soon see that when considering topological spaces we fix a topology on them once and for all and then do actual topology. For example with metric spaces we almost always consider the "balls" topology which has a lot of nice properties. Anyway, the best way to get your head around these issues is to be patient, you will get used to these notions and very soon everything will be clearer. $\endgroup$ – Nitrogen Oct 22 '15 at 23:06

The issue is that we don't need a metric to study continuity, because $f\colon X \to Y$ is continuous if and only if for every $U \subseteq Y$ open, $f^{-1}(U)\subseteq X$ is open (here $X$ and $Y$ are topological spaces). We also know that an arbitrary union of open sets is again open, same for finite intersections. Also, $X$ and $\varnothing$ are open.

We try to "axiomatize" that, saying that a topology $\tau$ is a collection of sets in $X$ verifying the conditions in the definition you have there.

Then we say that the sets in $\tau$ are "open", and as far as we go here, "open" is nothing more than a name.

You can check that if you have a metric space $(X,d)$, then the collection $\tau_d$ of "$d$-open" sets (defined with open balls, using the metric $d$) do verify conditions 1., 2. and 3. of the definition of topology.

  • $\begingroup$ So, as I understood, I can come up with topological spaces and "continuous mappings" on them where these continuous functions has nothing to do with my intuitive calculus-based understanding of continuous functions, ie. I can find continuous functions without geometric meaning. But when I consider standard euclidean metric, everything comes up nicely in terms of geometry? $\endgroup$ – meguli Jul 24 '17 at 12:22
  • $\begingroup$ @meguli Yes, that's the gist of it $\endgroup$ – Ivo Terek Jul 24 '17 at 14:29

Not every topology is induced by a metric. For example, let $X$ be an infinite set and $\mathscr T=\{U\subseteq X:X\setminus U\text{ is finite or } U=\emptyset\}$. You can easily check that $\mathscr T$ satisfies the definition of a topology. So, any set whose complement is finite is considered to be open in this topology.


What you've learned previously is that given a metric space, $X$, we can define a topology on $X$.

Note that it is quite possible for two different metrics on $X$ to define the same topology on $X$. Since continuity under a metric only depends on the topology, this means that the exact metric is much less important than the open sets that come from the metric. (Metrics have other purposes, such as defining uniform continuity, something which does not exist in general topology.)

Also, there are topologies on $X$ that do not come from any metric.

For example, if $X=\mathbb R$, we can define a topology I'll call $\Lambda$, for "left," where the open sets are of the form $(-\infty,\alpha)$, for $\alpha\in\mathbb R$, and the empty set and all of $\mathbb R$. This is a perfectly fine topology by the above definition, but it does not correspond to a metric.


There is interesting spaces where the topology cannot be given by a metric one.

A simple exemple would be the spaces of functions from $\Bbb R$ to $\Bbb R$ for the pointwise convergence (or, in other words, the product topology on $\Bbb{R}^\Bbb{R}$)


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