Squish, Crease, Tear: A Simple Intuition for Open and Continuous Maps in Topology

I think it would be fair to say that many students in the starting lessons of general topology get a bit puzzled by the definition of continuous functions when they first encounter it:

Any function $f : X \rightarrow Y $ is continuous if the pre-image $f^{-1}(A)$ of every open set $A$ in $Y$ is open in $X$.

Doesn't it feel like the definition is backward? Shouldn't a continuous function send open sets to open sets instead to preserve structure? Well, functions which do that are called open functions (commonly open maps).

When we see that continuity and openness are different conditions, some natural questions arise: What do they convey? And how are they different?

Even though we learn counterexamples to clarify that they do not imply each other, in many pedagogical settings, the lesson ends with a stern warning to not get confused between the two without providing any intuition as to why they are so different.

This post is a mostly non-rigorous attempt to do that.

We start with some thoughts on the simple real line and then see some general reasons as to why continuity is not just openness.

Continuity in Real Line

Take the good-old epsilon-delta definition of continuity: a function $f : X \rightarrow Y$ is continuous at a point $c$ if for each $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $$ |x - c| < \delta \implies |f(x) - f(c)| < \epsilon $$ Since this is the first rigorous definition of continuity we learn, let us see what this definition tries to ensure.

It first starts with a $\epsilon$ room around the image $f(c)$ and then says: no matter how small $\epsilon$ one chooses around image of $f(c)$ there must exist a $\delta$ room around $c$ such that every point in it must map inside the $\epsilon$ room.

We can see that this definition would prohibit any jumps. Take the step function which takes the value $-1$ at all $x < 0$ and $1$ at all $x \geq 0$. We can easily set a room $(1 - \epsilon, 1 + \epsilon)$ around $f(0)$ and easily verify that we can not draw any $\delta$ room around $0$ which would map completely inside this target.

The discontinuity occurs because the step function tears the domain at $0$ to map one half of the domain to $-1$ and rest of the domain to $1$. It is akin to tearing a strip of paper which represents the number line and mapping one part to $-1$ and another to $1$. At the risk of losing some rigour, it might be helpful to rely on this analogy as a ladder to develop intuition.

It is common in pedagogical parlance to hear that continuity means "close enough points in domain must map to close enough points in codomain" but we need to be careful about how we formalise this.

Imagine if we mistakenly represent this definition as follows: a function $f : X \rightarrow Y$ is continuous at a point $c$ if for each $\epsilon$ there exists a $\delta$ such that $$ |x - c| < \epsilon \implies |f(x) - f(c)| < \delta $$ This is the same definition as before but we've just switched the $\epsilon$ and $\delta$ in the condition. Now we choose an $\epsilon$ and try to say that if $x$ stays within the $\epsilon$ room then the $f(x)$ must stay in the $\delta$ room. But notice a function does not have to map to every point in the $\delta$ room, this distorted definition does not do anything useful we can always satisfy this definition by taking a big enough $\delta$.

This shows that task of finding a $\delta$ to constrain $x$ so that $f(x)$ is constrained really does tell us something about the function. But finding a $\delta$ to constrain $f(x)$ given that $x$ is constrained is not really a condition at all. This happens because a function by definition must map every point in its domain somewhere but is not obligated to cover any point in the codomain.

Now, if we look closely enough the definition of open map is also very similar to our wrong $\epsilon-\delta$ but even more restrictive. For a map to be open, every open set $A$ must map to some open set $B$ in codomain.

As mentioned before, this looks like a nice condition but note again what it proposes: image of every open set $A$ must be some open set $B$. This means that every point in the open set $B$ must be mapped by some point in $A$.

This has a few unintended consequences which make open maps quite restrictive but at the same time does not ensure continuity in general.

Open is Too Restrictive

If a function $f$ must map every single $A$ to some open map $B$, it must ensure that whatever $B$ happens to, it is covered in the fullest.

This immediately disallows any functions which stay constant in any small interval in their domain. If the function is constant in any interval $(a, b)$, then we can find smaller open interval $(a + \epsilon, b - \epsilon)$ ($\epsilon < b - a$) which would map to a single point in the codomain, which obviously happens to be not open.

But constant functions are as benign a function as possible. It certainly is continuous, but it looks like it can not be open. To borrow the analogy, this really means that for a function to be open, squishing to a point is not allowed, as the function must spread to whole of $B$ for every $A$ whatever open set $B$ happens to be.

Take another example, say, $f = x^2$. Though this is also a well-behaved continuous function on $\mathbb{R}$, it is not an open map. If we take an open interval around $0$, say $(-1, 1)$, we can see that the image of this open interval is $[0, 1)$, which is clearly not open in $\mathbb{R}$. This function is also continuous, but not open.

If we stretch the analogy further, we can see that $x^2$ creases the domain at $0$ in order to create a minima and revert back.

We can see that for a function to be open, both squishing and creasing of the domain are not allowed while mapping into the codomain.

This means that any nice continuous function which attains an extrema and changes direction in its domain ceases to be open. It is a common exercise in analysis to prove that any continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ which happens to be open must be strictly monotonic.

If $f$ is continuous and open on interval $I$ but not strictly monotonic, then there would exist $$ u < v < w $$ in $I$ such that $f(v)$ is not between $f(u)$ and $f(w)$. This means that $f$ would have a local extremum at some interior point in $I$. And we have already seen that we can take a sufficiently small open interval around an extremum which maps to a half-open interval. This would mean $f$ is not open.

So, it is clear that open map is not a useful way to ensure continuity in functions as it disallows many nicely behaved continuous functions.

But let's look at this question from another angle: does the restriction of openness rule out any discontinuities.

It turns out the answer is a yes, but mostly accidently, I think.

It is easy to see that any open map would avoid any simple isolated discontinuities like a step function. Since the step function we supplied above is constant, it is obviously not open. But if we look at a strictly monotonic map with a discontinuity, say

$$ f(x) = \begin{cases} x & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases} $$

We can see that it also fails to be open. We can see that for the interval $(-1, 1)$, the image is $(-1, 0) \cup [1, 2)$ which is clearly not open.

If we push this line further, we can convince ourselves that $f$ can not possibly have isolated discontinuities by which we mean that $f$ can not be discontinuous at any $c$ while still being continuous on some $(c-\epsilon, c)$ and $(c, c+\epsilon)$.

To see why that is true, we can see that $f(c)$ would lie either inside or outside the open images of these two intervals.

It can not lie outside as an isolated point, because then it would sit by itself in the image of $(c-\epsilon, c+\epsilon)$, making that image non-open as we have seen before.

But if it lies inside the image of either side intervals, say the left interval $(c - \epsilon, c)$, we can just find a smaller interval around $c$ to exclude that one point in $(c - \epsilon, c)$ which happened to also map to $f(c)$.

Strict monotonicity ensures that $f$ attains value $f(c)$ only once inside $(c - \epsilon, c)$. But with the smaller interval, $f(c)$ would be back to being an isolated point outside both (shrunk) images and we already ruled that out.

So $f(c)$ can only sit right at the shared edge between the open images which just means $f$ is continuous at $c$.

Open is Possibly Discontinuous

It looks like disallowing a good chunk of functions also disallows simple types of discontinuity. But, openness does not disallow all types of continuities.

We already saw that isolated discontinuities are disallowed if surrounded by continuity.

But the function can still diverge while being open. $\tan(x)$ is a simple example. It diverges at every $x = \frac{(2n + 1) \pi} {2}$ but still happens to be open as any open interval in the domain yields either an open interval or a union of two infinite open intervals in the domain.

So, we have seen that open maps are good at ensuring that the function does not squeeze or crease the domain in ways that introduce boundaries in the images. But open maps are not very helpful in ensuring that the function does not tear up the domain. While simple tearings like isolated jumps are disallowed because they inevitably introduce creases, there are ways to construct open maps which defy continuity in pathological ways.

Since open maps only ensure that open sets map to open sets, if we can somehow manage to map every single open set to the entire real number line, the function does become open but would not be continuous. A famous example is the Conway Base-13 function. It is based on an idea which roughly works like this:

In Conway's construction, the number $x$ is written in Base-13 with the usual Base-10 digits and other symbols like $+$, $-$, and $.$
Now every number $x$ can be mapped to either $0$ or some other real $r$ based on some arbitrary rule.
The rule might look like this: If the number when written in Base-13 had a tail which looks like a valid Base-10 real, then it is mapped to that real. If not, then it is mapped to $0$.
Take some number $x$ which when translated to Base-13 looks like: $$ +-492+1-2+3.14159... $$ then this number would get mappped to $\pi$.
But if the Base-13 expansion never had a tail which looked like a valid real say: $$ +-492+++... $$ it is mapped to $0$.

Now, somewhat strangely, this function does happen to be an open map. If we take any interval $(a, b)$, no matter how small, we can show that its image under the above function is the entire real line $\mathbb{R}$ (which is open in $\mathbb{R}$) Take any arbitrary real $r$, all we need to do is to prefix a valid Base-13 head to produce a number $c$ such that:

$c \in (a, b)$
tail of $c$ when written in Base-13 is equal to $r$ Since head is the most significant part of the number and can be arbitrarily long before the tail begins, we can always find a $c$ which satisfies these. But even though this function is open, it is obviously not continuous.

So, even though openness disallows simple enough tears in the number line because they introduce creases in some image and destroy openness. But if we tear every single point in every single open interval and scatter them like dust particles over the entire number line, then openness is not violated as the whole real number line does not have any boundaries.

Step Out of Reality

So, now we have some idea of what open maps must be like on the real line. But what about a general topology where there are no metrics or intervals? What does continuity and openness mean in a space with vague or no notion of distance or proximity?

Before we go there we must notice that open sets in a general topology do not represent any property unlike on the real line. In an unspecified general context, open sets are just sets which behave in a certain manner vis-a-vis set-theoretic operations. Any collection of sets can be open if they preserve under arbitrary unions and finite intersections. This is how the topology is defined in the first place.

Now, the curious thing about pre-image of a function is that it plays very very well with set-theoretic operations. For any function $f : X \rightarrow Y$ and any $A$ and $B$ in $Y$ $$ f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B) \\ f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B) \\ f^{-1}(Y - A) = X - f^{-1}(A) $$

Regardless of the function, pre-images faithfully preserve unions and intersections. This is largely due of the fact that a function by definition must map any point in the domain to a unique point in the codomain.

Note from the above equations that the set of pre-images in $X$ of all open sets in $Y$ of $f : X \rightarrow Y$, already fulfills the condition of being a topology on $X$. This means that even before we define continuity or openness, pre-images automatically induce a topology on $X$ just by the virtue of their properties.

Said differently, if we consider the sets of points which got transformed into open sets in the codomain $Y$, we already know that they would be a valid topology. Continuity only requires that they must have come from the topology already endowed in $X$ itself.

This begins to shine some light as to what continuity might be trying to portray. It shows that whatever structure $Y$ has in terms of open sets, if we try to pull back the structure along the function to $X$, we just need to ensure that a similar or finer structure is available in $X$.

But when we talk about images, things are not quite smooth. We started by thinking that images might propagate or push the structure from $X$ to $Y$, but alas that is not at all true. Images simply don't play well with set-theoretic constructions. Unions still work, but intersections don't work well. $f(A \cap B)$ is not necessarily equal to $f(A) \cap f(B)$ because functions can map multiple points in the domain to the same point in the codomain. Also $f(X - A)$ is not equal to $Y - f(A)$ in general.

So, even if we could ensure that open sets propagate from $X$ to $Y$ via images of open maps, we would never be able to ensure that they induce a topology on $Y$ from the topology of $X$ itself.

This is the reason that imposing conditions on the image of a function does not afford us any special behaviour when it comes to ensuring continuity.

If we try to round back to the analogy of tearing which continuity is supposed to disallow, we see that pre-images are the right choice. If we take a simple step function on $\mathbb{R}$ which jumps on some $c$ in domain to $f(c)$ in range, then the pre-image of any sufficiently small open interval $(f(c) - a, f(c) + a)$ would obviously be outside the usual topology of the domain $\mathbb{R}$ (as it is the half-closed ray $[c, \infty)$) This just means that the induced topology from all the pre-images of the step function contains at least one element different from the usual topology on $\mathbb{R}$. This makes the pulled-back set fundamentally alien to the original topology of $\mathbb{R}$. This is exactly the tear in the codomain which creates a set-theoretic boundary in the pre-image that the original domain topology never possessed. It is detected when the induced topology in the domain is found to be different and irreconcilable from the endowed topology.

Two Mindsets

When we first study topology, our natural impulse is to think of functions as transformations which push the structure of a space forward onto another. That is why we might think requiring open sets to map to open sets feels like the only logical way to preserve structure.

But as we see, open images as a condition is neither necessary nor sufficient and not even justified in general. It either disallows benign smooth behaviour like local extrema or constancy but at the same time allow pathological maps to tear and scatter open sets everywhere.

Continuity works precisely under a pullback mindset. Instead of trying to propagate the structure of the domain onto the codomain, we ask: does the domain possess enough structural resolution to make sense of the open sets in the codomain? If the answer is yes, the function is continuous. This is why if we have the finest topology in the domain or the coarsest topology in the codomain, all functions become continuous.

But this should not mean that open maps are useless in total. If a function which is both continuous and bijective (one-one onto) also happens to be open then its inverse becomes continuous which allows us to mark two topological spaces to be structurally identical (homeomorphism).

But taken separately, open and continuous maps measure two entirely different features of a transformation:

Continuity prevents tearing by ensuring the structure of codomain can be pulled back into the topology of the domain. Openness prevents squishing and creasing by ensuring the room of the domain is distributed adequately across the image.