Is f' continuous when removing elements from X and Y?

[radou]

Here's something that's bothering me a bit.

Let f : X --> Y be a continuous function, where X and Y are topological spaces.

i) is f' : X\{a} --> Y\{f(a)} continuous? (a is an element of X)
ii) if A is a countable subset of X, is f' : X\A --> Y\f(A) continuous?

 

[Landau]

The following definition might be useful for these things; it expresses the fact that continuity is a local property.

Definition: Let f:X\to Y be a set function, and x in X. We say f is continuous at x if for every neighbourhood V of f(x) in Y, there exists a neighbourhood U of x in X such that f(U) is contained in V.

Exercise: prove that f:X\to Y is continuous if and only if f is continuous at x, for all x in X. In calculus, this is the usual way of defining continuity (first define it at a point, and then say a function is continuous if it is so at every point.)

With this result your questions become trivial :)

Or you could use other characterizations of continuity, e.g. f is continuous iff it preserves converging nets.

 

[radou]

Yes, this is just what I thought it is, only, for some reason, something seemed suspicious about it. :)

Thanks for the replies.

 

[radou]

Actually, I needed this to prove that if A is a countable subset of R^2, then R^2 is path connected.

Since R^2 is path connected, for any x, y in R^2 there exists f : [0, 1] --> R^2 such that f(0) = x and f(1) = y. If we now consider R^2\A, and if some points in the image set of our "path functions" happen to be in A, which we removed from R^2, then f is still continuous.

 

[Landau]

In particular, you don't need to restrict to countable sets in the second part of the question.

Actually, I needed this to prove that if A is a countable subset of R^2, then R^2 is path connected.

You mean 'then R^2\A is path connected'?

 

[radou]

You mean 'then R^2\A is path connected'?

Yes, it was a mistype.

In particular, you don't need to restrict to countable sets in the second part of the question.

OK, this was an exercise from the book I'm working with. I assume what you wrote holds because what we concluded in the posts above, i.e. if f :X --> Y is continuous, and if we remove any proper subset of the domain of f, and its corresponding image set, then f is continuous.

 

[Jamma]

I don't understand your proof that R^2 \A is still path connected.

Nowhere in your proof did you use that A is countable, and obviously it doesn't work if you don't impose this; you could cut out a circle from the plane for example.

Your original assertion in post 1 is correct, but the problem with your proof is that, yes, the function is still continuous, but it is no longer a path, since you have removed points from the domain. You'd need to extend your map f: [0,1]\{f^-1(A)} --> R^2\A to a map f':[0,1] -->R^2\A for it to still be a path.

 

[Landau]

Intuitively the result is pretty clear. If we remove one point, it is totally obvious on a picture: you can just move around that point in two straight lines. In the same way, you can move around any finite number of points by 'zigzagging'. For a countable set of points it is a bit harder. It is still a discrete set, but if you remove e.g. $A:=\{(1/n,0):\ n\in\mathbb{N}\}$, the point accumulate at the origin. Still, it looks doable, but I'd have to think a bit for a formal proof.

 

[Jamma]

Well, your example doesn't look anywhere near as dodgy as to removing all of the rational points.

 

[Landau]

Hehe, true. Here is a proof. This proof even looks too simple to be true.

 

[radou]

I don't understand your proof that R^2 \A is still path connected.

Nowhere in your proof did you use that A is countable, and obviously it doesn't work if you don't impose this; you could cut out a circle from the plane for example.

Your original assertion in post 1 is correct, but the problem with your proof is that, yes, the function is still continuous, but it is no longer a path, since you have removed points from the domain. You'd need to extend your map f: [0,1]\{f^-1(A)} --> R^2\A to a map f':[0,1] -->R^2\A for it to still be a path.

Right, my fault. If I remove some points from a closed interval, obviously it isn't a closed interval any more, and hence we have no path (by definition) any more. :)

Thanks, I'll look at the later replies tomorrow in order to solve this.

 

[Jamma]

Yep, that 2nd proof seems fine to me (the first one didn't have a working link), it really is that simple, pretty much not topology involved =D

 

[Jamma]

For a countable set of points it is a bit harder. It is still a discrete set

Oh, just realized that you need to be a bit careful here. Q^2 is a countable subset of R^2 but it isn't discrete.

 

[Landau]

Yep, that 2nd proof seems fine to me (the first one didn't have a working link), it really is that simple, pretty much not topology involved =D

Sorry, the first one should have been http://planetmath.org/encyclopedia/MathbbR2SetminusCIsPathConnectedIfCIsCountable.html .

Oh, just realized that you need to be a bit careful here. Q^2 is a countable subset of R^2 but it isn't discrete.

Ah, you're right, thanks for the correction.

 

[Office_Shredder]

I think you could even construct an explicit new path from x to y. If your current path hits A, find the first point of A (based on whatever arbitrary enumeration of A you have) that lies on the path. You can construct a little semicircle around that point which doesn't intersect A at the point in question, or any new points. Repeat ad nauseum

 

[Jamma]

Yes, I think that is correct.

I'm pretty sure that you can even construct a smooth path; I can't be bothered to work out the general forumula, but the equation

c(x-a)^2+d(y-b)^2=r

for c=d=1 defines a sort of smooth ellipse. You can move this ellipse so that it ends up passing on the desired points. Then you can vary c,d and r so that this still holds for uncountably many values (I was thinking connecting (0,-1) to (0,1). You can start with x^2+y^2=1 but also (1/a)x^2+y^2=1 by varying a>1 for uncountably many values. A rotation and/or reflection of this situation works for any two points). Then one of them doesn't hit A.

 

[radou]

Here's another idea of how to prove it:

If X is locally path connected, then every connected open set in X is path connected. (This was an exercise I solved.)

R^2 is locally path connected, since for any x in R^2, and for any neighborhood U of x, there exists an open ball B(x, ε) contained in U, and open balls are path connected.

R^2\A is open, since for any x in R^2\A one can find an open ball B(x, ε) contained in R^2\A. It is connected too (I'm not so sure about this, but it seems obvious on an intuitive level, which may not be good enough), and hence path connected.

 

[Landau]

R^2\A is open, since for any x in R^2\A one can find an open ball B(x, ε) contained in R^2\A.

What about the earlier mentioned A=Q^2?

It is connected too (I'm not so sure about this, but it seems obvious on an intuitive level, which may not be good enough), and hence path connected.

This seems like a big non-trivial step in this proof. I don't see how it would be more obvious that it is connected than that it is path-connectedness.

 

[radou]

What about the earlier mentioned A=Q^2?

Oh, I oversaw this, sorry. What I wrote down would work if A was a discrete set, right?

This seems like a big non-trivial step in this proof. I don't see how it would be more obvious that it is connected than that it is path-connectedness.

Yes, that's what I feared.

OK, I'll stick to the proof you gave on the second link in post #10. Actually, it's overwhelmingly simple. :)

By the way, R^2\{a}, where a is in R^2, is connected, right? I used this fact in some other proof a while ago, but I didn't prove it. Frankly, I don't have an idea of how to prove it. It seems kind of obvious, though, but that doesn't mean anything.

 

[Landau]

By the way, R^2\{a}, where a is in R^2, is connected, right? I used this fact in some other proof a while ago, but I didn't prove it. Frankly, I don't have an idea of how to prove it. It seems kind of obvious, though, but that doesn't mean anything.

Well, here it is again the case that it is obviously path-connected. Take any two points, and you immediately see how to draw a path, just make sure it doesn't go through a :P

 

[radou]

Well, here it is again the case that it is obviously path-connected. Take any two points, and you immediately see how to draw a path, just make sure it doesn't go through a :P

OK, clearly if it's path connected, it's connected, but I meant how we could prove that it's connected without knowing it's path connected and that path connectedness implies connectedness. :)

 

[Landau]

Why would you want to? Is suppose you could try to write R^2-{a} as the union of two disjoint open subset, and get a contradiction.

 

[radou]

Why would you want to? Is suppose you could try to write R^2-{a} as the union of two disjoint open subset, and get a contradiction.

Yes, it occurred to me, but I didn't manage to, that's why I'm asking. Nevermind, this is off topic anyways.