Challenge Micromass' big counterexample challenge

[ResrupRL]

#2 counterexample
Let
<br /> A=\left[\begin{array}{cc}<br /> i &amp; 1 \\<br /> 1 &amp; -i<br /> \end{array} \right] <br />
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
<br /> \mathbf{x}=\left[\begin{array}{c}<br /> i \\<br /> 1<br /> \end{array}<br /> \right]<br />
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. :)