Challenge Micromass' big counterexample challenge

[micromass]

I adore counterexamples. They're one of the most beautiful things about math: a clevery found ugly counterexample to a plausible claim. Below I have listed 10 statements about basic analysis which are all false. Your job is to find the correct counterexample. Some are easy, some are not so easy. Here are the 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 Samy_A Any open set $G$ such that $\mathbb{Q}\subseteq G\subseteq \mathbb{R}$ has either $G=\mathbb{Q}$ or $G=\mathbb{R}$.
2. SOLVED BY ResrupRL Every symmetric matrix in $\mathcal{M}_n(\mathbb{C})$ is diagonalizable for each $n$
3. SOLVED BY stevendaryl Every $C^1$ function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is square-integrable, that is $\int_{-\infty}^{+\infty} |f(x)|^2 dx <+\infty$, has $\lim_{x\rightarrow +\infty} f(x) = 0$.
4. SOLVED BY andrewkirk There is no infinitely differentiable function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x) = 0$ if and only if $x\in \{1/n~\vert~n\in \mathbb{N}\}\cup\{0\}$.
5. SOLVED BY ProfuselyQuarky Every derivative of a differentiable function is continuous.
6. SOLVED BY Samy_A For any $A\subseteq \mathbb{R}$ open holds that if $f^\prime(x) = g^\prime(x)$ for each $x\in A$, then there is some $C\in \mathbb{R}$ such that $f(x) = g(x) + C$.
7. SOLVED BY fresh_42 If $(a_n)_n$ is a sequence such that for each positive integer $p$ holds that $\lim_{n\rightarrow +\infty} a_{n+p} - a_n = 0$, then $a_n$ converges.
8. SOLVED BY jbriggs444 There is no function $f:\mathbb{R}\rightarrow \mathbb{R}$ whose graph is dense in $\mathbb{R}^2$.
9. SOLVED BY andrewkirk If $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a function such that $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ exists and is finite, then both $\lim_{x\rightarrow 0}\lim_{y\rightarrow 0} f(x,y)$ and $\lim_{y\rightarrow 0}\lim_{x\rightarrow 0} f(x,y)$ exist and are finite.
10. SOLVED BY andrewkirk If $A$ and $B$ are connected subsets of $[0,1]\times [0,1]$ such that $(0,0),(1,1)\in A$ and $(1,0), (0,1)\in B$, then $A$ and $B$ intersect.

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!

I will give the source of each statement after the correct answer has been given.

 

[Samy_A]

Nice, some I know, some not.

Let me try 1:
Any open set $G$ such that $\mathbb{Q}\subseteq G\subseteq \mathbb{R}$ has either $G=\mathbb{Q}$ or $G=\mathbb{R}$.
Set $\mathbb{Q} = \{ q_1,q_2,q_3, ...\}$
Let's build an open set $G$ such that $\mathbb{Q}\subseteq G$ that can be written as $G=\cup ]q_n-\epsilon_n,q_n+\epsilon_n[$.
$\mathbb{Q}\subseteq G$ is trivially true. That $G$ is open is also trivial, as $G$ is the union of open sets.
Also, any open interval in $\mathbb R$ will contain irrational numbers, so that $\mathbb{Q} \neq G$.

Now, we don't want $G=\mathbb{R}$.
So let's build the $\epsilon_n$ such that some chosen non rational number is not in $G$.
$\pi$ is not a rational number.
Set $\displaystyle \epsilon_n=\frac{1}{2}\min_{m \leq n} |\pi -q_m|$.
Clearly $\epsilon_n>0$. Also $\pi \notin ]q_n-\epsilon_n,q_n+\epsilon_n[$, since $|\pi- q_n| \geq 2\epsilon_n \gt \epsilon_n$ (and this for all $n$).
Hence $\pi \notin G$.

EDIT: I think my definition of the $\epsilon_n$ is too complicated.
$\displaystyle \epsilon_n=\frac{1}{2}|\pi -q_n|$ would also work.

 

[fresh_42]

Don't hesitate to post any feedback in the thread!

The polite feedback: Great examples and at first glimpse rather tough ones.
The philosophical feedback: Do not take any obvious claims for granted unless proven (esp. in topology).
The optimist's feedback: An entertaining Alzheimer prophylaxis.
The pessimist's feedback: Hell, am I stupid.
The humorous feedback: You've had a topologist for breakfast, haven't you?
The practical feedback: I'm afraid of typing in one of those counterexamples even if I had found them.
The hairsplitter's feedback: I assume (0,0) denotes a point in the plane and not an open interval.
The personal feedback: Those things drive me crazy since I'm strongly attracted to find them out. If it only weren't such a time and paper killer.
The micromass' feedback: I know there is a solution. (ref. to the integral in Roman numbers)
The PF standard feedback: Wouldn't this be a hilarious insight?

 

[fresh_42]

Nice, some I know, some not.

Let me try 1:
Any open set $G$ such that $\mathbb{Q}\subseteq G\subseteq \mathbb{R}$ has either $G=\mathbb{Q}$ or $G=\mathbb{R}$.
Set $\mathbb{Q} = \{ q_1,q_2,q_3, ...\}$
Let's build an open set $G$ such that $\mathbb{Q}\subseteq G$ that can be written as $G=\cup ]q_n-\epsilon_n,q_n+\epsilon_n[$.
$\mathbb{Q}\subseteq G$ is trivially true. That $G$ is open is also trivial, as $G$ is the union of open sets.
Also, any open interval in $\mathbb R$ will contain irrational numbers, so that $\mathbb{Q} \neq G$.

Now, we don't want $G=\mathbb{R}$.
So let's build the $\epsilon_n$ such that some chosen non rational number is not in $G$.
$\pi$ is not a rational number.
Set $\displaystyle \epsilon_n=\frac{1}{2}\min_{m \leq n} |\pi -q_m|$.
Clearly $\epsilon_n>0$. Also $\pi \notin ]q_n-\epsilon_n,q_n+\epsilon_n[$, since $|\pi- q_n| \geq 2\epsilon_n \gt \epsilon_n$ (and this for all $n$).
Hence $\pi \notin G$.

EDIT: I think my definition of the $\epsilon_n$ is too complicated.
$\displaystyle \epsilon_n=\frac{1}{2}|\pi -q_n|$ would also work.

Can't it be that $\pi \in ]q_m - ε_m, q_m+ε_m [$ for some $m ≠ n$? One can find a rational number at each distance to $\pi$.

 

[Samy_A]

Can't it be that $\pi \in ]q_m - ε_m, q_m+ε_m [$ for some $m ≠ n$?

I don't understand what $n$ is in your question.

But I've been somewhat sloppy, let me rewrite it.

Any open set $G$ such that $\mathbb{Q}\subseteq G\subseteq \mathbb{R}$ has either $G=\mathbb{Q}$ or $G=\mathbb{R}$.
Set $\mathbb{Q} = \{ q_1,q_2,q_3, ...\}$
Let's build an open set $G$ such that $\mathbb{Q}\subseteq G$ that can be written as $G=\cup ]q_n-\epsilon_n,q_n+\epsilon_n[$.
$\mathbb{Q}\subseteq G$ is trivially true. That $G$ is open is also trivial, as $G$ is the union of open sets.
Also, any open interval in $\mathbb R$ will contain irrational numbers, so that $\mathbb{Q} \neq G$.

Now, we don't want $G=\mathbb{R}$.
So let's build the $(\epsilon_n)_n$ such that some chosen non rational number is not in $G$.
$\pi$ is not a rational number.
$\forall n \geq 1$, set $\displaystyle \epsilon_n=\frac{1}{2}|\pi -q_n|$
$\forall n \geq 1$, clearly $\epsilon_n>0$.
Also, $\forall n \geq 1$, $\pi \notin ]q_n-\epsilon_n,q_n+\epsilon_n[$, since $|\pi- q_n| = 2\epsilon_n > \epsilon_n$. Hence $\pi \notin G$.

 

[fresh_42]

Thanks. The ∀ quantor makes it clear. I was distrait by the minimum which made me somehow "fix" the n. Sorry. Nice solution. Do you have any idea about the square problem?

 

[TeethWhitener]

Maybe I'm thinking #1 is easier than it actually is, but why not something as simple as $G=(-\infty , \pi)\cup(\pi,\infty)$? In other words, $G = \mathbb{R}/\{\pi\}$. It's open because it's a union of open sets. Since $\mathbb{Q} \subset \mathbb{R}$ and the only number in $\mathbb{R}$ which is excluded from $G$ (that is, $\pi$) is irrational (proof not included ), the second definition shows that $\mathbb{Q} \subseteq G$. Since irrational numbers like $2\pi$ are in $G$, $G \neq \mathbb{Q}$. Since $\pi \in \mathbb{R}$ is explicitly excluded from $G$, $G \neq \mathbb{R}$. Is there something wrong with this example?

 

[micromass]

Maybe I'm thinking #1 is easier than it actually is, but why not something as simple as $G=(-\infty , \pi)\cup(\pi,\infty)$? In other words, $G = \mathbb{R}/\{\pi\}$. It's open because it's a union of open sets. Since $\mathbb{Q} \subset \mathbb{R}$ and the only number in $\mathbb{R}$ which is excluded from $G$ (that is, $\pi$) is irrational (proof not included ), the second definition shows that $\mathbb{Q} \subseteq G$. Since irrational numbers like $2\pi$ are in $G$, $G \neq \mathbb{Q}$. Since $\pi \in \mathbb{R}$ is explicitly excluded from $G$, $G \neq \mathbb{R}$. Is there something wrong with this example?

No, this is the easiest example, but Samy_A's example works too and he was first, so the credits go to him

 

[Samy_A]

Thanks. The ∀ quantor makes it clear. I was distrait by the minimum which made me somehow "fix" the n. Sorry. Nice solution. Do you have any idea about the square problem?

No.
I have a guess, but I will put it in a spoiler so as to not distract or wrongfoot people who want to try it.

Spoiler: Guess

Maybe the Topologist's sine curve can be used to build a counterexample.

 

[Samy_A]

Maybe I'm thinking #1 is easier than it actually is, but why not something as simple as $G=(-\infty , \pi)\cup(\pi,\infty)$? In other words, $G = \mathbb{R}/\{\pi\}$. It's open because it's a union of open sets. Since $\mathbb{Q} \subset \mathbb{R}$ and the only number in $\mathbb{R}$ which is excluded from $G$ (that is, $\pi$) is irrational (proof not included ), the second definition shows that $\mathbb{Q} \subseteq G$. Since irrational numbers like $2\pi$ are in $G$, $G \neq \mathbb{Q}$. Since $\pi \in \mathbb{R}$ is explicitly excluded from $G$, $G \neq \mathbb{R}$. Is there something wrong with this example?

Of course!

 

[ProfuselyQuarky]

The practical feedback: I'm afraid of typing in one of those counterexamples even if I had found them.

That's me I've got a killer counterexample for #5 but heck no way am I going to post it here.

 

[TeethWhitener]

No, this is the easiest example, but Samy_A's example works too and he was first, so the credits go to him

Absolutely. I just wanted to make sure I wasn't missing something fundamental.

 

[Samy_A]

That's me I've got a killer counterexample for #5 but heck no way am I going to post it here.

Why is that?

 

[ProfuselyQuarky]

Why is that?

Reason #1: I’ve got like 10 minutes before leaving for my foreign language class and I stink at LaTeX, so it’ll take me a long while to figure out how to type it all up. It would take even longer to scan my scratch paper and upload it.

Reason #2: I’m 99.9% sure my counterexample is legitimate, but my fear is always “I probably missed something and so it’s probably wrong and someone will probably point out the mistake in a sarcastic way that’s meant to be funny but I'll take the wrong way and won’t find it funny”

And now, I shall log off . . .

 

[fresh_42]

7. $a_n = \sum_{k=1}^{k=n} \frac{1}{k}$. But I have the same difficulty as TeethWhitener. What am I missing?

 

[micromass]

7. $a_n = \sum_{k=1}^{k=n} \frac{1}{k}$. But I have the same difficulty as TeethWhitener. What am I missing?

Nothing. That is the right counterexample.

 

[micromass]

Reason #1: I’ve got like 10 minutes before leaving for my foreign language class and I stink at LaTeX, so it’ll take me a long while to figure out how to type it all up. It would take even longer to scan my scratch paper and upload it.

Reason #2: I’m 99.9% sure my counterexample is legitimate, but my fear is always “I probably missed something and so it’s probably wrong and someone will probably point out the mistake in a sarcastic way that’s meant to be funny but I'll take the wrong way and won’t find it funny”

And now, I shall log off . . .

Come on, post it!

 

[jbriggs444]

There is no function $f:\mathbb{R}\rightarrow \mathbb{R}$ whose graph is dense in $\mathbb{R}^2$

Consider the set of canonical decimal expansions such that the first set of n digits after the decimal are arbitrary, the next n digits are all zero and the remaining digits are all non-zero for some strictly positive integer n. For any given decimal it is easy to decide whether it is of this form. Further, for any number of this form, there is only one value of n for which the constraint will hold. It is clear that reals with canonical decimals of this form are dense in the reals.

Consider the trailing non-zero digits in decimals of this form. This is an infinite string of digits in the range 1-9 and can be used to arbitrarily encode information of one's choice. The remainder of the challenge is simply dotting the i's and crossing the t's.

Let the first digit encode a sign. 1 => +1. 2 => -1. 3-9 => 0. Call this $S$.

Let the second digit encode a sign on the exponent. 1 => +1, 2 => -1. 3-9 => zero. Call this $s$

Let the next digits encode an exponent. 9 = end of sub-string, subtract 1 from each of the intervening digits to produce the octal expansion of a non-negative integer. If there is no 9 in the string, the exponent is taken as zero. If a 9 appears immediately, the exponent is taken as zero. Call this $e$

Subtract 1 from each of the remaining digits to produce the base nine fraction with the radix point assumed to the left of the first digit. If there are no remaining digits (e.g. if the trailing non-zero digits have no 9) then this mantissa is taken as zero. Call it $m$.

Let f(x) = x if the canonical decimal expansion of x fails to take the requisite form.
Let f(x) = $Sm\cdot9^{se}$ if the canonical decimal expansion of x does take the requisite form.

For any given (x,y) pair and any given tolerance it is clear that we can find infinitely many base 9 floating point values within that tolerance of y. And it is also clear that we can find infinitely many reals within the tolerance of x with canonical decimal expansions that encode such a base 9 floating point value. Which is to say that the graph of f is dense in R^2.

 

[micromass]

Consider the set of canonical decimal expansions such that the first set of n digits after the decimal are arbitrary, the next n digits are all zero and the remaining digits are all non-zero for some strictly positive integer n. For any given decimal is is easy to decide whether it is of this form. Further, for any number of this form, there one value of n for which the constraint will hold. It is clear that reals with canonical decimals of this form are dense in the reals.

Consider the trailing non-zero digits in decimals of this form. This is an infinite string of digits in the range 1-9 and can be used to arbitrarily encode information of one's choice. The remainder of the challenge is simply dotting the i's and crossing the t's.

Let the first digit encode a sign. 1 => +1. 2 => -1. 3-9 => 0. Call this $S$.

Let the second digit encode a sign on the exponent. 1 => +1, 2 => -1. 3-9 => zero. Call this $s$

Let the next digits encode an exponent. 9 = end of sub-string, subtract 2 from each of the intervening digits to produce the octal expansion of a non-negative. If there is no 9 in the string, the exponent is taken as zero. If a 9 appears immediately, the exponent is taken as zero. Call this $e$

Subtract 1 from each of the remaining digits to produce the base nine fraction with the radix point assumed to the left of the first digit. If there are no remaining digits (e.g. if the trailing non-zero digits have no 9) then this mantissa is taken as zero. Call it $m$.

Let f(x) = x if the canonical decimal expansion of x fails to take the requisite form.
Let f(x) = $Sm\cdot9^{se}$ if the canonical decimal expansion of x does take the requisite form.

For any given (x,y) pair and any given tolerance it is clear that we can find infinitely many base 9 floating point values within that tolerance of y. And it is also clear that we can find infinitely many reals within the tolerance of x with canonical decimal expansions that encode such a base 9 floating point value. Which is to say that the graph of f is dense in R^2.

Very beautiful solution!

As an alternative, every function $f:\mathbb{R}\rightarrow \mathbb{R}$ which satisfies $f(x+y) = f(x) + f(y)$ but is not linear/continuous/Lebesgue measurable/of the form $f(x) = \alpha x$ (these four are equivalent), has a graph dense in $\mathbb{R}^2$.
Source: Herrlich "Axiom of Choice" Chapter 5.1

 

[Samy_A]

For any $A\subseteq \mathbb{R}$ open holds that if $f^\prime(x) = g^\prime(x)$ for each $x\in A$, then there is some $C\in \mathbb{R}$ such that $f(x) = g(x) + C$.

Not sure I interpret this correctly, but anyway.
$A=]0,1[\cup ]2,3[$.
$f: A \to \mathbb R: x \mapsto x$
$g: A \to \mathbb R: x \mapsto x$ for $x \in ]0,1[$, $x \mapsto x + 1$ for $x \in ]2,3[$.
$\forall x\in A: f'(x)=g'(x)=1$, but obviously there is no $C\in \mathbb{R}$ such that $f(x) = g(x) + C$.

 

[micromass]

Not sure I interpret this correctly, but anyway.
$A=]0,1[\cup ]2,3[$.
$f: A \to \mathbb R: x \mapsto x$
$g: A \to \mathbb R: x \mapsto x$ for $x \in ]0,1[$, $x \mapsto x + 1$ for $x \in ]2,3[$.
$\forall x\in A: f'(x)=g'(x)=1$, but obviously there is no $C\in \mathbb{R}$ such that $f(x) = g(x) + C$.

Right!

 

[TheDemx27]

For number six, the fact that there is a counterexample is especially counter-intuitive. I'd like to see that one solved.
edit: oh wait it has been...

 

[Samy_A]

I'm curious for 10.

 

[micromass]

For number six, the fact that there is a counterexample is especially counter-intuitive. I'd like to see that one solved.

It just did. It's actually a pet peeve of mine. I see a lot of people writing
\int \frac{1}{x}dx = \log|x| + C
while this is completely incorrect. The most general primitive of $1/x$ is not $\log|x| + C$, but rahter
\log|x|+ \left\{\begin{array}{l} C_1 ~\text{if}~x&lt;0\\ C_2 ~\text{if}~x&gt;0 \end{array}\right.
Where is the issue? The usual statement for $6$ comes from the mean-value theorem which only holds for connected domains (i.e. intervals). It does not hold when the domain of the function is not an interval, like the maximal domain for $1/x$.

I really don't get why calculus books don't pay attention to this and even make errors against it. You'd think it's harmless, but I've seen a number of threads here by people who were confused by something that is essentially this.

 

[JorisL]

I really don't get why calculus books don't pay attention to this and even make errors against it. You'd think it's harmless, but I've seen a number of threads here by people who were confused by something that is essentially this.

Oh wow, I never knew that this "problem" exists. Never thought about it to be honest and indeed the calculus book I used (Adams) did that if I'm not mistaken.

 

[stevendaryl]

I think I know how to solve #3.

Spoiler: The answer...

Suppose we let f^2(x) = \sum_{n=0}^{\infty} e^{-\lambda_n (x-n)^2}, where \lambda_n is a sequence of constants. (So f(x) is the square-root of the sum). Then it would certainly not be the case that lim_{x \rightarrow +\infty} f(x) = 0. But the integral might still be finite:

\int dx f^2(x) = \sum_{n=0}^{\infty} \int dx e^{-\lambda_n (x-n)^2} = \sum_{n=0}^{\infty} \sqrt{\frac{\pi}{\lambda_n}}

This sum converges, provided that \lambda_n grows sufficiently rapidly. For example, \lambda_n = 2^n.

So my answer is: f(x) = \sqrt{ \sum_{n=0}^{\infty} e^{-2^n (x-n)^2}}

You'd have to prove that f(x) is well-defined and differentiable, but I think it's true.

 

[ProfuselyQuarky]

Well, *sniff* here you go . . .

(for #5)

All you need to do is find a derivative of a differentiable function that is discontinuous.

At first I had in mind $f(x)=|x|$, but $|x|$ is not differentiable at the point 0 and actually, this only occurs in isolated points apparently. So, here's a piecewise function:

$f(x) = \left\{ \begin{array}{ll} x^2\sin(\frac {1}{x}) & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$

For this, $f'(x)=2x\sin \frac {1}{x} - \cos \frac {1}{x}$ whenever $x\neq0$.

I find the derivative by solving for the limit. $1\geq|\sin x|$ per $x$, thus the simplified limit is $\lim_{x\to0} x\sin \frac {1}{x}$ or, simply, 0. This, in turn, gives us:

$f'(x) = \left\{ \begin{array}{ll} 2x\sin \frac {1}{x} - \cos \frac {1}{x} & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$

It's a basic sine curve with $\lim_{x\to0} f'(x)$. This holds true for real numbers but the function diverges. Thus, we have a derivative of a differentiable function that is discontinuous.

Okay, now you all can tell me what's wrong with it

 

[ProfuselyQuarky]

P.S. Can someone give me the code to properly type a limit without subscripts?

 

[Samy_A]

P.S. Can someone give me the code to properly type a limit without subscripts?

Do you mean this: $\displaystyle \lim_{x\to0} x\sin \frac {1}{x}$?

 

[ProfuselyQuarky]

Do you mean this: $\displaystyle \lim_{x\to0} x\sin \frac {1}{x}$?

Yeah, exactly! I can only get this: $\lim_{x\to0} x\sin \frac {1}{x}$ With the code:

\lim_{x\to0} x\sin \frac {1}{x}

 

[micromass]

Yeah, exactly! I can only get this: $\lim_{x\to0} x\sin \frac {1}{x}$ With the code:

\lim_{x\to0} x\sin \frac {1}{x}

Add \displaymath before the limit.

 

[Samy_A]

Yeah, exactly! I can only get this: $\lim_{x\to0} x\sin \frac {1}{x}$ With the code:

\lim_{x\to0} x\sin \frac {1}{x}

As you can see in my previous post, adding the \displaystyle tag makes some LaTeX expressions nicer.

Well, *sniff* here you go . . .
...
$f(x) = \left\{ \begin{array}{ll} 2x\sin \frac {1}{x} - \cos \frac {1}{x} & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$

It's a basic sine curve with $\lim_{x\to0} f'(x)$. This holds true for real numbers but the function diverges. Thus, we have a derivative of a differentiable function that is discontinuous.

Okay, now you all can tell me what's wrong with it

Looks good to me, let's see what micromass thinks. There is a little inconsequential typo, as you wrote $f(x)$ instead of $f'(x)$.

 

[Isaac0427]

Every derivative of a differentiable function is continuous.

$y=\frac{x}{x}$ is differentiable, and
it's derivative, using the quotient rule, is $\frac{0}{x^2}$, which is undefined at x=0 and thus not continuous.

 

[ProfuselyQuarky]

Add \displaymath before the limit.

As you can see in my previous post, adding the \displaystyle tag makes some LaTeX expressions nicer.

$\displaystyle \lim_{x\to0} x\sin \frac {1}{x}$ works but $\displaymath \lim_{x\to0} x\sin \frac {1}{x}$ doesn't. Thanks @Samy_A!

Looks good to me, let's see what micromass thinks. There is a little inconsequential typo, as you wrote f(x)f(x) instead of f′(x)f'(x).

Yeah, sorry

 

[micromass]

Well, *sniff* here you go . . .

(for #5)

All you need to do is find a derivative of a differentiable function that is discontinuous.

At first I had in mind $f(x)=|x|$, but $|x|$ is not differentiable at the point 0 and actually, this only occurs in isolated points apparently. So, here's a piecewise function:

$f(x) = \left\{ \begin{array}{ll} x^2\sin(\frac {1}{x}) & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$

For this, $f'(x)=2x\sin \frac {1}{x} - \cos \frac {1}{x}$ whenever $x\neq0$.

I find the derivative by solving for the limit. $1\geq|\sin x|$ per $x$, thus the simplified limit is $\lim_{x\to0} x\sin \frac {1}{x}$ or, simply, 0. This, in turn, gives us:

$f(x) = \left\{ \begin{array}{ll} 2x\sin \frac {1}{x} - \cos \frac {1}{x} & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$

It's a basic sine curve with $\lim_{x\to0} f'(x)$. This holds true for real numbers but the function diverges. Thus, we have a derivative of a differentiable function that is discontinuous.

Okay, now you all can tell me what's wrong with it

That is correct!

 

[micromass]

$y=\frac{x}{x}$ is differentiable, and
it's derivative, using the quotient rule, is $\frac{0}{x^2}$, which is undefined at x=0 and thus not continuous.

Undefined at $0$ doesn't mean discontinuous at $0$

 

[micromass]

I think I know how to solve #3.

Spoiler: The answer...

Suppose we let f^2(x) = \sum_{n=0}^{\infty} e^{-\lambda_n (x-n)^2}, where \lambda_n is a sequence of constants. (So f(x) is the square-root of the sum). Then it would certainly not be the case that lim_{x \rightarrow +\infty} f(x) = 0. But the integral might still be finite:

\int dx f^2(x) = \sum_{n=0}^{\infty} \int dx e^{-\lambda_n (x-n)^2} = \sum_{n=0}^{\infty} \sqrt{\frac{\pi}{\lambda_n}}

This sum converges, provided that \lambda_n grows sufficiently rapidly. For example, \lambda_n = 2^n.

So my answer is: f(x) = \sqrt{ \sum_{n=0}^{\infty} e^{-2^n (x-n)^2}}

You'd have to prove that f(x) is well-defined and differentiable, but I think it's true.

Cool idea, but I can't really accept this solution until some gaps are filled:

• Prove $f(x)$ is well-defined
• Prove $f(x)$ is differentiable.
• Show why $\lim_{x\rightarrow +\infty} f(x) =0$
• Show that the interchange of series and integral is justified.

 

[Isaac0427]

Undefined at $0$ doesn't mean discontinuous at $0$

lim_x→0f(x)≠f(0) If f(x)=$\frac{0}{x^2}$.
I know that @ProfuselyQuarky solved it before me, just want to know why I'm wrong.

 

[micromass]

lim_x→0f(x)≠f(0) If f(x)=$\frac{0}{x^2}$.
I know that @ProfuselyQuarky solved it before me, just want to know why I'm wrong.

Because $0$ is not in the domain of your function. It's true that $f$ is continuous at $0$ iff $\lim_{x\rightarrow 0}f(x) = f(0)$, but that is only when $0$ is actually in the domain of your function.

 

[Isaac0427]

Because $0$ is not in the domain of your function. It's true that $f$ is continuous at $0$ iff $\lim_{x\rightarrow 0}f(x) = f(0)$, but that is only when $0$ is actually in the domain of your function.

Ah ok.

 

[stevendaryl]

Cool idea, but I can't really accept this solution until some gaps are filled:

• Prove $f(x)$ is well-defined
• Prove $f(x)$ is differentiable.
• Show why $\lim_{x\rightarrow +\infty} f(x) =0$
• Show that the interchange of series and integral is justified.

Oh, bother! The idea is to have a function that is basically a bunch of "spikes" that always have height 1, but whose widths decrease. Gaussians have that property, but working with them is a little bit of a pain. But here's another choice:

First define g(x) as follows:

g(x) = 0 if x &lt; -\frac{1}{2}
g(x) = 1 + cos(2\pi x) if -\frac{1}{2} &lt; x &lt; +\frac{1}{2}
g(x) = 0 if x &gt; \frac{1}{2}

Then g(x) is continuous and once-differentiable.

Then \int_{-\infty}^{+\infty} g(\lambda x) dx = \frac{1}{\lambda} \int_{-\frac{1}{2}}^{+\frac{1}{2}} (1+cos(2\pi x)) dx = \frac{1}{\lambda}

Let \lambda_n be any sequence of real numbers greater than 1 such that \sum_n \frac{1}{\lambda_n} &lt; \infty

Define g_n(x) = g(\lambda_n (x-n)). Then for any n \neq m, either g_n(x) = 0 or g_m(x) = 0 or both. Furthermore, g_n(n+\frac{1}{2}) = g_{n+1}(n+\frac{1}{2}) = 0.

Let N[x] = the smallest n such that n+\frac{1}{2} &gt; x

So now define q(x) = g_{N[x]}(x), where n is the smallest number such that n+\frac{1}{2} &gt; x. Then q(x) is continuous: For x in the range n-\frac{1}{2} &lt; x &lt; n+\frac{1}{2}, q(x) = g_n(x), which is continuous. At x=n+\frac{1}{2}, we have:

q(x-\epsilon) = g_n(x-\epsilon) = 0
q(x+\epsilon) = g_{n+1}(x+\epsilon) = 0

q(x) is also once-differentiable, since g_n(x) is, and the derivatives are zero at the boundary.

Now, \int_{-\infty}^{+\infty} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} g_n(x) dx = \sum_n \frac{1}{\lambda_n} &lt; \infty

 

[micromass]

Oh, bother! The idea is to have a function that is basically a bunch of "spikes" that always have height 1, but whose widths decrease. Gaussians have that property, but working with them is a little bit of a pain. But here's another choice:

Alright, I can accept that answer!

 

[andrewkirk]

Here's my attempt at item 10. It uses two topologists' sine curves pasted back to back, and takes out the midpoint (0,0) from the central strut in order to connect the complement of the curve.

Consider the quadrilateral Q in $\mathbb{R}^2$ with vertices a=(-1,0),c=(1,0),b=(0,2),d=(0,-2).

Let M be the set $\{(x,y)\ :\ x\in[-1,1]\wedge \ y=(|x|-1)\sin\frac1x\}$. This is the wiggly part of a topologists' sine curve, together with its reflection in the y axis. The factor $(|x|-1)$ scales it down away from the pathological region, to make it fit inside the corners of Q.
Let N be the set $\{0,y)\ :\ y\in[-1,1]-\{0\}\}$. This is the 'terminal strut' of both sine curves, which becomes the 'central strut' of the pair.
Let $A =M\cup N$.
Then I think A and its complement are both connected subsets of Q such that A contains diagonally opposite corners a and c, while ~A contains diagonally opposite corners b and d.
Applying the natural homeomorphism mapping Q to the quadrilateral (0,0)(0,1)(1,1)(1,0) (multiply vertical coordinate by 0.5, rotate anti clockwise around origin by 45 degrees, divide all coordinates by $\sqrt 2$, then add (0.5,0.5) to all coords) gives us the desired counterexample.

 

[stevendaryl]

Oh, bother! The idea is to have a function that is basically a bunch of "spikes" that always have height 1, but whose widths decrease. Gaussians have that property, but working with them is a little bit of a pain. But here's another choice:

First define g(x) as follows:

g(x) = 0 if x &lt; -\frac{1}{2}
g(x) = 1 + cos(2\pi x) if -\frac{1}{2} &lt; x &lt; +\frac{1}{2}
g(x) = 0 if x &gt; \frac{1}{2}

Then g(x) is continuous and once-differentiable.

Then \int_{-\infty}^{+\infty} g(\lambda x) dx = \frac{1}{\lambda} \int_{-\frac{1}{2}}^{+\frac{1}{2}} (1+cos(2\pi x)) dx = \frac{1}{\lambda}

Let \lambda_n be any sequence of real numbers greater than 1 such that \sum_n \frac{1}{\lambda_n} &lt; \infty

Define g_n(x) = g(\lambda_n (x-n)). Then for any n \neq m, either g_n(x) = 0 or g_m(x) = 0 or both. Furthermore, g_n(n+\frac{1}{2}) = g_{n+1}(n+\frac{1}{2}) = 0.

Let N[x] = the smallest n such that n+\frac{1}{2} &gt; x

So now define q(x) = g_{N[x]}(x), where n is the smallest number such that n+\frac{1}{2} &gt; x. Then q(x) is continuous: For x in the range n-\frac{1}{2} &lt; x &lt; n+\frac{1}{2}, q(x) = g_n(x), which is continuous. At x=n+\frac{1}{2}, we have:

q(x-\epsilon) = g_n(x-\epsilon) = 0
q(x+\epsilon) = g_{n+1}(x+\epsilon) = 0

q(x) is also once-differentiable, since g_n(x) is, and the derivatives are zero at the boundary.

Now, \int_{-\infty}^{+\infty} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} g_n(x) dx = \sum_n \frac{1}{\lambda_n} &lt; \infty

Actually, I didn't give the answer, which is not q(x) but \sqrt{q(x)}.

 

[andrewkirk]

Why do I think my answer to 10 works? Well the argument as to why a topologists' sine curve is connected is known. The wiggly bit is clearly connected and any neighbourhood of any point on the strut must contain infinitely many points on the wiggly bit. So every point on the strut must be connected to the wiggly bit. Taking away the point (0,0) doesn't change this argument. Sticking another wiggly bit on the other side, we see that any neighbourhood of any point on the strut contains infinitely many points of both wiggly bits. So the two wiggly bits must each be in the same component as points on the strut, and hence in the same component as each other, so the set is connected.

As for the complement, consider the set X consisting of all points in Q below (having lower y coordinate than) a point on a wiggly bit, and the set Y consisting of all points in Q above (having greater y coordinate than) a point on a wiggly bit.
Define $X'=X\cup\{(0,y)\ :\ y\in[-2,-1)\}$ and $Y'=Y\cup\{(0,y)\ :\ y\in(1,2]\}$
It is easy to see that X' and Y' are connected, and that the complement of A is $X'\cup Y'\cup\{(0,0)\}$.
But it is easily seen that any neighbourhood of (0,0) contains infinitely many points of both X' and Y', so those three parts must all be in the same component, and hence the union, which is the complement of A, is connected.

 

[stevendaryl]

Why do I think my answer to 10 works? Well the argument as to why a topologists' sine curve is connected is known.

Yeah, but to me it means that the definition of "connected" isn't quite right. An alternative definition is "path connected", which is that you can from any point to any other point through a continuous one-dimensional path.

 

[andrewkirk]

Yeah, but to me it means that the definition of "connected" isn't quite right. An alternative definition is "path connected", which is that you can from any point to any other point through a continuous one-dimensional path.

Well that's the point of the counterexample (and the point of the topologists' sine curve): to exemplify the difference between path connectedness and mere connectedness. I think it would be straightforward to prove that the proposition is true (ie there is no counterexample) if one replaces 'connected' by 'path-connected'.

But I agree that 'path-connected' is the best match to the non-technical, everyday notion of what 'connected' means.

 

[andrewkirk]

Here's my attempt at number 4

4.There is no infinitely differentiable function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x) = 0$ if and only if $x\in \{1/n~\vert~n\in \mathbb{N}\}\cup\{0\}$

The function $f:\mathbb{R}\to\mathbb R$ such that, for $x>0$:
$f(x)=e^{-1/x^2}\sin\frac{\pi}{x}$
We set $f(0)=0$ and, for $x<0$ we set
$f(x)=\int_0^{|x|}B(t)dt$ where $B$ is a Bump Function with support (0,1).

This is clearly infinitely differentiable away from 0. Examination of cases shows that it is infinitely differentiable at 0 (I'll expand on that later when I have more time). It has no negative roots, and it has positive roots at values of 1/n for natural n.

 

[andrewkirk]

And lastly, my attempt at 9 - the only one without an attempted or confirmed solution yet.

9. If $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a function such that $\lim_{(x,y)\rightarrow (0,0)} f(x,y)$ exists and is finite, then both $\lim_{x\rightarrow 0}\lim_{y\rightarrow 0} f(x,y)$ and $\lim_{y\rightarrow 0}\lim_{x\rightarrow 0} f(x,y)$ exist and are finite.

Define $f:\mathbb{R}^2\to\mathbb{R}$ by
$f(x,y)=\sqrt{x^2+y^2}$ if $x,y$ are both rational, and $2\sqrt{x^2+y^2}$ otherwise.

Then it's easy to show that $\lim_{(x,y)\to(0,0)}f(x,y)=0$
But $\lim_{x\to0}f(x,y)$ and $\lim_{y\to0}f(x,y)$ do not exist except for where $y=0$ and $x=0$ respectively, because the limit of the former over the rationals is $|y|$ and its limit over the irrationals is $2|y|$. The former failure prevents $\lim_{y\rightarrow 0}\lim_{x\rightarrow 0} f(x,y)$ from existing and the latter failure prevents $\lim_{x\rightarrow 0}\lim_{y\rightarrow 0} f(x,y)$ from existing.

 

[andrewkirk]

I think it would be straightforward to prove that the proposition is true (ie there is no counterexample) if one replaces 'connected' by 'path-connected'.

A proof occurred to me. At first I thought we could use the intermediate value theorem on the two paths, until I realized that they don't necessarily map out the locus of functions from $\mathbb R$ to $\mathbb R$. We can adapt the curves to remove 'turn-backs' so they become functions, but we lose continuity, which is necessary to use the IVT.

So instead we use the Jordan Curve Theorem. Say there's a continuous path in the unit square (0,0)(0,1)(1,1)(1,0) from (1,0) to (0,1), that does not pass through (1,1). Then combining that with the circular arc formed by the three-quarters of the unit circle that falls outside the unit square, we get a simple closed curve in the number plane that has the origin as an interior point and (1,1) as an exterior point. Jordan tells us that this curve divides the number plane into two disjoint open sets. Since the union U of those sets is not connected it is not path-connected (which is a stronger condition) either. So there can be no path from (0,0) to (1,1) that lies entirely inside U. Hence any path must cross the boundary of U at some point. If it crosses it on the circular part then that path is disqualified from consideration as it lies outside the unit square. So it must cross it on the part inside the square, which is the path between the two other diagonally-opposite vertices. So any two paths, within the square, between pairs of opposite vertices must cross.

Now say we have subsets A and B of the square, both path-connected, with each containing a different pair of diagonally-opposite vertices of the square. Then A and B each contain a path within the square between its pair of diagonally-opposite vertices. But those paths must cross. So A and B must share a point.