Can a Continuous Function Map Reals to Rationals?

[JasonRox]

Can there exist a continuous function from the real numbers to the rationals?

Where we are considering the usually topology for the real numbers and the relative topology for the rationals with respect to the topology of real numbers.

At first my intuition said no. Quickly going through random functions in my head, I couldn't create one.

Is my intuition right?

I've been thinking about a contradiction if a continuous function did exist, but can't find one that seals the deal.

Can someone atleast confirm or correct my intuition without giving a solution?

Hopefully I get a solution before tomorrow night. I'll have time at work to think about it some more, and try to use different approaches. I'll try looking at invariants and what not.

 

[AKG]

I suppose you mean a continuous function from the reals onto the rationals. The answer is "no." What are some important topological properties that the reals have and that the rationals don't?

 

[JasonRox]

Yeah, I got it using the limit point definition of continuity.

I'm reading two different books where one uses open sets as the main focus and the other using limit points.

I used the idea that atleast one of the points in the rationals will have an inverse image of an uncountable number of elements in R. It must exist. And since that one point is closed, then the inverse image must also be closed by definition of continuity. Then I went from there.

 

[AKG]

Where did you go from there? I was thinking about using connectedness.

 

[matt grime]

Yep, me too, connectedness seemed to be the important thing, nay the only thing that sprang to mind, and I too would like to see the sequential proof.

 

[JasonRox]

Here is the proof, if it's valid.

Now assume a continuous function does exist.

We know the inverse image of an element in Q (call it p) will map to an uncountable number of elements.

Since f is continuous and p is closed, then f^-1(p) is closed. Now, f^-1(p) has uncountable many elements, and it is closed (call it set S in R). Now, we know that S will contain a limit point call it x. By definition of continuity, if x is a limit point of S, then f(x) is a limit point of f. But f(x)=p and f={p}, but p isn't a limit point to the set {p}, which contradicts the fact that f is continuous.

Note: If it's not making any sense, let me know. I'll write it more formally.

It could be false, but it seems fine.

Note: I guess what is left to prove is that a set with uncountable many elements in R has atleast one limit point. Now, if the set (call it S) did not have a limit point, then for some interval for every point in S, there is a distance e such that an interval about that point where it does not contain a point in S. Take the minimum e amongst all the elements. So, all the elements are separated by atleast that much. That makes it countable now, contradicting the fact that it was uncountable.

 

[cogito²]

Hey Jason about your proof that there is a limit point. You must take into account that the points may repeat. In fact you may have points repeating for an entire sequence without having limit points. For example

1,1,2,2,3,3,4,4,...

Then your distance between points would be 0, but you'd have no limit point. Also you wouldn't be able to throw away any finite number of points to make your process work.

What you're trying to prove though is still true. Remember all you need to prove is that there is some bounded subset of the real numbers which contains an infinite (countable suffices) number of elements from f^-1(p). Consider if there is no such bounded subset (ie ever bounded subset contains a finite number of elements from f^-1(p)) and what that would mean about f^-1(p).

 

[matt grime]

Here is the proof, if it's valid.

Sorry it's not. In fact your proof indicates that there can never be a continuous map from the reals to anyone point set, which surely you can see is wrong.

Now assume a continuous function does exist.

this is unnecessary, just let f be a map and show f cannot be continuous, there is no need to go for a contradiction here.

We know the inverse image of an element in Q (call it p) will map to an uncountable number of elements.

Since f is continuous and p is closed, then f^-1(p) is closed. Now, f^-1(p) has uncountable many elements, and it is closed (call it set S in R). Now, we know that S will contain a limit point call it x. By definition of continuity, if x is a limit point of S, then f(x) is a limit point of f. But f(x)=p and f={p}, but p isn't a limit point to the set {p}

p is a limit point, the only one, of {p}, sequences are allowed to be constant. I think you're confusing this with subsequences where you can't pick the same element of the sequence repeatedly, ie x_1,x_1,x_1,... is not a subsequence of x_1,x_2,x_3,...Note also how you say '{p} is closed' and then that {p} has no limit points. I thought your preferred definition of closed was that it contained all its limit points.

 

[JasonRox]

p is a limit point, the only one, of {p}, sequences are allowed to be constant. I think you're confusing this with subsequences where you can't pick the same element of the sequence repeatedly, ie x_1,x_1,x_1,... is not a subsequence of x_1,x_2,x_3,...

Yeah, big mistake.

Something else now. Using connectedness mentionned earlier.

By construction of the topology of the rationals, it will contain open and closed sets. This is what I "see".

Let a and b be irrational.

The interval $(a,b) \cap Q$ is an open set in Q call it S.

The interval $[a,b] \cap Q$ is a closed set in Q, which is precisely the set S from above. Therefore, S is closed and open.

Is that right?

That means Q is not connected.

 

[JasonRox]

Hey Jason about your proof that there is a limit point. You must take into account that the points may repeat. In fact you may have points repeating for an entire sequence without having limit points. For example

1,1,2,2,3,3,4,4,...

Then your distance between points would be 0, but you'd have no limit point. Also you wouldn't be able to throw away any finite number of points to make your process work.

I'm not getting what your saying here.

 

[matt grime]

Q is not connected because, well, it isn't. The open sets S={ s in Q : s<sqrt(2)} and T={t in Q :t> sqrt(2)} disconnect it. As do uncountably many other choices of irrational number.

 

[JasonRox]

Q is not connected because, well, it isn't. The open sets S={ s in Q : s<sqrt(2)} and T={t in Q :t> sqrt(2)} disconnect it. As do uncountably many other choices of irrational number.

Yes, but is my example valid?

 

[matt grime]

What do you think? Have you produced two open non-empty disjoint sets?

 

[cogito²]

I'm not getting what your saying here.

You said to take "e" as being the minimum width of the intervals. I'm saying that that distance can be 0 even without having a limit point. Then you can't use that to conclude it's countable.

 

[matt grime]

One more thing: you can't take the 'minimum of the e' since that will not exist. the inf will, but it is not guaranteed to be any of the e and there is nothing to suppose that the inf is not zero.

 

[JasonRox]

What do you think? Have you produced two open non-empty disjoint sets?

Finding an open and closed set within the set other than Q and the null set is basically finding two open sets that are disjoint.

So, I would say yes I did find two open sets that are disjoint.

 

[JasonRox]

One more thing: you can't take the 'minimum of the e' since that will not exist. the inf will, but it is not guaranteed to be any of the e and there is nothing to suppose that the inf is not zero.

I see what's going on now.

Good thing I posted the question and "proof".

 

[JasonRox]

p is a limit point, the only one, of {p}, sequences are allowed to be constant.

At first this got me confused, but in fact I am right.

{p} does not have a limit point.

Sequences can be constant, but can not "finish" as a contant of the point p.

An equivalent statement is that every neighbourhood of p has a point in the set A pther than p. For the set {p}, that is impossible.

http://mathworld.wolfram.com/LimitPoint.html

 

[JasonRox]

I almost forgot.

As an added note, any finite subset of a T1 space does not contain any limit points.

 

[mathwonk]

do you know the intermediate value theoirem? it syas that the continuous image of any intervalk is another interval. so a continuous map from the interval of all reals to the rsati0onals would have image an interval of the reals which contains only rationals. what are msuch intervals?

 

[matt grime]

At first this got me confused, but in fact I am right.

{p} does not have a limit point.

Sequences can be constant, but can not "finish" as a contant of the point p.

An equivalent statement is that every neighbourhood of p has a point in the set A pther than p. For the set {p}, that is impossible.

http://mathworld.wolfram.com/LimitPoint.html

yes, my mistake. and it makes me wonder what the point of defining limit points like that is. i don't know where i got the slogan: a set is closed if and only if it contains all its limit points.

They also can't be much good for proving things about continuous functions. There is obvisouly a continuous function from R to {p} so the limit point stuff not preserved.

 

[JasonRox]

They also can't be much good for proving things about continuous functions. There is obvisouly a continuous function from R to {p} so the limit point stuff not preserved.

That's what is interesting.

I'll give it some thought to what happens to the limit points.

 

[JasonRox]

They also can't be much good for proving things about continuous functions. There is obvisouly a continuous function from R to {p} so the limit point stuff not preserved.

It is preserved if f is a homeomorphism.

Therefore, that was my mistake.

 

[mathwonk]

havew you read my post? it reduces the whole question to the IVT of calc I.

 

[JasonRox]

do you know the intermediate value theoirem? it syas that the continuous image of any intervalk is another interval. so a continuous map from the interval of all reals to the rsati0onals would have image an interval of the reals which contains only rationals. what are msuch intervals?

There doesn't exist such an interval. Unless you count things like [0,0] as an interval.

I'm assuming the Intermediate Value Theorem applies to all metric spaces.

Can we create such a topology on the rationals such that a continuous function exists?

Sure we can point out the obvious and point to the Indiscrete Topology, but what about a metric topology on the rationals? This seems like an interesting thing to think about. Of course, we'd have to construct it so that it is connected along with other continuous invariants.

Note: After re-reading my posts, it seems unlikely a metric topology for the rationals exists so that f is continuous from the reals to the rationals. I'll give it some more thought, and if I think its true, I'll try and prove it.

 

[AKG]

The point is that there does exist such an interval, and it is the interval [x,x], so if mathwonk's argument is correct, then the only continuous function is the constant function.

The general Intermediate Value Theorem in my book says:

Let f : X -> Y be a continuous map, where X is connected and Y is an ordered set in the order topology. If a and b are two points in X, and r is a point in Y between f(a) and f(b), then there is a point c of X such that f(c) = r.

----------------

Now, you want to look at the rationals, but give them a new topology that isn't indiscrete. You also want this space to be metrizable (i.e. this topology should be inducible from a metric). So we're just looking at any old denumerable set, and now we have to figure out a metric on it.

Look at the natural numbers with the co-finite topology (that is, the only open sets are the empty set, and those sets with finite complment). This space is connected, and T1, but not Hausdorff, so it's not metrizable. Perhaps we can prove that a denumerable Hausdorff space is separable. Actually, we can prove that a denumerable metric space is separable:

First, consider the following argument: Let p be a point in the plane, and suppose you scatter denumerably many points in any way you please about the plane. You can prove that there is a circle centered at p which contains none of these points. Why? Because there are uncountably many circles (one for each choice of radius, and we can choose any positive real number for our radius), and only countably many points, so there must be some circle that contains none of these points. Note that by circle, I mean a circle like S¹, not a disk. The stuff inside the circle is an open set, and the stuff outside the circle is an open set, so regarding our scattered points as a subspace of the plane, it is disconnected. Now, unless I'm overlooking something, this argument can be generalized to any denumerable metric space.

 

[mathwonk]

yes, that's the proof. i.e. by my argumenmt, the only possible image for a continuous function from reals to rationals is a trivial interval of form [a,a], i.e. such a function is a constant map.

 

[JasonRox]

yes, that's the proof. i.e. by my argumenmt, the only possible image for a continuous function from reals to rationals is a trivial interval of form [a,a], i.e. such a function is a constant map.

That's essentially what I implied when I said...

Unless you count things like [0,0] as an interval.

For AKG, I can't answer that question until I've gave it some thought.

 

[AKG]

What question?

 

[JasonRox]

What question?

My own question that I asked earlier. You seem to have went on about a discussion about a metric topology for the rationals to create a continuous function from the reals to the rationals.

I'll give it some thought before I go on to read your discussion, whether or not it has a solution.

 

[mathwonk]

theorem: every continuous function from the reals to the rationals is constant.
proof the image of the map is an interval containing only rationals, hence of form [c,c].

 

[JasonRox]

theorem: every continuous function from the reals to the rationals is constant.
proof the image of the map is an interval containing only rationals, hence of form [c,c].

Yes, but doesn't this theorem only hold for the rationals with the relative topology with respect to the real numbers with the usual topology?

If you give the rationals the indiscrete topology, isn't every function continuous?

 

[DeadWolfe]

Yes, but of course, when asking if there is a continus map between sets X and Y we always need to think of the given topologies on sets, since the question is ill-defined otherwise.

 

[JasonRox]

Yes, but of course, when asking if there is a continus map between sets X and Y we always need to think of the given topologies on sets, since the question is ill-defined otherwise.

Well yeah, but my question a few posts ago asks about a possible metric topology on the rationals such that you can create a continuous function from the reals (usual topology) to the rationals (with new metric topology).

Haven't gave much thought yet. I should write a list of things to think about.

 

[matt grime]

Of course there are continuous functions from the Reals to the rationals, as has been repeatedly pointed out (constant maps). Perhaps you mean continuous *surjections*, or even just non-constant continuous maps.

Yes if f:X\to Y is map either from the discrete or to the indiscrete topology it is trivially continuous.

Mind you, what metric d'ya think gives the indsicrete topology?