I am now trying to understand what a topology and a topological space is. Yes, I know the "formal" or "mathematical definition" of it, it is in my notes so it's easy for me to reiterate that. Please bear with me as I am trying my best to express my confusion, it's a little hard to even do that in words.

Here's the definition I am sticking to

$X$ 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. $\phi \in \tau$ and $X \in \tau$

$i$) So my problem is,for each 1,2,3 conditions, I understand what they mean. So each subset in $\tau$ satisfies 1,2,3, which as a whole, is some subset of $X$ and we'll just name it "a topology" on $X$...yes?

My issue is, I don't see what this "collectively" gives. So if you are to explain it to someone who does not really do math, and explain it casually, what would you say? what does this "set of subsets" give? Is it nothing more than "just the set of subsets that satisfy 1,2,3, end of story"? That is probably one of the reason why I cannot find a topology for any specific given set. I just don't know how.

$ii$) And also, the notion of induced topology by a metric is unclear to me. The "discrete topology" as I hear very often, is apparently one of the most common topology which is induced by the discrete metric.

In my notes, it says

The discrete topology of $X$ is the collection of all subsets of $X$, which is the largest possible topology on $X$.

So, if I take a set of positive integers $X=\mathbb{Z}^+$ then the discrete topology is $\{\mathbb{Z}^+, \phi,\{1\},\{2\},\{3\},...,\{1,2\},\{1,3\},...,\{2,3\},\{2,4\},...,\{1,2,3\},\{1,2,4\},...\}$? But even if it is, how is this "induced by the discrete metric? How is it relevant? I could have just blindly followed the definition of a topology to get all these subsets...right? without using or referring to any metric.

$iii$) And what is a topological space? And yes, I "know" it's $(X, \tau)$ where $\tau$ is a topology on $X$ but again, I can't help but confuse when I think of "metric spaces"

Metric space so far makes a lot more sense to me. I see $d(x,y)$, a metric as a function, so when we say a metric space $(X,d)$ just as how we phrase $(X,\tau)$,I understand it as a set $X$ with some function$d(x,y)$ which can be "applied" to the elements of $X$ to "give a value", which is the "distance." So it's a set $X$ where I have given a "method" to tell how far any 2 elements are in it.

Now, $(X, \tau)$ doesn't sink in to me because $\tau$, unlike $d(x,y)$ doesn't do anything to $X$ (does it?). Meaning, it's just a set in a sense(with some special features) but does not allow me to "pick some value in $X$" to give me "some value"(whatever it is). So it comes to, as stupid as it may sound, "what is the point of having $\tau$"? Ultimately...just what is a "topological space"? A pair of sets $X$ and $\tau$? If it is analogous to the metric space, $d(x,y)$ "defined" on $X$, what would it mean that $\tau$ "defined" on $X$?

I hope expert topologists would understand what I am confused with an where I have the wrong way of thinking. I really need someone to enlighten me here, it's just getting so abstract to me..

I really appreciate your reading until here, it was indeed a long question, I am sorry. Thank you in advance for you help

  • 1
    $\begingroup$ There are a lot of questions here. I will address just two of them. (1) The axioms for a topology are motivated by the fact that these properties are satisfied by the collection of open sets in, say, $\mathbb R$, or any metric space. (2) The discrete metric generates the topology you indicated (where all subsets of $X$ are open), because for each point $x \in X$, the ball with radius $1/2$ centered at $x$ consists of just the point $x$. Therefore, all the singleton sets $\{x\}$ are open in this topology. Then property (1) of the definition of topology implies that all subsets of $X$ are open $\endgroup$ – Bungo Oct 21 '15 at 17:24
  • $\begingroup$ I am not an expert, but the sets defined by the topology, i.e. the open sets, allow the definition of continuous functions. Similarly, the definitions of connectedness and compactness. You can see what these are for say $\mathbb{R}^n$ and see how the abstract definition captures these. $\endgroup$ – Aravind Oct 21 '15 at 17:26
  • $\begingroup$ I wouldn't say this is well researched. $\endgroup$ – Alec Teal Oct 21 '15 at 20:15
  • $\begingroup$ Regardng question ii), in your attempt at writing down the discrete topology, in addition to $\mathbb{Z}^+$ itself you appear to be only listing the finite subsets of $\mathbb{Z}^+$. Appearances may be deceiving me. Nonetheless, I'll point out that the discrete topology on $\mathbb{Z}^+$ consists of the collection of all subsets of $\mathbb{Z}^+$, not just the finite subsets. In fact, you could perhaps give yourself the exercise to show that the collection of subsets consisting of $\{\mathbb{Z}^+, \emptyset, \ldots$ (all finite subsets)$ \ldots \}$ is not a topology. $\endgroup$ – Lee Mosher Oct 21 '15 at 20:15
  • $\begingroup$ (+1) great question, i'm having EXACTLY THE SAME type of problem from when i first opened munkres $\endgroup$ – user153330 Feb 3 '16 at 23:25

A topology is any collection of sets that satisfies the given three conditions. That means that most sets have more than one possible topology that can be defined on them. For example, there are several possible topologies you can define on $\mathbb R$.

That means that your comment of

I cannot find a topology for any specific given set.

Makes no sense, since if I simply give you a set, there is no way for you to discover the topology.

To improve your intuition of topology, it's easiest to first look at topologies induced by metric spaces. If $(X,d)$ is a metric space, then the set of all open sets of $X$ (where open is defined using only $d$), is a topology on $X$. Remember, a set $A$ is open in a given metric if for every element $a\in A$, there exists a ball that contains $a$ and is itself contained in $A$.

For example, if $X$ is equipped with a discrete metric, then every singleton set $\{x_0\}$ is an open set because it is actually an open ball:

$$\{x_0\}=\{x\in X: d(x,x_0) < \frac12\}.$$

Furthermore, that means that every set $A\subseteq X$ is open, because for each $a\in A$ you can find a ball that contains it and is contained in $A$. Namely, that ball is $\{a\}$.

That means that the set of all open sets given the discrete metric is just the set of all sets. Therefore, the topology induced by the discrete metric is the discrete topology.

And finally, you ask what a topological space is. It is simply a pair of sets $(X,\tau)$. Where in metric spaces, the $d$ is a function that tells you the distances between elements of $X$, here, $\tau$ is a family of sets that tells you the properties of subsets of $X$ (i.e., it tells you what elements are open).

  • $\begingroup$ Hi there, thanks for answering. I'll need some time to understand all this so please allow me to be very slow in commenting back...But, in the where I said " I cannot find a topology for any specific set" I meant something like this; Q. Prove that there is a topology on a set X consisting of he cofinite subsets together with $\phi$. See, it sort of is asking me to find a topology, but I don't know "how". I don't know where to start. I don't what set X is ($\mathbb{R}$?$\mathbb{C}$?) so I cannot indicate any subset that might make its compliment finite(i.e. cofinite). $\endgroup$ – Melba1993 Oct 22 '15 at 21:18
  • $\begingroup$ @Melba1993 What is $\phi$ in that case, and what does "together with" mean? Does it mean "all cofinite sets that contain $\phi$?" $\endgroup$ – 5xum Oct 22 '15 at 21:21
  • $\begingroup$ $\phi$ as in the "empty set", sorry I don't know the latex symbol command for it... I just copied the exact wordings from my worksheet from the lecture... it says "this is called the cofinite topology" after that, does this "cofinite topology" contain all cofinite sets that contain the empty set? I hope i'm being clearer now... $\endgroup$ – Melba1993 Oct 22 '15 at 21:25
  • $\begingroup$ @Melba1933 In future, the emptyset symbol is obtaned with the command \emptyset. Also, I believe you misunderstand the question. The question is not "find me the topology that has these sets in it". The question is "Here is a family of sets. Prove to me that this family of sets is a topology." To do that, you need to prove that the family satisfies the three axioms that it needs to satisfy. $\endgroup$ – 5xum Oct 22 '15 at 21:29
  • $\begingroup$ @Melba1993 In your case, the family of sets is "all cofinite sets, plus the emtpy set", or in other words, $\tau = \{A\subseteq X| A\text{ is finite}\}\cup\{\emptyset\}$ is given, and you need to prove that $\tau$ is a topology. $\endgroup$ – 5xum Oct 22 '15 at 21:30

Think first about functions of a single real variable, as in basic first-year calculus. The intuition behind the definition of continuity is that a function $f: \mathbb{R} \to \mathbb{R}$ is continuous at a point $x_0$ if and only if points that are "close" to $x_0$ map to points that are "close" to $f(x_0)$. That's an imprecise statement, but it captures the basic idea.

This same (informal, impecise) definition also captures what it means for a function $g: \mathbb{R}^n \to \mathbb{R}^m$ to be continuous. We look for points that are "close together" and we require that points that start out "close" don't get mapped "apart".

All of the above can be made more rigorous and formal, and if we do that we get the usual $\epsilon - \delta$ definition of continuity. But what if we step away from that definition? What if we want to talk about a function from one set to another, where the sets don't necessarily have to be thought of as points in some $n$-dimensional space? Is it still possible to talk about "continuity" if we lack a way to talk about "distance"?

The definition you quote does that. Each "open set" in a general topology defines a kind of "nearness", in the sense that given any two points and any open set we can ask whether those points belong to the set, and if they do, we can say that those two points are "close" to each other relative to that set.

The condition that the entire space $X$ is open means that there is a "maximally loose" kind of "nearness" that does not care which points you choose, because (relative to $X$) any two points are "nearby".

The condition that the empty set is also open means that there is a "maximally strict" kind of "nearness" that does not care which points you choose, because (relative to $\emptyset$) no two points are "nearby".

In most topological spaces, there are more open sets that just those two. In general open sets nest within each other, so that two points might be "nearby" relative to one open set, but not "nearby" relative to a smaller open set. Thus being in an open set together generalizes and abstracts the idea of "being close together" without requiring that we have a way of assigning a numerical measure to the distance between two points.

But sometimes there is a way to measure the distance between two points: a metric. And when there is, that metric induces a natural topology, in which we can define "$x$ and $y$ are close" to mean "$x$ and $y$ are less than some distance $\epsilon$ apart from each other."

  • $\begingroup$ typo : care not car ^-^, btw great answer (+1) $\endgroup$ – user153330 Feb 3 '16 at 23:36
  • $\begingroup$ @user153330 Thanks for noticing, I have fixed the typo. $\endgroup$ – mweiss Feb 4 '16 at 0:56

Let's pretend there's a such thing as "infinitely close." (There isn't, generally, but let's pretend.) We care about this, because ideas like "continuity" depend on an intuitive notion of "infinitely close," even though it doesn't actually exist (usually). So, how would we describe this notion?

Well, with metric spaces, if we want to say that $x$ and $y$ are infinitely close, we can try saying something like "$d(x,y)<1$, and $d(x,y)<0.1$, and $d(x,y)<0.01$, and…" and so on. That is, they approach it with the idea of distance.

Topological spaces approach this from a more set-theoretic perspective. If we want to say $x$ is infinitely close to $0$, we could say "$x\in(-1,1)$, and $x\in(-0.1,0.1)$, and $x\in(-0.01,0.01)$, and…"

This line of thinking leads you to the idea of a "neighborhood." (A set $N$ is a neighborhood of $x$ if there's an open set $O$ with $x\in O\subseteq N$.) In this line of thinking, a neighborhood of $x$ is expected to contain everything infinitely close to $x$. Most people introduce topology in terms of open sets, mostly out of convenience. (An open set is a set that's a neighborhood of each of its points.) However, I've also seen it introduced in terms of neighborhoods. It doesn't really matter, since you can define them in terms of each other.

So, the topology of a topological space is the set of open sets in it. The axioms of the topology are the properties that we'd expect open sets to have. (There are some properties that almost made the list, but people opted for generality and decided that they aren't necessary for topologies. I'm thinking of the so-called separation axioms, which you may learn about later.)

  • $\begingroup$ great answer, really liked it gives some insight on the relation between neighborhood in the classic analysis sense and with topology (+1) $\endgroup$ – user153330 Feb 3 '16 at 23:40
  • $\begingroup$ could you please explain how we could use topology to prove for example that $\lim_{M\to+\infty}\frac{1}{M}=0$? $\endgroup$ – user153330 Feb 4 '16 at 22:45
  • $\begingroup$ @user153330 I'm not sure why you're saying that "neighborhoods" are more analysis than topology. They seem to be purely topological ideas to me. $\endgroup$ – Akiva Weinberger Feb 4 '16 at 22:47
  • $\begingroup$ but how could we use them to prove for example that $\lim_{M\to+\infty}\frac{1}{M}=0$ $\endgroup$ – user153330 Feb 4 '16 at 22:49
  • $\begingroup$ @user153330 Well, let a deleted neighborhood of $x$ be a set of the form $N_x\setminus\{x\}$ where $N_x$ is a neighborhood of $x$. Let $\overline{\Bbb R}:=\Bbb R\cup\{-\infty,+\infty\}$ be the extended reals, with the order topology, and let $f:\Bbb R\to\Bbb R$ be the function defined by $f(M)=\frac1M$. $\endgroup$ – Akiva Weinberger Feb 4 '16 at 22:53

Topology is a game.

You can make up whatever rules you'd like as long as they satisfy your three listed properties. The open sets can be nice balls, like in a nice metric space like $R$, or they can be crazy weird.

The point of the study of topology is to realize and use theorems that don't depend on other properties of the space, just on what it means to be open and closed in that space.

To someone outside of math I say that for most fields: Make up whatever rules you like. SOMETIMES these are interesting or useful. Mostly not.

  • $\begingroup$ then the question would become why do we play topology instead of another game? this approach of just a game actually just begs the question $\endgroup$ – user153330 Feb 4 '16 at 22:44
  • $\begingroup$ I think topologists play because, to them, it is interesting. Many others have disagreed. Clearly it is a matter of opinion. $\endgroup$ – amcalde Feb 5 '16 at 14:00

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