Yet another counterexample challenge

[micromass]

Well, the last thread of counterexamples was pretty fun. So why not do it again! Again, I present you a list with 10 mathematical statements. The only rub now is that only $9$ are false, thus one of the statements is true. Provide a counterexample to the false statements and a proof for the correct one!

Rules:

• For a counterexample to count, the answer must not only be correct, but a detailed argumentation must also be given as to why it is a counterexample.
• Any use of outside sources is allowed, but do not look up the question directly. For example, it is ok to go check analysis books, but it is not allowed to google the exact question.
• If you previously encountered this statement and remember the solution, then you cannot participate in this particular statement.
• All mathematical methods are allowed.

Here you go:

1. SOLVED BY PeroK $\mathbb{R}$ is the disjoint countable union of closed intervals
2. SOLVED BY mfb Any open set in $\mathbb{R}^2$ is the disjoint countable union of open balls, where an open ball is a set of the form $B(\mathbf{a},r) = \{\mathbf{x}\in \mathbb{R}^2~\vert~(a_1 - x_1)^2 + (a_2 - x_2)^2 < r^2\}$ for $\mathbf{a}\in \mathbb{R}^2$ and $r>0$.
3. SOLVED BY andrewkirk Let $\pi:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}$ be a bijection and let $\sum_{n} a_n$ be an absolutely convergent series in $\mathbb{R}$. Then $\sum_n a_n = \sum_{k=0}^\infty \sum_{l=0}^\infty a_{\pi(k,l)}$.
4. SOLVED BY PeroK Every continuous bounded function $\mathbb{R}\rightarrow \mathbb{R}$ is uniformly continuous.
5. SOLVED BY Samy_A Every function $\mathbb{R}\rightarrow \mathbb{R}$ is the derivative of some function.
6. SOLVED BY PeroK If $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and sends open sets to open sets, then it also sends closed sets to closed sets.
7. SOLVED BY mfbThere is no monotonic function $\mathbb{R}\rightarrow \mathbb{R}$ whose set of discontinuities is $\mathbb{Q}$.
8. SOLVED BY Samy_A Every nonconstant function $\mathbb{R}\rightarrow \mathbb{R}$ that is periodic has a smallest period.
9. SOLVED BY andrewkirk For any infinitely differentiable monotonic function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\lim_{x\rightarrow +\infty} f(x) = 0$, also holds that $\lim_{x\rightarrow +\infty} f^\prime(x) = 0$.
10. SOLVED BY andrewkirk Every subset of $[0,1]$ of measure zero is the set of points of discontinuity of a Riemann-integrable function $[0,1]\rightarrow \mathbb{R}$.

Thank you all for participating! I hope some of these statements were surprising to some of you and I hope some of you have fun with this! Don't hesitate to post any feedback in the thread!

 

[Samy_A]

Every nonconstant function $\mathbb{R}\rightarrow \mathbb{R}$ that is periodic has a smallest period.

Define $f:\mathbb{R}\rightarrow \mathbb{R}$ as follows
$\left\{\begin{array}{l} x \in \mathbb Q : x \mapsto 1\\ x \notin \mathbb Q : x \mapsto 2 \end{array}\right.$
Every rational number is a period, since for every rational number $q$, $x+q$ will be rational if and only if $x$ is rational. That implies that for every rational number $q$, $\forall x \in \mathbb R: f(q+x)=f(x)$.

 

[Samy_A]

Every bounded function $\mathbb{R}\rightarrow \mathbb{R}$ is uniformly continuous.

That seems trivial. Did you mean continuous bounded function?

 

[micromass]

That seems trivial. Did you mean continuous bounded function?

Yes, sorry!

 

[Samy_A]

Every function $\mathbb{R}\rightarrow \mathbb{R}$ is the derivative of some function.

I feel there must be an easier one, but this should work.
The derivative of a function must be Lebesgue measurable.

Proof: assume $f=g'$
That means that $\displaystyle \forall x \in \mathbb R: \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g'(x)=f(x)$.
Define $f_n:\mathbb{R}\rightarrow \mathbb{R}: x \mapsto \frac{g(x+1/n)-g(x)}{1/n}$
Then $\displaystyle \forall x \in \mathbb R: \lim_{n \rightarrow +\infty}f_n(x)=g'(x)=f(x)$.
Since $g$ is differentiable, it is also Lebesgue measurable. Hence the functions $f_n$ are also Lebesgue measurable.
$f$ being the pointwise limit of Lebesgue measurable function is itself Lebesgue measurable.

So a counterexample is a function $f$ that is not Lebesgue measurable.

 

[Samy_A]

I feel there must be an easier one, but this should work.
The derivative of a function must be Lebesgue measurable.

Proof: assume $f=g'$
That means that $\displaystyle \forall x \in \mathbb R: \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g'(x)=f(x)$.
Define $f_n:\mathbb{R}\rightarrow \mathbb{R}: x \mapsto \frac{g(x+1/n)-g(x)}{1/n}$
Then $\displaystyle \forall x \in \mathbb R: \lim_{n \rightarrow +\infty}f_n(x)=g'(x)=f(x)$.
Since $g$ is differentiable, it is also Lebesgue measurable. Hence the functions $f_n$ are also Lebesgue measurable.
$f$ being the pointwise limit of Lebesgue measurable function is itself Lebesgue measurable.

So a counterexample is a function $f$ that is not Lebesgue measurable.

An easier one:

Define $f:\mathbb{R}\rightarrow \mathbb{R}$ as follows
$\left\{\begin{array}{l} x \in \mathbb Q : x \mapsto 1\\ x \notin \mathbb Q : x \mapsto 0 \end{array}\right.$

Notice $f=0$ almost everywhere.

If $f=g'$, then $\displaystyle \int_0^a g'(x)dx=g(a)-g(0)$. (EDIT: I wonder: for this I probably need to first prove that g' is measurable, so we need the proof in the previous post anyway.)
But $\displaystyle \int_0^a g'(x)dx =\int_0^a f(x)dx =0$.
Hence $g$ is constant, and its derivative must be 0. We have reached a contradiction.

 

[PeroK]

Number 4 seems quite easy. $f(x) = sin(x^2)$

Let $x_n = (n\pi)^{1/2}$ and $y_n = [(n + 1/2)\pi]^{1/2}$

$\forall n \ \ f(x_n) = sin(n\pi) = 0$ and $f(y_n) = sin(n\pi + \pi/2) = \pm 1$

hence $|f(x_n) - f(y_n)| = 1$

But $\lim_{n \rightarrow \infty}|x_n - y_n| = 0$ (*)

Therefore, $\forall \delta > 0$ we can choose $n$ such that $|x_n - y_n| < \delta$ but $|f(x_n) - f(y_n)| = 1$

Hence $f$ is not uniformly continuous.

(*) $|x_n - y_n|(x_n + y_n) = \pi/2$

$|x_n - y_n| = \frac{\pi}{2(x_n + y_n)} < \frac{\pi}{2\sqrt{n\pi}}$

 

[PeroK]

I feel there must be an easier one, but this should work.
The derivative of a function must be Lebesgue measurable.

Proof: assume $f=g'$
That means that $\displaystyle \forall x \in \mathbb R: \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g'(x)=f(x)$.
Define $f_n:\mathbb{R}\rightarrow \mathbb{R}: x \mapsto \frac{g(x+1/n)-g(x)}{1/n}$
Then $\displaystyle \forall x \in \mathbb R: \lim_{n \rightarrow +\infty}f_n(x)=g'(x)=f(x)$.
Since $g$ is differentiable, it is also Lebesgue measurable. Hence the functions $f_n$ are also Lebesgue measurable.
$f$ being the pointwise limit of Lebesgue measurable function is itself Lebesgue measurable.

So a counterexample is a function $f$ that is not Lebesgue measurable.

Does that make this equivalent to the axiom of choice?

 

[Samy_A]

Does that make this equivalent to the axiom of choice?

Think so (for the existence of not Lebesgue measurable functions). But you don't need the axiom of choice to prove that a derivative is Lebesgue measurable. So my second example probably works without invoking the axiom of choice.

 

[micromass]

I feel there must be an easier one, but this should work.
The derivative of a function must be Lebesgue measurable.

Proof: assume $f=g'$
That means that $\displaystyle \forall x \in \mathbb R: \lim_{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g'(x)=f(x)$.
Define $f_n:\mathbb{R}\rightarrow \mathbb{R}: x \mapsto \frac{g(x+1/n)-g(x)}{1/n}$
Then $\displaystyle \forall x \in \mathbb R: \lim_{n \rightarrow +\infty}f_n(x)=g'(x)=f(x)$.
Since $g$ is differentiable, it is also Lebesgue measurable. Hence the functions $f_n$ are also Lebesgue measurable.
$f$ being the pointwise limit of Lebesgue measurable function is itself Lebesgue measurable.

So a counterexample is a function $f$ that is not Lebesgue measurable.

Here is another proof: https://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)

 

[micromass]

If $f=g'$, then $\displaystyle \int_0^a g'(x)dx=g(a)-g(0)$. (EDIT: I wonder: for this I probably need to first prove that g' is measurable, so we need the proof in the previous post anyway.)

This is a very interesting remark, and highly depends on what kind of integral you're looking at. Let's first look at the classical fundamental theorem of calculus (quoted from wiki):

Let $f$ and $F$ be real-valued functions defined on a closed interval $[a,b]$ such that $F$ is continuous and the derivative of $F$ is $f$. That is, $f$ and $F$ are functions such that for all $x$ in $(a, b)$,
$$F^\prime (x) = f(x)$$
If $f$ is Riemann integrable on $[a,b]$ then
$$\int_a^b f(x)dx = F(b) - F(a)$$

So we see in your counterexample, that you will first need to prove that $g^\prime(x)$ is Riemann integrable. This is obviously not the case. The classical fundamental theorem is not strong enough to let you do what you want.

Luckily, there is an extension of the Riemann integral, called the Henstock integral. This integral has a very similar definition to the Riemann integral, but has an improved fundamental theorem. Its theory is explained in Bartle "A modern theory of integration" https://www.amazon.com/dp/0821808451/?tag=pfamazon01-20 Let me quote the relevant part

If $f : [a, b] \rightarrow \mathbb{R}$ has a primitive $F$ on $[a,b]$, then $f$ is Henstock integrable and
$$\int_a^b f(x)dx = F(b) - F(a)$$

(So this fundamental theorem actually proves Henstock integrability of $f$. No further assumptions are necessary other than it being a derivative. Not even measurability. (Note that Henstock integrability implies measurable though). So this theorem does exactly what you want. Note that this isn't even the most general theorem of this kind, you can get away with a lot less than what the quoted part assumed, all of that is in Bartle.

Now I could be very justified to put in a rant on why Henstock integrals are not at all taught in schools. They are extremely easy to define, and are more general than Lebesgue integrals (for functions $\mathbb{R}\rightarrow \mathbb{R}$), and they have way better properties than the silly Riemann integral. Sadly, they are ignored by the vast majority of books (one notable exception is DePree Swartz https://www.amazon.com/dp/0471853917/?tag=pfamazon01-20)

 

[PeroK]

I assume for number 1 the statement should be "not" and then the counterexample is R given as a disjoint countable union of closed intervals. I'll do non-negative R.

The first set of intervals are $[0, 1], [2, 3], [4, 5] \dots$

The second set of intervals are $[1\frac{1}{4}, 1\frac{3}{4}], [3\frac{1}{4}, 3\frac{3}{4}] \dots$

The third set of intervals are $[1\frac{1}{16}, 1\frac{3}{16}], [1\frac{13}{16}, 1\frac{15}{16}], [3\frac{1}{16}, 3\frac{3}{16}], [3\frac{13}{16}, 1\frac{15}{16}] \dots$

Etc.

 

[jbriggs444]

I assume for number 1 the statement should be "not" and then the counterexample is R given as a disjoint countable union of closed intervals. I'll do non-negative R.

But.

The only rub now is that only 9 are false, thus one of the statements is true. Provide a counterexample to the false statements and a proof for the correct one!

It seems that you have provided at least the core of a proof for the correct one.

 

[jbriggs444]

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

Spoiler

The idea is that one wants to come up with a kind of smoothed step function where the steps get smaller and smaller but retain a similar slope. If one had as a building block a monotone increasing function that is continuous and infinitely differentiable on the unit interval [0,1] with all derivatives equal to zero at both endpoints then one could use that to construct the counter example by splining such building blocks together. Unfortunately, my toolbag only contains a function with the required properties at one endpoint.

Wikipedia calls these "smooth transfer functions". And I think at this point I think I've been spoiled too badly to compete.

 

[micromass]

I assume for number 1 the statement should be "not" and then the counterexample is R given as a disjoint countable union of closed intervals. I'll do non-negative R.

The first set of intervals are $[0, 1], [2, 3], [4, 5] \dots$

The second set of intervals are $[1\frac{1}{4}, 1\frac{3}{4}], [3\frac{1}{4}, 3\frac{3}{4}] \dots$

The third set of intervals are $[1\frac{1}{16}, 1\frac{3}{16}], [1\frac{13}{16}, 1\frac{15}{16}], [3\frac{1}{16}, 3\frac{3}{16}], [3\frac{13}{16}, 1\frac{15}{16}] \dots$

Etc.

So, the easiest is to write this in base $4$ notation. Let me focus on the cover of $(1,2)$ which according to you is:

$$[0.1, 0.3]$$
$$0.01, 0.03],~[0.31, 0.33]$$
$$[0.001, 0.003],~[0.031, 0.033],~[0.0301, 0.303],~[0.331, 0.333]$$
And so on.

So the first step covers points of the form $0.1xxxx...$ and $0.2xxx...$ (assume terminating "decimals" are forbidden). No point with $1$ and $2$ in the first place remains in the remainder.
The second step covers points of the form $0.01xxx...$ and $0.02xxx...$ and $0.31xxx...$ and $0.32xxx$. Removing this, we see that no point with $1$ and $2$ in the first two places remain.

All in all, in the $n$th step, we see that no point with $1$ and $2$ in the first $n$ decimals places remain in the remainder.

So what your collection of closed intervals covers are all points with $1$ and $2$ in some decimal place. But from this description it is easy to see that the point $0.030303030303...$ is not covered by your collection of intervals. So your counterexample is not valid.

 

[micromass]

And I think at this point I think I've been spoiled too badly to compete.

Don't worry, you can still compete with this question if you want to. After all, checking wikipedia or textbooks for specific subjects is allowed, as long as you don't find a direct answer to the question.

 

[Samy_A]

Here is another proof: https://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)

Aahh, that's the one!
I knew there was some well known property that all derivatives had, but couldn't remember which one.

This is a very interesting remark, and highly depends on what kind of integral you're looking at. Let's first look at the classical fundamental theorem of calculus (quoted from wiki):
So we see in your counterexample, that you will first need to prove that $g^\prime(x)$ is Riemann integrable. This is obviously not the case. The classical fundamental theorem is not strong enough to let you do what you want.

Luckily, there is an extension of the Riemann integral, called the Henstock integral. This integral has a very similar definition to the Riemann integral, but has an improved fundamental theorem. Its theory is explained in Bartle "A modern theory of integration" https://www.amazon.com/dp/0821808451/?tag=pfamazon01-20 Let me quote the relevant part
(So this fundamental theorem actually proves Henstock integrability of $f$. No further assumptions are necessary other than it being a derivative. Not even measurability. (Note that Henstock integrability implies measurable though). So this theorem does exactly what you want. Note that this isn't even the most general theorem of this kind, you can get away with a lot less than what the quoted part assumed, all of that is in Bartle.

Now I could be very justified to put in a rant on why Henstock integrals are not at all taught in schools. They are extremely easy to define, and are more general than Lebesgue integrals (for functions $\mathbb{R}\rightarrow \mathbb{R}$), and they have way better properties than the silly Riemann integral. Sadly, they are ignored by the vast majority of books (one notable exception is DePree Swartz https://www.amazon.com/dp/0471853917/?tag=pfamazon01-20)

Very cool. Henstock integral: something to look at.

 

[micromass]

Aahh, that's the one!
I knew there was some well known property that all derivatives had, but couldn't remember which one.

Well, we have detailed some of its properties already: derivatives are Lebesgue measurable and satisfy the conclusion of the intermediate value theorem. One could ask whether this completely characterizes derivatives. But that's for a next set of counterexamples (the answer is no).

 

[fresh_42]

6. If $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and sends open sets to open sets, then it also sends closed sets to closed sets.

I define $f(x) = x^{-2}$ for $x\neq 0$ and $f(0)=0$.
My "fraud (?)" is this: I take the discrete topology on both $\mathbb{R}$. Then $f$ is clearly open.
With $f^{-1}(\{x\}) = \{\frac{1}{\sqrt{x}}\}$ for $x > 0$, $f^{-1}(\{0\})=\{0\}$ and $f^{-1}(\{x\})=∅$ for $x < 0$ $f$ is also continuous.
Now $f(\mathbb{R} - \{0\}) = \mathbb{R^+} = ∪_{x>0} \{x\}$ maps a closed set onto an open.

 

[micromass]

I define $f(x) = x^{-2}$ for $x\neq 0$ and $f(0)=0$.
My "fraud (?)" is this: I take the discrete topology on both $\mathbb{R}$. Then $f$ is clearly open.
With $f^{-1}(\{x\}) = \{\frac{1}{\sqrt{x}}\}$ for $x > 0$, $f^{-1}(\{0\})=\{0\}$ and $f^{-1}(\{x\})=∅$ for $x < 0$ $f$ is also continuous.
Now $f(\mathbb{R} - \{0\}) = \mathbb{R^+} = ∪_{x>0} \{x\}$ maps a closed set onto an open.

OK, but your map still takes closed sets to closed sets.

(And I kind of meant $\mathbb{R}$ to have the Euclidean topology)

 

[fresh_42]

OK, but your map still takes closed sets to closed sets.

(And I kind of meant $\mathbb{R}$ to have the Euclidean topology)

So you want to have all closed sets not to be closed? Even $f(\mathbb{R})$?

 

[PeroK]

It seems to me that for number 6, you have to exclude $\mathbb{R}$ itself as an allowable closed set. Any monotonic bounded function such as $arctan$ would map $\mathbb{R}$ to an open interval. That seems a trivial counterexample otherwise.

 

[micromass]

So you want to have all closed sets not to be closed? Even $f(\mathbb{R})$?

No, I just want one image of a closed set that is not closed.

 

[micromass]

It seems to me that for number 6, you have to exclude $\mathbb{R}$ itself as an allowable closed set. Any monotonic bounded function such as $arctan$ would map $\mathbb{R}$ to an open interval. That seems a trivial counterexample otherwise.

Yes, that's a right solution. Not all of them need to be hard

 

[PeroK]

Hopefully proof for number 1: all sequences of closed intervals are essentially the same. Focus on $(0, 1)$ and show that can't be covered:

Let's assume we have a valid sequence of closed disjoint intervals covering $(0,1)$. The first interval is $I_1 = (x_1, y_1)$ and the set remaining is:

$R_1 = (0, x_1) \cup (y_1, 1)$

Assume, wlog, that the second interval is a subset of the first interval: $I_2 = (x_2, y_2) \subset (0, x_1)$

$R_2 = (0, x_2) \cup (y_2, x_1) \cup (y_1, 1)$

We're actually going to take a subsequence of the original sequence, so to save on subscripts I'll take $I_3 = (x_3, y_3) \subset (y_2, x_1)$ (the next interval that splits this remaining interval):

$R_3 \supset (y_2, x_3) \cup (y_3, x_1)$

And, take $I_4 = (x_4, y_4) \subset (y_2, x_3)$

$R_4 \supset (y_2, x_4) \cup (y_4, x_3)$

And, take $I_5 = (x_5, y_5) \subset (y_4, x_3)$

$R_5 \supset (y_4, x_5) \cup (y_5, x_3)$

Continuing in this way, we can generate a sequence of open subsets:

$(0, x_1) \supset (y_2, x_3) \supset (y_4, x_5) \dots$

And the intersection of the $R_n$ sets contains the intersection of these sets.

Finally, by the completeness of $\mathbb{R}$ there must be a point in this intersection. We can generate a monotone sequence from the $y_n$ endpoints that must converge to some $y$. If $y$ is not in the intersection, then it must be outside one of the sets. But, the next set has no endpoints in common with the previous one, so it must be a finite distance outside that and subsequent sets, which is a contradiction.

This non-empty intersection is outside the union of all the $I_n$, which is a contradiction to the original assumption.

(Note that the same argument can apply to every interval generated, so there must be (at least) a countable number of points outside the union.)

 

[fresh_42]

I hope I didn't miss anything.

2. Any open set in R^2 is the disjoint countable union of open balls, where an open ball is a set of the form $B(\mathbf{a},r) = \{\mathbf{x}\in \mathbb{R}^2~\vert~(a_1 - x_1)^2 + (a_2 - x_2)^2 < r^2\}$ for $\mathbf{a}\in \mathbb{R}^2$ and $r>0$.

Assuming this to be true there are (pairwise) disjoint open balls $B(a_n,r_n)$ with $U= ]0,1[ \times ]0,1[= \cup_{n∈ℕ}B(a_n,r_n)$.
Therefore for all $ε>0$ $\cup_{n∈ℕ}B(a_n,r_n+ε)$ is an open cover of the compact set $\overline{U}$.
This means there is a finite subcover, so
$$\overline{U} ⊆ \cup_{n=1}^{n=N}B(a_n,r_n+ε) ⊆ \overline{\cup_{n=1}^{n=N}B(a_n,r_n+2ε)}$$
Building the limit $ε→0$ means $[0,1] \times [0,1] = \overline{U} = \cup_{n=1}^{n=N}\overline{B(a_n,r_n)}$ where at most finitely many points share the boundaries of several closed balls. With finitely many closed balls within $\overline{U}$, however, it is impossible to build the boundary of U.

 

[micromass]

I hope I didn't miss anything.Assuming this to be true there are (pairwise) disjoint open balls $B(a_n,r_n)$ with $U= ]0,1[ \times ]0,1[= \cup_{n∈ℕ}B(a_n,r_n)$.
Therefore for all $ε>0$ $\cup_{n∈ℕ}B(a_n,r_n+ε)$ is an open cover of the compact set $\overline{U}$.
This means there is a finite subcover, so
$$\overline{U} ⊆ \cup_{n=1}^{n=N}B(a_n,r_n+ε) ⊆ \overline{\cup_{n=1}^{n=N}B(a_n,r_n+2ε)}$$
Building the limit $ε→0$ means $[0,1] \times [0,1] = \overline{U} = \cup_{n=1}^{n=N}\overline{B(a_n,r_n)}$ where at most finitely many points share the boundaries of several closed balls. With finitely many closed balls within $\overline{U}$, however, it is impossible to build the boundary of U.

Isn't the finite subcover and the amount of balls in the subcover dependend on $\varepsilon$? So if you change $\varepsilon$, you might get very different balls, or more balls. Doesn't that make your entire $\varepsilon\rightarrow 0$argument a bit fishy?

 

[fresh_42]

Isn't the finite subcover and the amount of balls in the subcover dependend on $\varepsilon$? So if you change $\varepsilon$, you might get very different balls, or more balls. Doesn't that make your entire $\varepsilon\rightarrow 0$argument a bit fishy?

Thanks, that's right. Da... I've already discarded another ε-Argument with rational $r_n$ for similar reasons. Seems that kept me from looking more carefully at the rest.

 

[mfb]

I'm confused by the "disjoint" in (2). Wikipedia would suggest the disjoint union of balls in R^2 is not a subset of R^2 any more.

7: Let $(a_n)_n$ be an enumeration of all rational numbers. Define $w_n=2^{-n}$. Note that $\sum_{n=0}^{\infty} w_n = 2$ is finite and all summands are positive.

Define $$f(x)= \sum_{n=0}^{\infty} w_n H(x-a_n)$$ with the Heaviside step function.
The sum converges as it sums over a subset of the previous sum.
It is monotonic as every summand is monotonic (it is even strict monotonic, as the set of rational numbers is dense)
It has discontinuities of step size w_n at every rational number a_n.
It is continuous at every irrational number x as for every $\epsilon$, there is an N such that $w_n < \epsilon$ for n>N, and the set of a_n with n<=N is finite and does not include x, so there is a suitable $\delta$ for continuity.

 

[micromass]

I'm confused by the "disjoint" in (2). Wikipedia would suggest the disjoint union of balls in R^2 is not a subset of R^2 any more.

I mean that the set involved are pairswise disjoint.

 

[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.