Aren't you tired of counterexamples already?

[micromass]

And we continue our parade of counterexamples! Most of them are again in the field of real analysis, but I put some other stuff in there as well.

This time the format is a bit different. We present 10 statements that are all of the nature $P$ if and only if $Q$. As it turns out, only one of those implications is really true, the other is not. The objective is to both prove the true implication and provide a counterexample to the false one. There are two catches however:

1. There is one statement where both implications are completely true. In this case, you must prove both statements.
2. There is one statement where both implications are completely false. In this case, you must provide a counterexample to both statements.

Rules:

• For an answer to count, the answer must not only be correct, but a detailed argumentation must also be given as to why it is correct.
• 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.
• The first person to provide a complete answer will be credited. Other people may be credited for the answer as well, depending on the contribution they have given.

Here you go:

1. SOLVED BY mfb Given a series $\sum a_n$. Then the ratio test can be used to establish convergence of the series if and only if the root test can be used to establish convergence of the series.
2. SOLVED BY andrewkirk In any Banach space $X$ and given any series $\sum a_n$ in $X$, then the series converges absolutely (that is: $\sum \|a_n\|$ converges) if and only if the series converges unconditionally (that is: for any bijection $\pi:\mathbb{N}\rightarrow \mathbb{N}$ holds that $\sum_n a_{\pi(n)}$ converges to the same number).
3. SOLVED BY Samy_A A set $A\subseteq \mathbb{R}^2$ is closed if and only if it is the topological boundary of some set. That is: there is some $B\subseteq \mathbb{R}^2$ such that $\partial B = A$. https://en.wikipedia.org/wiki/Boundary_(topology)
4. SOLVED BY fresh_42 For a number $n\in \mathbb{N}\setminus \{0,1\}$ holds that there exists (up to isomorphism) only one group of order $n$ if and only if $n$ is prime.
5. SOLVED BY mfb A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous if and only if it is almost everywhere equal to a continuous function.
6. SOLVED BY Samy_A A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is constant if and only if it is differentiable almost everywhere and $f^\prime = 0$ almost everywhere.
7. SOLVED BY Samy_A A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is Borel measurable if and only if it is the pointwise limit of continuous functions.
8. SOLVED BY Samy_A A compact topological space $X$ is separable if and only if each collection of pairswise disjoint open sets is countable.
9. SOLVED BY mfb A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is measurable if and only if it is somewhere differentiable of order $2$.
10. SOLVED BY fresh_42 A field $F$ is infinite if and only if it has zero characteristic.

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]

A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is constant if and only if it is differentiable almost everywhere and $f^\prime = 0$ almost everywhere.

A constant function on $\mathbb{R}$ is differentiable everywhere and has derivative 0 everywhere. The "only if" is trivially true.
Conversely, the function $f: \mathbb{R} \to \mathbb{R}$ that maps strictly positive numbers to 1 and negative numbers and 0 to 0 is differentiable almost everywhere (not in 0), and the derivative is 0 almost everywhere (again not in 0). But it is not constant. The "if" is therefore false.

 

[micromass]

A constant function on $\mathbb{R}$ is differentiable everywhere and has derivative 0 everywhere.
Conversely, the function $f: \mathbb{R} \to \mathbb{R}$ that maps strictly positive numbers to 1 and negative numbers and 0 to 0 is differentiable almost everywhere (not in 0), and the derivative is 0 almost everywhere (again not in 0). But it is not constant. The "if" is therefore false.

Hmm, I forgot to add that the function should be continuous But ok, I'll accept your answer. For the people interested in what I had in mind: https://en.wikipedia.org/wiki/Cantor_function

 

[Samy_A]

Hmm, I forgot to add that the function should be continuous But ok, I'll accept your answer. For the people interested in what I had in mind: https://en.wikipedia.org/wiki/Cantor_function

I did wonder: damn that's an easy one.

 

[Samy_A]

Deleted

 

[fresh_42]

10. A field $F$ is infinite if and only if it has zero characteristic.

A field of characteristic zero contains his prime field ℚ and is therefore infinite.
On the other hand is the quotient ring $\mathbb{Z_p}(x)$ of the polynomial ring $\mathbb{Z_p}[x]$ infinite of characteristic $p$ an example of an infinite field of nonzero characteristic.

 

[fresh_42]

4. For a number $n\in \mathbb{N}\setminus \{0,1\}$ holds that there exists (up to isomorphism) only one group of order $n$ if and only if $n$ is prime.

If there is only one group of order $p$ (prime) it has to be $ℤ_p$.
For non primes it is not true:
$Z_4$ and Klein's four-group $V_4 = K_4 = Z_2 \times Z_2$.
All squares in $V_4 = <a, b | a^2 = b^2 = (ab)^2 = 1>$ equal the neutral element $1$ whereas $3+3 = 2 ≠ 0$ in $Z_4$. Therefore they cannot be isomorphic.

Now let $G$ be a group of prime order $p$ and $a ≠ 1$ an element of $G$.
Then $a$ generates a subgroup $S$ of $G$ which order divides $p$. Since $p$ is prime $S=G≅ℤ_p$.
(The fact that the orders of subgroups $S$ divide the order of the group $G$ can be proven by considering $G/S$. This partitions $G$ in equivalence classes of the same order (although they in general do not form a group again since $S$ was not assumed to be normal).)

 

[micromass]

If there is only one group of order $p$ (prime) it has to be $ℤ_p$.
For non primes it is not true:
$Z_4$ and Klein's four-group $V_4 = K_4 = Z_2 \times Z_2$.
All squares in $V_4 = <a, b | a^2 = b^2 = (ab)^2 = 1>$ equal the neutral element $1$ whereas $3+3 = 2 ≠ 0$ in $Z_4$. Therefore they cannot be isomorphic.

That doesn't answer the question. You answered correctly that if $p$ is prime, then there is only one group of order $p$ up to isomorphism. I asked whether there are nonprimes $n$ such that there is only one group of order $n$ up to isomorphism.

 

[fresh_42]

A set $A\subseteq \mathbb{R}^2$ is closed if and only if it is the topological boundary of some set. That is: there is some $B\subseteq \mathbb{R}^2$ such that $\partial B = A$.

Since the boundary of a set is its closure without its inner points it is closed.
On the other hand is the upper half plane (including the "x"-axis) of $ℝ^2$ a closed set which is no boundary of any other set since it contains inner points.

 

[micromass]

Since the boundary of a set is its closure without its inner points it is closed.

Right.

On the other hand is the upper half plane (including the "x"-axis) of $ℝ^2$ a closed set which is no boundary of any other set since it contains inner points.

I agree that the boundary of a set $A$ cannot contain inner points of the set $A$. But why does that necessarily imply that the boundary itself has empty interior?

 

[JonnyG]

The upper half plane is the boundary of $A = \{ (x,y) \in \mathbb{R}^2: y \geq 0 \text{ and } x,y \in \mathbb{Q}\}$

 

[fresh_42]

I agree that the boundary of a set $A$ cannot contain inner points of the set $A$. But why does that necessarily imply that the boundary itself has empty interior?

If $A=\{(x,y) \in ℝ^2 | y>0\}$ were a boundary of some set $B$ then $\{(x,y)\in ℝ^2 | (x-1)^2 + (y-1)^2 < ¼\}$ would be a open neighborhood of $(1,1)$ which is completely in $\partial B$ which contradicts the boundary definition you quoted from Wiki. Therefore $A$ is a closed set which isn't a boundary of any (other or not) set $B$.

 

[micromass]

If $A=\{(x,y) \in ℝ^2 | y>0\}$ were a boundary of some set $B$ then $\{(x,y)\in ℝ^2 | (x-1)^2 + (y-1)^2 < ¼\}$ would be a open neighborhood of $(1,1)$ which is completely in $\partial B$ which contradicts the boundary definition you quoted from Wiki. Therefore $A$ is a closed set which isn't a boundary of any (other or not) set $B$.

Did you see post 11?

 

[fresh_42]

That doesn't answer the question. You answered correctly that if $p$ is prime, then there is only one group of order $p$ up to isomorphism. I asked whether there are nonprimes $n$ such that there is only one group of order $n$ up to isomorphism.

Part 2: There is only one group of order 15, $ℤ_{15} ≅ℤ_3 \times ℤ_5$.
Proof: Sylow's theorems. https://en.wikipedia.org/wiki/Sylow_theorems

 

[micromass]

Part 2: There is only one group of order 15, $ℤ_{15} ≅ℤ_3 \times ℤ_5$.
Proof: Sylow's theorems. https://en.wikipedia.org/wiki/Sylow_theorems

Yep. Some interesting results:

From the numbers $\{1,...,2015\}$, there are $656$ numbers which have only one group of that order. There are $305$ primes smaller than $2015$.
There are $393$ numbers which have only $2$ groups of that order. There are $11$ numbers which have $3$ groups of that order.

Of the groups from order $1$ to $2015$, more than $99\%$ of them have order $1024$. There are $49,487,365,422$ groups of that order. It is unkown how many groups there are of order $2048$. In comparison, the median groups per order is $2$.

 

[fresh_42]

Yep. Some interesting results:

From the numbers $\{1,...,2015\}$, there are $656$ numbers which have only one group of that order. There are $305$ primes smaller than $2015$.
There are $393$ numbers which have only $2$ groups of that order. There are $11$ numbers which have $3$ groups of that order.

Of the groups from order $1$ to $2015$, more than $99\%$ of them have order $1024$. There are $49,487,365,422$ groups of that order. It is unkown how many groups there are of order $2048$. In comparison, the median groups per order is $2$.

And I thought the monster groups alone were mind blowing. (I'm still totally perplexed of how in the world they found them.) ... back to my pathologic problem with those double dense sets ... (set as well as complement dense)

 

[Samy_A]

A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is measurable if and only if it is the pointwise limit of continuous functions.

Continuous functions are Lebesgue measurable, so the pointwise limit of continuous functions is itself Lebesgue measurable. That proves the "if" part.

However, continuous functions are also Borel measurable, so the pointwise limit of continuous functions is Borel measurable.
Hence a function that is Lebesgue measurable but not Borel measurable is a counterexample for the "only if" part.

 

[micromass]

Continuous functions are Lebesgue measurable, so the pointwise limit of continuous functions is itself Lebesgue measurable. That proves the "if" part.

However, continuous functions are also Borel measurable, so the pointwise limit of continuous functions is Borel measurable.
Hence a function that is Lebesgue measurable but not Borel measurable is a counterexample for the "only if" part.

Right, but to make it less easy, assume I put Borel measurable instead of measurable in the statement

 

[Samy_A]

Right, but to make it less easy, assume I put Borel measurable instead of measurable in the statement

Indeed, but I will leave that to someone else. Just now I have been reading about it.

 

[fresh_42]

The upper half plane is the boundary of $A = \{ (x,y) \in \mathbb{R}^2: y \geq 0 \text{ and } x,y \in \mathbb{Q}\}$

Doesn't this contain the proof? (You know my weakness with densities, so I apologize if I overlooked something again.)
Let $A$ be any closed set and $A_ℚ = A ∩ ℚ^2$.
Then ${A_ℚ}^° = ∅$ and $\partial A_ℚ = \overline{A_ℚ} \backslash {A_ℚ}^° = \overline{A} \backslash ∅ = A$.

 

[micromass]

Doesn't this contain the proof? (You know my weakness with densities, so I apologize if I overlooked something again.)
Let $A$ be any closed set and $A_ℚ = A ∩ ℚ^2$.
Then ${A_ℚ}^° = ∅$ and $\partial A_ℚ = \overline{A_ℚ} \backslash {A_ℚ}^° = \overline{A} \backslash ∅ = A$.

That was my first attempt too, but take $A = \{0\}\times (\mathbb{R}\setminus \mathbb{Q})$.

 

[fresh_42]

That was my first attempt too, but take $A = \{0\}\times (\mathbb{R}\setminus \mathbb{Q})$.

But $\mathbb{R}\setminus \mathbb{Q}$ is not closed here, so isn't $A$.

 

[micromass]

But $\mathbb{R}\setminus \mathbb{Q}$ is not closed here, so isn't $A$.

Sorry, what was I thinking. Silly me

Take $A = \{\pi\} \times \mathbb{R}$.

 

[fresh_42]

Sorry, what was I thinking. Silly me

Take $A = \{\pi\} \times \mathbb{R}$.

... which is the boundary of $\{(x,y) ∈ ℝ^2 | x < π\}$.

 

[micromass]

... which is the boundary of $\{(x,y) ∈ ℝ^2 | x < π\}$.

Right... But that means your proof fails.

 

[Samy_A]

Does this work?
$A \subset \mathbb R²$ is separable.
That means that there exist a countable $B \subset A$ such that $\bar B =A$.
$B$ being countable means that $\mathring B = \varnothing$, so that $\partial B = \bar B \setminus \mathring B = A$.

 

[micromass]

Does this work?
$A \subset \mathbb R²$ is separable.
That means that there exist a countable $B \subset A$ such that $\bar B =A$.
$B$ being countable means that $\mathring B = \varnothing$, so that $\partial B = \bar B \setminus \mathring B = A$.

Why is $A$ separable? I'll accept that $\mathbb{R}^2$ is separable.

 

[Samy_A]

Why is $A$ separable? I'll accept that $\mathbb{R}^2$ is separable.

The reasoning was: $\mathbb R²$ is second-countable, and I thought that implied that every subspace also is second-countable. And second-countable implies separable.

 

[micromass]

The reasoning was: $\mathbb R²$ is second-countable, and I thought that implied that every subspace also is second-countable. And second-countable implies separable.

OK, that's good enough for me.

 

[fresh_42]

8. A compact topological space $X$ is separable if and only if each collection of pairwise disjoint open sets is countable.

I have a question here: Shall the collection of pairwise disjoint open sets itself be countable or the index set of it?
I think I have a counterexample for the "only if" case in which both are uncountable showing compact and separable isn't strong enough.
However, I'm asking for the reverse direction to clear the valid condition. (And it's really not easy to find a compact non separable space leave alone one with that crude condition on any open sets.)

 

[micromass]

I have a question here: Shall the collection of pairwise disjoint open sets itself be countable or the index set of it?

What is the difference?

 

[fresh_42]

What is the difference?

$O_x ∪ O_y$ are countably many open sets whereas each single one must not contain countably many elements. I suppose the index set, here $\{x,y\}$ is supposed to be countable.

 

[micromass]

$O_x ∪ O_y$ are countably many open sets

What you wrote down is one open set. I don't get it.

 

[fresh_42]

What you wrote down is one open set. I don't get it.

$]0,1[$ is a union of countably many open sets (1), whereas $]0,1[$ has uncountably many elements so its union isn't countable, too. So does the "countable" in 8. refer to the open sets itself or the cardinality of the set which indexes the union which I assume?
(There are certainly topologies in which open sets can have countably many elements.)

 

[micromass]

$]0,1[$ is a union of countably many open sets (1), whereas $]0,1[$ has uncountably many elements so its union isn't countable, too. So does the "countable" in 8. refer to the open sets itself or the cardinality of the set which indexes the union which I assume?
(There are certainly topologies in which open sets can have countably many elements.)

It's the cardinality of the index set.