# Challenge Aren't you tired of counterexamples already?

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1. May 2, 2016

### 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!

Last edited: May 20, 2016
2. May 2, 2016

### Samy_A

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.

3. May 2, 2016

### micromass

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

4. May 2, 2016

### Samy_A

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

5. May 2, 2016

### Samy_A

Deleted

Last edited: May 2, 2016
6. May 2, 2016

### Staff: Mentor

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.

7. May 2, 2016

### Staff: Mentor

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).)

8. May 2, 2016

### micromass

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.

9. May 2, 2016

### Staff: Mentor

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.

10. May 2, 2016

### micromass

Right.

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?

11. May 2, 2016

### JonnyG

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

12. May 2, 2016

### Staff: Mentor

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 $B$.

13. May 2, 2016

### micromass

Did you see post 11?

14. May 2, 2016

### Staff: Mentor

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

15. May 2, 2016

### micromass

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 $2$.

16. May 2, 2016

### Staff: Mentor

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)

17. May 2, 2016

### Samy_A

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.

18. May 2, 2016

### micromass

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

19. May 2, 2016

### Samy_A

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

20. May 2, 2016

### Staff: Mentor

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 $\partial A_ℚ = \overline{A_ℚ} \backslash {A_ℚ}^° = \overline{A} \backslash ∅ = A$.

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