Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $Y$ be topological spaces. A function $f: X \rightarrow Y$ is defined as continuous if for each open set $U \subset Y$, $f^{-1}(U)$ is open in $X$. This definition makes sense to me when $X$ and $Y$ are metric spaces- it is equivalent to the usual $\epsilon-\delta$ definition. But why is this a good definition when $X$ and $Y$ are not metric spaces? How should we think about this definition intuitively?

share|cite|improve this question
I added the tag intuition, seems fitting. – Asaf Karagila Dec 31 '10 at 15:44
up vote 16 down vote accepted

One abstract way to think about continuity (in the sense that it generalizes to non-metric spaces) is that it is about error. A function $f : X \to Y$ is continuous at $x$ precisely when $f(x)$ can be "effectively measured" in the sense that, by measuring $x$ closely enough, we can measure $f(x)$ to any desired precision. (In other words, the error in our measurement of $f(x)$ can be controlled. "Precision" here means "to within an arbitrary neighborhood of $f(x)$," so it does not depend on any metric notions.) This is an abstract formulation of one of the most basic assumptions of science: that (most of) the quantities we try to measure ($f(x)$) depend continuously on the parameters of our experiments ($x$). If they didn't, science would be effectively impossible.

If you like thinking about limits, a function is continuous if and only if it preserves limits of filters or, equivalently, nets. These are two ways to generalize converge of sequences to spaces which are not first-countable.

share|cite|improve this answer
If you don't want to have to learn all the machinery of nets and filters, you can also think about topological spaces in terms of the Kuratowski closure axioms: . It is instructive to prove that these axioms are equivalent to the usual ones. – Qiaochu Yuan Dec 31 '10 at 16:41

Maybe it's just me, but I've never thought that the usual $\epsilon$-$\delta$ definition of continuity is intuitive at all. Why should a function be continuous at $x$ if every ball of radius $\epsilon$ around $f(x)$ contains the image under $f$ of a ball of radius $\delta$ around $x$?

Instead, in metric spaces, I think of a function as continuous if it preserves limits, which can be intuitively (and generalizably) be phrased by saying that $f$ is continuous if and only if whenever $x$ is in the closure of a set $A$, then $f(x)$ is in the closure of the set $f(A)$.

(Take a piece of paper and draw out the arguments of the next two paragraphs)

To see that the $\epsilon$-$\delta$ continuity implies 'closure' continuity, suppose that $f$ is not closure continuous at $x$, that is $x$ is in the closure of some set $A$ but $f(x)$ is not in the closure of $f(A)$. Then there exists an $\epsilon$-ball around $f(x)$ that does not intersect $f(A)$ even though every $\delta$-ball intersects $A$. Hence, some $\epsilon$-ball around $f(x)$ contains no image of a $\delta$-ball around $x$, and so $f$ is also not $\epsilon$-$\delta$ continuous.

To see that 'closure' continuity implies $\epsilon$-$\delta$ continuity, suppose that $f$ is not $\epsilon$-$\delta$ continuous. Then there exists an $\epsilon$-ball around $f(x)$ that contains no image of a $\delta$-ball around $x$. In other words, the preimage of the $\epsilon$-ball around $f(x)$ contains no $\delta$-ball around $x$, so let $A$ be the collection of points that are not in the preimage of the $\epsilon$-ball. Then $x$ is in the closure of $A$ since any $\delta$-ball around $x$ has a point outside the preimage of the $\epsilon$-ball and hence in $A$, but $f(x)$ is not in the closure of $f(A)$ since the $\epsilon$-ball around $f(x)$ is disjoint from $f(A)$.

Now, the cool thing to notice is that the above equivalence of definitions works perfectly fine if you replace $\delta$-balls and $\epsilon$-balls with open sets in the appropriate topological spaces, so really what you should care about is how to make sense of 'closure' continuity in a space that is not a metric space, and the answer is given by the wiki:Kuratowski Closure Axioms.

You might also find useful the answers to this mathoverflow question, specifically this one by sigfpe and this one by Vectornaught. The first one talks about how open sets can be thought of as rulers that try to measure imprecisely things in the vector space (but which doesn't explain why continuity it as it is)), while the second phrases the Kuratwoski Closure Axioms in terms of the intuitive notion of 'nearness' of points (which does account for continuity).

share|cite|improve this answer

You have it exactly right. It works well when $X$ and $Y$ are metric spaces, and has proved useful in more general contexts. When you think of open sets as points "near" another, this is the proper translation of the usual $\epsilon-\delta$ definition.

share|cite|improve this answer

It's true that this definition generalizes that for metric spaces, but there are other generalizable definitions (e.g. takes convergent sequences to convergent sequences), and perhaps implicit in the OP's question is: why this particular definition?

Of course part of the answer is that it turns out to work well, but this is not too satisfying. The following are just a couple of things that just occurred to me.

More generally, given a class of mathematical objects like a topological space (vector space, group, ring, etc.) it is natural to ask: what are the "structure-preserving maps" between such objects? Vector spaces are sets equipped with a scalar multiplication map; groups are sets equipped with a group operation, etc. and the notions of linear map and homomorphism are precisely defined to preserve this structure.

Now with a topological space, of course, the structure comes as a set of "open sets." Here the generalization from concepts of metric spaces is especially clear. Therefore, a continuous map, a structure-preserving map on a topological space, should be one that "preserves open sets." At first you might think that such a map should take open sets to open sets (i.e. an open map), but examining the conditions on open sets shows that this is bad. The point is that if $f$ is a set map, then $f^{-1}$ is actually much "nicer" than $f$ in terms of how it commutes with unions, intersections, etc.

In fact, I might make the following observation. A structure-preserving map from $X$ to a set $Y$ should prescribe a natural structure for $Y$.

share|cite|improve this answer
If you want to think in these terms, then the structure that continuous maps preserve is the closure operator: – Qiaochu Yuan Dec 31 '10 at 16:42

Suppose $x$ is a point in some topological space, then we can define $\mathcal{N}_x$ to be the set of (not necessarily open) neighbourhoods of $x$. Then "$f$ is continuous at $x$" is defined as

$$\forall V \in \mathcal{N}_{f(x)} \exists U \in \mathcal{N}_x : f(U) \subseteq V.$$

Or more colourfully: Whenever "the enemy" comes with a cleverly chosen and "small" neighbourhood of $f(x)$, we must be able to find a neighbourhood of $x$ that maps into said neighbourhood.

What is nice about this is that it is (obviously?) the topological version of the $\varepsilon$-$\delta$ definition from $\mathbb{R}$ and metric spaces that we know (and love(?)), and it's relatively easy to prove that $f: X \to Y$ is continuous at $x$ for all $x \in X$ if and only if $f^{-1}(U)$ is open in $X$ for all open $U \subseteq Y$.

What I'm trying to say is that I don't have much intuition for the "preimage of open sets is open" definition either, but it's not clear to me that you really need that. We take this as the definition because it's simple, it's entirely written in terms of the topologies of $X$ and $Y$ (i.e. the collections of their respective open sets) and easily shown to be equivalent to something which we do have an intuition about (assuming that one finds $\varepsilon$-$\delta$ intuitive, obviously).

share|cite|improve this answer

Let me also add one little bit (perhaps this may even seem backwards).

Topology is defined using sets satisfying some set of axioms, that we call Open sets. However, the collection of subsets we choose (respecting some set of axioms) to call a Topology on the space may vary with the underlying set. But, we can go from one such collection to another using a special class of maps. But, how do we define such maps?

Now, properties that are to be 'intrinsic' to the Topology should not depend on this choice, so should be invariant across these maps. But, these properties (whatever they be) are defined via set operations.

Perhaps we want the assignment to be direct: i.e. $U$ is open and thus $f(U)$ should be open. But, notice that since the map is general: $f(A \cap B)$ is contained and not equal to $f(A) \cap f(B)$ etc. However, with the assignment $f^{-1}$ (which need not even be well-defined) all is fine with unions and intersections, i.e., we have equalities, so, the properties follow through almost trivially (see for example the proofs that continuous maps preserve compactness, connectedness etc.)

share|cite|improve this answer
It is somewhat missing the point to think about open sets as a structure that we assign to a set in the same way that a multiplication operation is a structure we assign to a set to make it a monoid. The thing we're assigning structure to is the lattice of subsets of the set, and so the reason f^{-1} is relevant is that the natural functor from the category of sets to the category of Boolean algebras is contravariant. In fact the category of sets contravariantly embeds as a certain subcategory, and from there it makes perfect sense to think about inverse images. – Qiaochu Yuan Dec 31 '10 at 16:47

Part of the problem is that courses in analysis are not presented geometrically enough. If you present a few pictures of continous and non continuous (partial) functions $f: \mathbb R \to \mathbb R$ and see how the condition $f(M) \subseteq N$ works out for neighbourhoods $M$ of $x$, $N$ of $f(x)$, then you begin to see how the definition works out. What can be confusing is that $\epsilon$ and $\delta$ are measurements of the sizes of nieghbourhoods rather than the actual neighbourhoods themselves, and so one step further away from intuition. Of course for many calculations you do need the sizes, as well as the understanding.

share|cite|improve this answer
I think these intuitions are important but do not give an honest picture of what a general topological space looks like (not necessarily metrizable, not necessarily even Hausdorff, etc.). The general picture is to me less about geometry and more about... I don't have a good word here. See, for example, Dan Piponi and Justin Hilburn's answers here:… – Qiaochu Yuan Aug 10 '12 at 14:17
An answer on this involves the character of the questioner: one gives a different answer to an undergraduate than to a second year postgraduate. There is a difference between how a subject starts and how it develops. See also the discussion by Grothendieck in Section 5 of "Esquisse d'un Programme", where he is seeking definitions more closely related to geometry, rather than analysis, and so with more structure. Also I would naturally prefer to talk about the character and convenience of the category of spaces and continuous maps; but that is quite sophisticated. – Ronnie Brown Oct 5 '12 at 15:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.