Yet another counterexample challenge

[mfb]

Then I just see a few examples, and nearly everything should be a counterexample?
Try to find a countable set of disjoint balls that covers R^2. Take an arbitrary ball and consider its boundary: it won't be part of any ball.

 

[fresh_42]

Then I just see a few examples, and nearly everything should be a counterexample?
Try to find a countable set of disjoint balls that covers R^2. Take an arbitrary ball and consider its border: it won't be part of any ball.

The point is that there must not be any union by countably many balls of an open set. The idea is clear, it cannot be done. But why? I mean a sketch shows why, but with countably many balls available it's not clear how. IMO one has to get rid of the infinity somehow or apply a diagonal counting argument. (With closed balls it's obvious but with open?)

 

[mfb]

The point is that there must not be any union by countably many balls of an open set.

Where did the disjoint go? That is the part that confused me.

 

[micromass]

Then I just see a few examples, and nearly everything should be a counterexample?
Try to find a countable set of disjoint balls that covers R^2. Take an arbitrary ball and consider its border: it won't be part of any ball.

Yes, the way I worded it makes it trivial to disprove. I'll just accept your answer, and leave the actual intended problem for a next thread.

 

[fresh_42]

Where did the disjoint go? That is the part that confused me.

That's part of the problem. Beneath open balls you can always place another open ball in between such that they are still countable and disjoint, i.e. no point is part of two balls but exactly in one ball.

 

[fresh_42]

Yes, the way I worded it makes it trivial to disprove. I'll just accept your answer, and leave the actual intended problem for a next thread.

If I may make a suggestion: I think it might be interesting to gather examples of well-known theorems that fail to hold in prime characteristic. I've seen those now and then but never payed a lot of attention to $\mathbb{F_p}$.

 

[mfb]

That's part of the problem. Beneath open balls you can always place another open ball in between such that they are still countable and disjoint, i.e. no point is part of two balls but exactly in one ball.

Not in such a way that it covers the boundary of one of them. If a boundary point would be part of a different ball, it would need an open set around it that is still in the ball, but that would overlap with the other ball.

 

[micromass]

If I may make a suggestion: I think it might be interesting to gather examples of well-known theorems that fail to hold in prime characteristic. I've seen those now and then but never payed a lot of attention to $\mathbb{F_p}$.

The idea of this (and the previous) counterexample thread was to provide a list of counterexamples whose statement most lower-level students (like calculus students) could understand. In the future though, I will definitely put up counterexample challenges with more advanced mathematical notions, but I chose not to do it now.

 

[fresh_42]

Not in such a way that it covers the boundary of one of them. If the boundary point would be part of a set, it would need an open set around it that is still in the ball, but that would overlap with the other ball.

Yes, but the ANY in the question is meant as: there is NO union of ... In the boundary case you just might have the wrong cover. Of course balls cannot cover squares, but ℕ many of them?

 

[mfb]

Yes, but the ANY in the question is meant as: there is NO union of ... In the boundary case you just might have the wrong cover. Of course balls cannot cover squares, but $N$ many of them?

I showed that there is no such union. I started with "take an arbitrary set of balls that cover R^2" and lead that to a contradiction. Every such cover will have a ball, and that ball has an uncovered boundary. This is true for every attempt to cover R^2.

To generalize, you cannot cover R^2 with any union (even an uncountable one) of disjoint open sets (don't have to be balls), unless you cover it with a single open set = R^2.

 

[fresh_42]

I showed that there is no such union. I started with "take an arbitrary set of balls that cover R^2" and lead that to a contradiction. Every such cover will have a ball, and that ball has an uncovered boundary. This is true for every attempt to cover R^2.

To generalize, you cannot cover R^2 with any union (even an uncountable one) of disjoint open sets, unless you cover it with a single open set = R^2.

The "like" translate: "got it"

 

[andrewkirk]

9. For any infinitely differentiable monotonic function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\lim_{x\rightarrow +\infty} f(x) = 0$, also holds that $\lim_{x\rightarrow +\infty} f^\prime(x) = 0$.

My old friend the bump function will help us out again.

Let $h_a^b$ be the bump function with support $a,b$ and maximum value 1. Let $B=\int_0^1 h_0^1(x)dx$. Then $0<\int_a^b h_a^b(x)dx<(b-a)$.

Define $g:\mathbb R\to\mathbb R$ that is zero except when $x\in[n,n+2^{-n}]$ for some $n\in\mathbb N$ (here I take the natural numbers to start at 1, not 0), in which case $g(x)=h_n^{n+2^{-n}}(x)$.

Note that
$$B\equiv\int_{-\infty}^\infty g(x)dx=\sum_1^\infty \int_n^{n+2^{-n}} h_n^{n+2^{-n}}(t)dt
<\sum_{k=1}^\infty 2^{-n}=1
$$Now define $f:\mathbb R\to\mathbb R$ such that $$f(x)=-B-e^{-x}+\int_{-\infty}^x g(t)dt$$

This is infinitely differentiable and monotonic increasing. The latter follows from the fact that $g$ is non-negative.
$$\lim_{x\to +\infty}f(x)=-B-0+\int_{-\infty}^\infty g(x)dx
=-B-0+B=0$$
But for every integer $n>0$ there exists $x\in [n,n+2^{-n}]$ where the bump function makes $g$ attain its maximum of 1, so that
$$f'(x)=e^{-x}+\frac d{dx}\int_{-\infty}^x g(t)dt
=e^{-x}+g(x)
=e^{-x}+1>1
$$
So we cannot have $\lim_{x\to+\infty}f'(x)=0$. In fact, since outside those little intervals $f'(x)$ goes to zero, there is no limit of $f'(x)$ as $x\to\infty$.

 

[micromass]

Let $h_a^b$ be the bump function with support $a,b$ and maximum value 1. Let $B=\int_0^1 h_0^1(x)dx$. Then it's straightforward to show that $\int_a^b h_a^b(x)dx=B(b-a)$.

I'm not sure what you mean here since there are infinitely many different bump functions on $[a,b]$ with maximum $1$. Did you mean to single out a specific one, like the one in wiki?

 

[andrewkirk]

Yes, let's use the one in wiki. We need to identify a specific one, so that we can fix on a value of $B$.

I'd better write it out, to be sure:

$$h_a^b(x)=\exp\left(\frac{-1}{\left(\frac{a-b}2\right)^2-\left(x-\frac{a+b}2\right)^2}\right)$$

for $x\in[a,b]$ and zero elsewhere.

 

[micromass]

Yes, let's use the one in wiki. We need to identify a specific one, so that we can fix on a value of $B$.

OK, seems right then!

 

[fresh_42]

Another question: Why isn't $tan \frac{x\pi}{2}$ or any other unbounded function on $[0,1]$ already a counterexample to 10?

 

[micromass]

Another question: Why isn't $tan \frac{x\pi}{2}$ or any other unbounded function on $[0,1]$ already a counterexample to 10?

Can you elaborate?

 

[fresh_42]

Can you elaborate?

The only point of discontinuity of $tan (½ x\pi) : [0,1] → ℝ$ is $\{1\}$ of measure zero. But the function isn't Riemann-integrable on $[0,1]$.
Are we looking for a set $Z$ of Lebesgue measure zero for which there cannot exist ANY Riemann-integrable function with $Z$ as points of discontinuity?

 

[micromass]

The only point of discontinuity of $tan (½ x\pi) : [0,1] → ℝ$ is $\{1\}$ of measure zero. But the function isn't Riemann-integrable on $[0,1]$.
Are we looking for a set $Z$ of Lebesgue measure zero for which there cannot exist ANY Riemann-integrable function with $Z$ as points of discontinuity?

Yes. For $\{1\}$, there is a Riemann integrable function that has that as points of discontinuity.

 

[fresh_42]

But with if $Z$ is a any set of measure zero then $f(x)=1$ for $x\notin Z$ and $0$ elsewhere is Riemann-integrable. Sorry, if I ask something stupid, it's late here.

 

[micromass]

But with if $Z$ is a any set of measure zero then $f(x)=1$ for $x\notin Z$ and $0$ elsewhere is Riemann-integrable. Sorry, if I ask something stupid, it's late here.

Is that true for $Z = \mathbb{Q}$? And in that case, is the set of dicontinuity $\mathbb{Q}$?

 

[fresh_42]

Is that true for $Z = \mathbb{Q}$? And in that case, is the set of dicontinuity $\mathbb{Q}$?

Lebesgue's criterion for Riemann integrability says that a bounded function is Riemann integrable if and only if its points of discontinuity is of (Lebesgue) measure zero. In the case of ℚ I thought of mfb's construction with the heaviside function only that I'd let it decrease again somewhere. This way I'd get a function that is Riemann-integrable but discontinuous at all rational points.

 

[micromass]

Lebesgue's criterion for Riemann integrability says that a bounded function is Riemann integrable if and only if its points of discontinuity is of (Lebesgue) measure zero. In the case of ℚ I thought of mfb's construction with the heaviside function only that I'd let it decrease again somewhere. This way I'd get a function that is Riemann-integrable but discontinuous at all rational points.

Yes, but my point is that your construction fails for $Z=\mathbb{Q}$.

 

[fresh_42]

Yes, but my point is that your construction fails for $Z=\mathbb{Q}$.

Without having worked through it yet, in Wiki is a proof (maybe a sketch) for the general statement of and in the integrability section:
https://en.wikipedia.org/wiki/Riemann_integral#Integrability

 

[micromass]

Without having worked through it yet, in Wiki is a proof (maybe a sketch) for the general statement of and in the integrability section:
https://en.wikipedia.org/wiki/Riemann_integral#Integrability

Yes, it is well known that any Riemann integrable function has a set of discontinuity of measure zero. What I'm asking about is not that but the converse.

 

[fresh_42]

Yes, it is well known that any Riemann integrable function has a set of discontinuity of measure zero. What I'm asking about is not that but the converse.

Therefore my function $f: [0,1] → ℝ$ with $f(x) = 1$ if $x\notin ℚ$ and $0$ elsewhere. The only way out is an unbounded function which brings me back to the tan.

 

[micromass]

Therefore my function $f: [0,1] → ℝ$ with $f(x) = 1$ if $x\notin ℚ$ and $0$ elsewhere. The only way out is an unbounded function which brings me back to the tan.

What about that function? It's not Riemann integrable and it's set of discontinuity is $[0,1]$?

 

[fresh_42]

What about that function? It's not Riemann integrable and it's set of discontinuity is $[0,1]$?

If Lebesgue's criterion is correct then it is Riemann-integrable. (I'll better have another look tomorrow, right now I can't see any loophole.)

 

[micromass]

If Lebesgue's criterion is correct then it is Riemann-integrable. (I'll better have another look tomorrow, right now I can't see any loophole.)

It's not since its set of discontinuity points is uncountable.

 

[fresh_42]

It's not since its set of discontinuity points is uncountable.

But it's constant almost everywhere, the holes are only at the rationals.