most recent 30 from http://mathoverflow.net 2013-05-24T03:29:48Z http://mathoverflow.net/feeds/user/10876 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines trutheality 2010-11-21T20:11:50Z 2012-12-17T06:34:33Z

<p>Let $f: \mathbb R^n \rightarrow \mathbb R^n$, where $n> 1$ be a bijective map such that the image of every line is a line.</p> <p>Is $f$ continuous?</p> <p>I think it is, but the proof isn't immediately obvious to me. Related to <a href="http://math.stackexchange.com/q/11232/3638" rel="nofollow">this question on math.SE</a>.</p> <p>Feel free to retag.</p>

http://mathoverflow.net/questions/46520/sampling-with-non-uniform-costs/46535#46535 trutheality 2010-11-18T20:33:29Z 2010-11-18T20:33:29Z

<p>This looks like it might be related to reinforcement learning (in machine learning).</p> <p>In the typical reinforcement learning problem an agent has a set of actions available to it and an unknown cost associated with each action. In your case the actions are to sample (and which element to sample), or to stop sampling and commit to the estimate. You could also view the action of committing to an estimate as having a cost depending on the quality of the resulting estimate.</p> <p>Maybe you might find some useful literature there.</p>

http://mathoverflow.net/questions/46448/geometric-interpretation-of-componentwise-linear-fractional-transformationlft/46456#46456 trutheality 2010-11-18T04:24:05Z 2010-11-18T05:03:50Z

<p>As mentioned, this is a central projection to the $x+y+z=1$ plane, or, in the positive orthant, you can also view it as an $L_1$-norm normalization.</p> <p>Preservation of circles:</p> <ul> <li>Circles in planes parallel to $x+y+z=1$ are preserved.</li> <li>Circles (and any other shapes) in planes <s>normal to $x+y+z=1$</s> <i>passing through the origin</i> are projected to line segments.</li> <li>Other circles are projected to ellipses.</li> </ul> <p>Cross ratios:<ul> <li>Component-wise cross ratios are obviously preserved.</li> <li>Geometric cross-ratios (i.e. for 4 collinear points the cross-ratios of distances between them) are also preserved, as they would be with any conic or linear projection (with the exception of the case when the points are on a line <s>normal to the $x+y+z=1$ plane</s> <i>passing through the origin</i>.)</li> </ul></p> <p>If you're interested in some other kind of cross-ratio, you have to define how you intend to "multiply" and "divide" the differences of points.</p> <p><b>Edit:</b> See corrections.</p>

http://mathoverflow.net/questions/46312/closed-connected-subset-of-a-connected-set trutheality 2010-11-17T01:11:02Z 2010-11-17T02:47:02Z

<p>Let $A$ be a closed set and let $B$ be a connected set such that $A \subset B$.</p> <p>Does there always exist a closed connected subset $C$ of $B$ that contains $A$?</p> <p>What if $B$ is path connected, is there always a path-connected $C$? A connected $C$?</p>

http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed/46292#46292 trutheality 2010-11-16T22:23:39Z 2010-11-16T22:30:37Z

<p>I suspect that Sergei Ivanov's proof can be extended to the case when $\mathbb R^2 \backslash Q$ is connected. I also suspect that the only case when $\mathbb R^2 \backslash Q$ is disconnected is precisely when $Q$ is a continuum.</p> <p>(Consider this "answer" a comment, but with the details filled in, it would imply that $\kappa = 2^\omega$.)</p> <p>Edit: I'm not so sure about the second point as the first: What is the cardinality of the smallest set $Q$ such that $\mathbb R^2 \backslash Q$ is disconnected?</p>

http://mathoverflow.net/questions/46866/is-napkin-conjecture-open-origami trutheality 2010-11-21T22:24:21Z 2010-11-21T22:24:21Z

<a href="http://en.wikipedia.org/wiki/Napkin_folding_problem" rel="nofollow">en.wikipedia.org/wiki/Napkin_folding_problem</a> Has some relevant information, you were probably thinking of the &quot;sea urchin&quot; solution, however, it does require some stretching of the paper.

http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines trutheality 2010-11-21T20:26:51Z 2010-11-21T20:26:51Z

I think that the standard terminology is that lines go off to infinity, while [0,1] and (0,1) would be segments.

http://mathoverflow.net/questions/46854/continuity-in-terms-of-lines trutheality 2010-11-21T20:23:01Z 2010-11-21T20:23:01Z

Thank you Willie Wong for editing accordingly.

http://mathoverflow.net/questions/46716/numbers-divisible-by-at-least-one-of-many-numbers/46717#46717 trutheality 2010-11-20T18:18:50Z 2010-11-20T18:18:50Z

@Fedor: Okay, I was just wondering if you had an accidental collision of notation. I'm deleting my answer since you made it explicit in your edit.

http://mathoverflow.net/questions/46520/sampling-with-non-uniform-costs/46535#46535 trutheality 2010-11-19T03:42:46Z 2010-11-19T03:42:46Z

Then it depends on what you know about how $b_i$ and $c_i$ are related. If they are independent then the estimate using $k$ lowest cost samples is as good as an estimate using a uniform sample. Otherwise you have to learn the relationship between $c_i$s and $b_i$s anyway. (There's also the case when the relationship is known, but then you can get an estimate of your statistic without sampling at all.)

http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed/46292#46292 trutheality 2010-11-17T00:36:46Z 2010-11-17T00:36:46Z

Uh... I was a little too careless. @Guillaume: Dense sets aren't closed, so $F$ can't be $\mathbb Q^2$, or that other set I constructed. Sergei's proof used the closedness property implicitly by using the open ball cover.

http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed/46292#46292 trutheality 2010-11-17T00:26:57Z 2010-11-17T00:26:57Z

@Guillaume: Good point, I overlooked infinite $F$s. Unfortunately, this doesn't bode well for Sergei's proof either: Let $Q=\mathbb Q^2$ and let $F = \{ 2a/b : a \in \mathbb Z, b \in \mathbb N \} \times \mathbb Q$. Then, because $F$ is dense, it's impossible to have an open ball that covers a point of $Q$ but no points of $F$. @Ori: On the other point, I was letting my path self-intersect, but you could be more careful and use unions of paths.

http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed/46292#46292 trutheality 2010-11-16T23:13:53Z 2010-11-16T23:13:53Z

* I meant $G$ is closed.

http://mathoverflow.net/questions/46279/if-q-is-a-subset-of-the-plane-of-size-less-than-continuum-then-does-every-closed/46292#46292 trutheality 2010-11-16T23:12:56Z 2010-11-16T23:12:56Z

The way I see the proof is that the countability and the covering of $Q \backslash F$ with disks was merely a means to avoid the &quot;bad&quot; points in $Q$ when path-connecting the points in $F$. Perhaps this works: For any $Q$ that has cardinality less than a continuum $\mathbb R^2 \backslash (Q \backslash F)$ is path connected. Let $G$ be a path in this set that passes through every point in $F$. Conveniently, paths are closed, so $F$ is closed.