I looked for a lot of explanations for the idea of defining a topology for a set to make a topological space, I found a lot of clever explanations but most of them seem like just "concrete explanations that may seem like the original idea but actually has nothing to do with it", and the worst of them who try to make an analogy between a topological space and a metric space by explaining them with a notation of semi-measures? which I guess has nothing to do with the actual idea behind the topology.

I think "but not quite sure that's why I'm asking" that the core idea of topology is in its definition of continuous functions, like if we take the donut and cup definition of two shapes: two shapes are the same topologically if there is no cuts needed to transform one of them to the other, and put two 2D surfaces in $\mathbb{R}^3$ , it's easy to see that they will satisfy this property if there is a kind of a continuous function and a continuous inverse between them, so in order to generalize that you need a minimal structure to define for sets so that you can define continuous functions, so is that it?

EDIT: It seems that the easiest way to make sense of topologies and there axioms at the beginning of studying the subject is basically through a generalization of continuous functions $f:\mathbb{R}^n\to \mathbb{R}^m$, in the simplest case of them $f:\mathbb{R}\to \mathbb{R}$ we can easily see that a function is continuous everywhere if the preimage of any open interval is also an open interval, if $f$ is discontinuous at some point say $x$ it's easy to see that you can make an open interval containing $f(x)$ but it's preimage will be half closed half open interval so it fails the condition of making the preimage of any open subset open, then the axioms of a topology can be seen as the properties of the set of open intervals, the union of 2 opens is open ...etc, so a topology is a generalization of the set of open intervals so we can define continuous functions for any set with such a structure topological spaces, this intuition seems also to be the direction of Munkres's topology book when he first introduce continuous functions.


I like Asvin's answer a lot, but I would like to add an answer of a slightly different flavor.

I see the answer as "yes," because a topology is the minimum amount of structural information about a space that is generally used in mathematics to decide whether a function is continuous.

The definition of "continuous" was probably first given with respect to functions from $\mathbb{R}$ to $\mathbb{R}$ (or subsets thereof), and defined with respect to $\mathbb{R}$'s metric structure, although the definition predated the idea of "$\mathbb{R}$'s metric structure" so nobody would have described it that way. A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at a point $c\in \mathbb{R}$ if for all $\varepsilon >0$, there exists a $\delta >0$ such that $|x-c|<\delta\Rightarrow |f(x)-f(c)|<\varepsilon$. (Notice that the metric structure of $\mathbb{R}$ is playing the key role here, as $|x-c|$ and $|f(x)-f(c)|$ are both measurements given by the metric.) A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous (no reference to a specific point) if it is continuous at each point $c\in \mathbb{R}$.

In this context, at some point it was noticed (and proven as a theorem) that $f$ being continuous on all of $\mathbb{R}$ is actually equivalent to the property that for any open set $U\subset\mathbb{R}$, the inverse image $f^{-1}(U)$ is also open. Working out this equivalence (for functions $\mathbb{R}\rightarrow \mathbb{R}$) is a very satisfying exercise.

Once this theorem was articulated and proven, it became possible to make the following profound observation: While the original definition of continuity made explicit use of distances like $|x-c|$ and $|f(x)-f(c)|$, actually it would be possible to determine whether $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous without being able to access these distances, as long as we still have access to whether or not any given set is open. In other words, the knowledge of which sets are open is enough information to determine continuity. Furthermore, if we changed how distances are measured but in such a way that it didn't change which sets are the open ones (for example by replacing $|x-y|$ with $2|x-y|$ as the distance between $x$ and $y$), it would not change which functions are the continuous ones. To summarize, while it initially appeared that the metric was intrinsically involved in determining if a function was continuous, the above theorem means that really it's just that the metric determines which sets are open, and the collection of open sets is really what determines continuity.

This is what paved the way for the definition of a topology. If knowing which sets are open is really what determines whether functions are continuous, then we can drop the metric from the definition completely and use the conclusion of the above theorem as a definition: if $X,Y$ are any spaces such that I know how to determine whether their subsets are open, then $f:X\rightarrow Y$ is defined to be continuous if for each open $U\subset Y$, the inverse image $f^{-1}(U)$ is open in $X$. Thus the information about which sets are open (i.e., the topology) is really the only info you need to determine continuity.

(I see this as analogous to the progression in learning trigonometry from (a) originally defining $\sin \theta$ as the ratio of certain sides in a right triangle, which only makes sense when $0<\theta<\pi/2$, but then (b) learning as a theorem that this definition of $\sin \theta$ coincides with the $y$-coordinate of a certain point on the unit circle, to finally (c) replacing the ratio of sides definition with this $y$-coordinate of the unit circle point as the definition, so that $\sin\theta$ can be defined even when $\theta$ is not in the interval $(0,\pi/2)$. Similarly, continuity is (a) first defined with respect to a metric, but then (b) it is proven as a theorem that the definition coincides with a criterion based only on open sets, and finally (c) the criterion based on open sets replaces the metric-based definition as the definition, allowing continuity to be defined even when there is no metric, as long as you know which sets are open.)

Let me add a comment relating this train of thought to William Elliot's answer, because it pulls in a somewhat different direction. William Elliot's answer foregrounds limits as the key idea of a topology, whereas this answer, and Asvin's, foreground continuity. I see this difference as essentially one of taste. Limits and continuity are deeply interlinked concepts. Indeed, the classical $\varepsilon$-$\delta$ definitions of the two notions (for functions $\mathbb{R}\rightarrow \mathbb{R}$) look almost identical. One can define either in terms of the other. So while my tendency (per the above) is to view continuity as the fundamental property that a topology abstracts from a metric space, I'm not really arguing against William Elliot's decision to view limits as the key thing being abstracted. It seems to me one can choose which way one prefers to look at it, and develop a complete theory from either starting point.

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  • $\begingroup$ a topology is the minimum amount of structural information about a space that is used in mathematics to decide whether a function is continuous --- Continuity can be defined for much more general "topological-like" spaces than topological spaces. In fact, one can define continuity for functions from $X$ to $Y,$ where the closure operator on each of these sets is an arbitrary mapping from the power set of that set to the power set of that set. See the bottom of p. 3 of this manuscript. $\endgroup$ – Dave L. Renfro Feb 5 at 16:26
  • $\begingroup$ Lol, I had originally written that sentence with a hedge --- "generally used" instead of just "used" --- to hedge against this exact sort of thing! But I deleted the "generally" b/c I thought it made the sentence unwieldy. Perhaps I'll add it back. $\endgroup$ – Ben Blum-Smith Feb 5 at 18:17
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    $\begingroup$ Two or three decades ago, I would have said that I never fail to be surprised by all the arcane and essentially unknown byways that exist in mathematics (Example 1 AND Example 2 AND Example 3 AND Example 4 AND Example 5 AND Example 6). I've seen too much now for this to still be the case! $\endgroup$ – Dave L. Renfro Feb 5 at 18:39

I think the answer to your question is a qualified "Yes!". A topology is indeed in many ways secondary to the continuous functions that are defined on it and it often convenient to make the continuous functions primary and induce the topology using that.

A great example of this is the Zariski topology on, say, $\mathbb C^n$. The topology is defined in the following way: It is the weakest topology that makes all polynomial functions $f: \mathbb C^n \to \mathbb C$ continuous where we give the target the cofinite topology which corresponds to the intuition that we can tell when the polynomial takes a particular value. So closed sets are precisely generated by zero sets of polynomials.

This idea can be made more sophisticated by talking about sheaves and such things but the fundamental point always is that we care about the functions first, topology second.

In case of the standard topology on Euclidean space, it is perhaps not so clear what the continuous functions should be a priori which is why the answer to your question is a qualified yes. The other way of thinking about closed and open sets is that closed sets are the "answers to questions you can ask about the space".

That is, in the Euclidean topology, we can always know if a particular number lies in say $[0,1]$. We just compute the first few significant digits and check. On the other hand, we can never be sure that a particular number lies in $(0,1)$. Even if the number looks like $0.00000\dots$ with a million leading zeros, the next digit might be non zero. Of course this is superficially related to the representation of the number we are using but any "computable" representation will have this property.

This is related to the continuous functions point of view because the function "n-th digit of a number" is a function we would very much like to be continuous.

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  • $\begingroup$ Your characterization of the Zariski topology is not correct. For example, it would immediately imply that the Zariski topology on $\mathbb{C}^1$ is discrete, which is not true. You could instead say it's the weakest topology that makes all polynomial functions $f\colon \mathbb{C}^n\to \mathbb{C}$ continuous, where we give the target the cofinite topology... But it's probably better just to characterize it as the weakest topology that makes all zero sets of polynomials closed. $\endgroup$ – Alex Kruckman Jan 27 at 15:32
  • $\begingroup$ Yes, you are right of course! I should have said we give the target the topology where singletons are closed, aka the cofinite topology. $\endgroup$ – Asvin Jan 27 at 15:35
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    $\begingroup$ Let f:[0,1] -> N be the n-th digit function. Give [0,1] the topology generated by the base { f$^{-1}$(n) : n in { 0,1,2,3,4,5,6,7,8,9 } }. Behold,, as you wished @Asvin a continuous n-th digit function. $\endgroup$ – William Elliot Jan 27 at 19:50
  • $\begingroup$ Of course the $n$-th digit of a number is not a continuous function on $[0,1]$ nor the reals!, at least in its standard topology. $\endgroup$ – Henno Brandsma Jan 27 at 23:16

Topology is an abstraction of nearness of two points which is used to define limits. whereupon one may define limits of sequences and continuity of functions. In metric spaces the distance function measures neatness. Within topological spaces, descending nests open sets are used.

However there is more to topology than continuous functions such as connected and disconnected spaces, bounded or unbounded spaces, topologically equal spaces. $[0,1]$ and $[0,2]$ are homeomorphic, ie topologically equal. Within metric spaces a space is bounded if its diameter is finite. Within topological spaces a space is bounded if it is compact.

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  • $\begingroup$ Manual line breaks look quite bad on most screens. I edited to remove them. $\endgroup$ – Alex Kruckman Jan 27 at 15:14
  • $\begingroup$ Makes sense to me, but weren't defining continuous functions for sets the main motive for defining topologies? like imagine if we are in some pre-topological mathematical era, I think topologies would come up naturally from trying to define continuous functions between sets right? $\endgroup$ – Ziad H. Muhammad Jan 28 at 12:38
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    $\begingroup$ As continuity @ZiadH.Muhammad, is based upon f(x) getting near to f(a) as x gets near to a, nearness is defined first. Adherence points are an example of being very near. $\endgroup$ – William Elliot Jan 28 at 14:03

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