User dejan govc - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:47:12Z http://mathoverflow.net/feeds/user/16447 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78129/a-brouwer-fixed-point-theorem-on-finite-sets/78137#78137 Answer by Dejan Govc for A Brouwer fixed point theorem on finite sets Dejan Govc 2011-10-14T13:48:59Z 2011-10-14T13:48:59Z <p>I believe the following is a counter-example:</p> <p>$f: \lbrace -1,0,1\rbrace^2 \to \lbrace -1,0,1\rbrace^2$</p> <p>$\forall x:$</p> <p>$f(-1,x) = (1,x)$</p> <p>$f(0,x)=(1,x)$</p> <p>$f(1,x)=(0,x)$</p> http://mathoverflow.net/questions/77446/what-is-known-about-ulams-problem-of-metric-spaces-with-isometric-squares What is known about Ulam's problem of metric spaces with isometric squares? Dejan Govc 2011-10-07T12:17:15Z 2011-10-08T22:39:37Z <h2>Background</h2> <p>In the book Problems in Modern Mathematics, S. Ulam asks the following question:</p> <p>Suppose $A$ and $B$ are metric spaces, such that $A^2$ and $B^2$ equipped with the 2-metric $d((x_1, y_1),(x_2, y_2)) = \sqrt{d(x_1, x_2)^2 + d(y_1, y_2)^2}$ are isometric. Does it follow that $A$ and $B$ must also be isometric themselves?</p> <p>The answer to this general question is negative. Take for example the space of the rational numbers $\mathbb{Q}$ and the space $\mathbb{Q} \sqrt{2}$ of rational multiples of $\sqrt{2}$. These two spaces are obviously not isometric, but their squares are, isometry being just the rotation by $\frac{\pi}{4}$. See <a href="http://www.ams.org/journals/proc/1971-029-03/S0002-9939-1971-0278262-5/S0002-9939-1971-0278262-5.pdf" rel="nofollow">http://www.ams.org/journals/proc/1971-029-03/S0002-9939-1971-0278262-5/S0002-9939-1971-0278262-5.pdf</a> for the proof.</p> <h2>Question</h2> <p>What is the status of this problem for slightly less general metric spaces? I am especially interested in the following cases:</p> <ol> <li>What is known about this problem for complete metric spaces?</li> <li>What is known for finite metric spaces?</li> </ol> <p>A good survey if it exists would be more than welcome.</p> http://mathoverflow.net/questions/74604/topological-space-with-some-conditions/74628#74628 Answer by Dejan Govc for Topological space with some conditions Dejan Govc 2011-09-06T01:40:17Z 2011-09-06T01:57:01Z <p>I believe such a space cannot exist for the following reason:</p> <p>Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then $f$ is unbounded on a compact set, which is a contradiction.</p> http://mathoverflow.net/questions/73129/interpolating-between-piecewise-linear-functions-with-a-family-of-smooth-functio/73506#73506 Answer by Dejan Govc for Interpolating between piecewise linear functions, with a family of smooth functions Dejan Govc 2011-08-23T16:07:15Z 2011-08-23T16:07:15Z <p>Interesting problem! This is not an actual answer, just a comment. I believe some minor condition may be missing, since it seems to me that the problem as posed does not necessarily have a solution. Set for example $F:[0,2)\to\mathbb R$, $F|_ {[0,1)}\equiv 1$, $F|_ {[1,2)}\equiv 0$ and $G:[0,2)\to\mathbb R$, $G|_ {[0,1)}\equiv 2$, $G|_ {[1,2)}\equiv 1$. These two functions satisfy conditions 1-4 but a smooth family you're looking for would have to contain some functions discontinuous at point $x=1$ because of condition 2 which asks that the family be strictly increasing. Maybe the inequalities for $L$ should be strict? Or maybe you should add to $E$ those points where $F(x)$ equals the left or right limit of $G$ there? Maybe I'm just missing something. In the latter case, I kindly ask the moderators to delete this comment.</p> http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/70333#70333 Answer by Dejan Govc for Can a continuous, nowhere differentiable function have specified "shape" at every point? Dejan Govc 2011-07-14T15:21:33Z 2011-07-14T15:21:33Z <p>I'm new here, so I hope my answer is of any use and not too late.</p> <p>I was wondering: Wouldn't it be perhaps more natural to consider limits of the form</p> <p>$\lim_{y\to x}\frac{f(y)-f(x)}{\phi(y)-\phi(x)}$?</p> <p>If for example we take $f(x) = |x|$ for $x\in \bf{R}$ and $\phi = f$, the "derivative" would be equal to one everywhere, which makes sense, since $f$ and $\phi$ are really the same, therefore their shape should be the same, right?</p> <p>By the previous definition which uses $\phi(y-x)$ in the denominator, we would get limits like</p> <p>$\lim_{y\to x}\frac{|y|-|x|}{|y-x|}$</p> <p>which exists only for $x = 0$, even though our intuition tells us the shape is supposed to be the same everywhere. Any comments?</p> http://mathoverflow.net/questions/5243/why-is-it-a-good-idea-to-study-a-ring-by-studying-its-modules/5742#5742 Comment by Dejan Govc Dejan Govc 2012-03-24T23:36:34Z 2012-03-24T23:36:34Z +1. Beautifully written. http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62793#62793 Comment by Dejan Govc Dejan Govc 2011-11-14T15:14:31Z 2011-11-14T15:14:31Z Not in characteristic 2. =) http://mathoverflow.net/questions/79442/number-of-distinct-values-taken-by-xx-x-with-parentheses-inserted-in-all-pos/79446#79446 Comment by Dejan Govc Dejan Govc 2011-11-02T21:30:49Z 2011-11-02T21:30:49Z A similar example I find interesting is to take the unique positive solution of $x^x=x+1$. Then you can easily check that $x^{(x^{(x^x)})} =(x^x)^{(x^x)}$ using the defining equation and the usual exponentiation rules. Although the transcendentality doesn't seem obvious anymore ... http://mathoverflow.net/questions/78129/a-brouwer-fixed-point-theorem-on-finite-sets/78137#78137 Comment by Dejan Govc Dejan Govc 2011-10-16T01:59:10Z 2011-10-16T01:59:10Z It is a beautiful problem, I must say. http://mathoverflow.net/questions/78129/a-brouwer-fixed-point-theorem-on-finite-sets/78137#78137 Comment by Dejan Govc Dejan Govc 2011-10-16T01:58:24Z 2011-10-16T01:58:24Z I have thought about it some more and $d(f(x_0),x_0)\leq 1$ won't always work, but I am quite sure you can prove that there always exists a point $x_0$ such that $d(f(x_0),x_0)\leq C$ where $C$ is a constant depending only on the dimension $d$. This is because you can extend your function by piecewise linear interpolation to a continuous function on $[-n,n]^d \subseteq \mathbb{R}^d$ and then use the original Brouwer fixed point theorem obtaining a fixed point. Now choose the closest lattice point. This point has the property we seek (this follows from your property and piecewise linearity). http://mathoverflow.net/questions/78129/a-brouwer-fixed-point-theorem-on-finite-sets/78137#78137 Comment by Dejan Govc Dejan Govc 2011-10-15T16:50:03Z 2011-10-15T16:50:03Z I am wondering ... Maybe it is still possible to get a version of Brouwer's fixed point theorem. The condition $f(A(x,y))\subseteq A(f(x,y))$ looks a lot like continuity. If you equip $X$ with the Euclidean metric, it just says that whenever $d(x,y) \leq 1$ it follows that $d(f(x),f(y)) \leq 1$. My intuition suggests that it should still be possible to prove that there is an almost-fixed point $x_0$ in the sense that $d(f(x_0),x_0) \leq 1$. This certainly works for the one-dimensional analog of the problem ($X = \lbrace 1,2, ..., n \rbrace$). http://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical Comment by Dejan Govc Dejan Govc 2011-10-08T16:28:42Z 2011-10-08T16:28:42Z Max Euwe showed using the Thue-Morse sequence that the rules of chess in his time did not exclude the possibility of an infinite game. I'd say that's interesting mathematically. http://mathoverflow.net/questions/77446/what-is-known-about-ulams-problem-of-metric-spaces-with-isometric-squares/77461#77461 Comment by Dejan Govc Dejan Govc 2011-10-08T10:45:03Z 2011-10-08T10:45:03Z Thank you! After reading the first article, it seems the problem for complete metric spaces is still wide open. http://mathoverflow.net/questions/18319/can-a-continuous-nowhere-differentiable-function-have-specified-shape-at-every/70333#70333 Comment by Dejan Govc Dejan Govc 2011-07-14T20:31:56Z 2011-07-14T20:31:56Z Oh, I think I get it now. You're looking for a function that is locally of the same shape as $\phi$ is around zero. My answer is probably a bit off-topic then ...