Micromass' big counterexample challenge

[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 $x < -\frac{1}{2}$
$g(x) = 1 + cos(2\pi x)$ if $-\frac{1}{2} < x < +\frac{1}{2}$
$g(x) = 0$ if $x > \frac{1}{2}$

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 $\sum_n \frac{1}{\lambda_n} < \infty$

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 $n+\frac{1}{2} > x$

So now define $q(x) = g_{N[x]}(x)$, where $n$ is the smallest number such that $n+\frac{1}{2} > x$. Then $q(x)$ is continuous: For $x$ in the range $n-\frac{1}{2} < x < n+\frac{1}{2}$, $q(x) = g_n(x)$, 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, $\int_{-\infty}^{+\infty} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} g_n(x) dx = \sum_n \frac{1}{\lambda_n} < \infty$

 

[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 $x < -\frac{1}{2}$
$g(x) = 1 + cos(2\pi x)$ if $-\frac{1}{2} < x < +\frac{1}{2}$
$g(x) = 0$ if $x > \frac{1}{2}$

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 $\sum_n \frac{1}{\lambda_n} < \infty$

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 $n+\frac{1}{2} > x$

So now define $q(x) = g_{N[x]}(x)$, where $n$ is the smallest number such that $n+\frac{1}{2} > x$. Then $q(x)$ is continuous: For $x$ in the range $n-\frac{1}{2} < x < n+\frac{1}{2}$, $q(x) = g_n(x)$, 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, $\int_{-\infty}^{+\infty} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} q(x) dx = \sum_n \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} g_n(x) dx = \sum_n \frac{1}{\lambda_n} < \infty$

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.

 

[ResrupRL]

#2 counterexample
Let
$$A=\left[\begin{array}{cc} i & 1 \\ 1 & -i \end{array} \right]$$
where #i# is the imaginary value. Note that $A^T=A$ and has the characteristic equation$\lambda^2=0$,
thus the eigenvalue of $A$ is $\lambda=0$ with algebraic multiplicity of $2$. The eigenvector of $A$ is
$$\mathbf{x}=\left[\begin{array}{c} i \\ 1 \end{array} \right]$$
with the geometric multiplicity of $1$ therefore $A$ is not diagonalizable.

 

[micromass]

They're all solved! Thanks everybody for participating. I hope you had fun!

As for sources:
(3) is from Gieres: http://arxiv.org/abs/quant-ph/9907069
(4) is from http://math.stackexchange.com/quest...eq-mathbbrn-is-the-zero-point-set-of-a-smooth
(7), (8), (9) are from Gelbaum, Olmsted "Counterexamples in Analysis"
(10) was first formulated by HallsofIvyhttps://[URL="https://www.physicsforums.com/threads/connected-sets.109700/"]www.physicsforums.com/threads/connected-sets.109700/[/URL]

 

[mfb]

I missed this thread :(.

Can you add links to the posted solutions to the first post?

 

[fresh_42]

I missed this thread :(.

Can you add links to the posted solutions to the first post?

There's a second part: https://www.physicsforums.com/threads/yet-another-counterexample-challenge.869304/

 

[andrewkirk]

Really great thread by the way Micromass! Those counterexamples were fun.

 

[Abdullah Naeem]

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.

Misconception alert!

Let $a_n$ be such that for each positive integer $p$, $\lim_{n\rightarrow +\infty} a_{n+p} - a_n = 0$
Since the modulus operation is a norm on the real numbers, we have equivalence with $| \; \lim_{n\rightarrow +\infty} a_{n+p} - a_n \;| = 0$. Since the modulus operation on the real numbers is a continuous function, we can write this as $\lim_{n\rightarrow +\infty} | \; a_{n+p} - a_n \;| = 0$. Is this the same as "for all $\epsilon >0$, there exists $N,p$ such that $d(x_{n+p},x_n)< \epsilon$ whenever $n \geq N$ "? Because if it is, then what you're asking for is when does a Cauchy sequence not converge. I'm a little confused since we know that the Harmonic series is not Cauchy but it satisfies your counter-example. What am I missing? And why do you have an extra $n$ in the subscript? I'm used to $(a_n)$.

 

[mfb]

Is this the same as "for all $\epsilon >0$, there exists $N,p$ such that $d(x_{n+p},x_n)< \epsilon$ whenever $n \geq N$ "?

It is not. The order of $\epsilon$, N and p is different, and this difference matters.

And you got the logic wrong for the Cauchy convergence.

 

[WWGD]

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.

Doesn't this work? : Consider any irrational x. As singletons ( and finite sets ) are open, then consider its complement in the Real line: $\mathbb R$- { $x$}, which is open as the complement of the closed set { $x$}? And, since we only removed an irrational, $\mathbb Q$ is contained in it. This gives us uncountably-many (counter) examples.

 

[WWGD]

Doesn't the case for no function $f: \mathbb R \rightarrow \mathbb R$ follow from Sard's theorem , assuming the graph is the set { $(x,f(x)$}? Sard's says that the image of the set of critical points of a map $f: \mathbb R^n \rightarrow \mathbb R^m$ has measure zero. If n < m, ( as, here, we are mapping the Reals into the subset $(x,f(x) \subset \mathbb R^2$, then the image $f(\mathbb R )$ has measure zero in $\mathbb R^2$? This means that differentiable functions are out, which may be somewhat obvious.

 

[mfb]

Doesn't this work? : Consider any irrational x. As singletons ( and finite sets ) are open, then consider its complement in the Real line: $\mathbb R$- { $x$}, which is open as the complement of the closed set { $x$}? And, since we only removed an irrational, $\mathbb Q$ is contained in it. This gives us uncountably-many (counter) examples.

That is basically what @Samy_A constructed, just with a much shorter argument.

 

[WWGD]

That is basically what @Samy_A constructed, just with a much shorter argument.

Didn't he remove whole intervals? I am saying:
Take x irrational, then $\mathbb R$ -{$x$} is your desired open set .

EDIT: It is open as the complement of the closed set {x}, and clearly $\mathbb Q \subset G \subset \mathbb R$.

 

[mfb]

Didn't he remove whole intervals?

If you take the union of all his sets (to get the final set) you should get the same with both approaches.

 

[zwierz]

I hope some of these statements were surprising to some of you

o yes, especially for those who did not read Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis. :)