User gerald edgar - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:14:15Z http://mathoverflow.net/feeds/user/454 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43148/basic-results-with-three-or-more-hypotheses/59576#59576 Answer by Gerald Edgar for Basic results with three or more hypotheses Gerald Edgar 2011-03-25T16:02:31Z 2013-05-16T21:53:22Z <p><a href="http://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem" rel="nofollow">Kolmogorov's three series theorem</a>.</p> http://mathoverflow.net/questions/130549/principal-value-of-integral/130606#130606 Answer by Gerald Edgar for Principal value of integral Gerald Edgar 2013-05-14T16:42:55Z 2013-05-15T01:22:36Z <p>Yes the value $\pi/2$ can be obtained like this.</p> <p>Let $$f(x):=\frac {1}{x^2}-\frac{\cot(x)}{x}$$</p> <p>We may compute $$\int_0^{\pi/2} f(x)\;dx + \sum_{k=1}^\infty \int_0^{\pi/2}\big(f(k\pi+x)+f(k\pi-x)\big)\;dx=\frac{\pi}{2} \tag{1}$$ and this converges. </p> <p>We can think of (1) as a "rearrangement" of the required integral. But the integrands in (1) are positive: Use $\cot x > 0$ for $0 &lt; x &lt; \pi/2$ and $\cot(k\pi+x) = \cot x$ and $\cot(k\pi-x) = -\cot x$. Also $(1/x) - \cot x$ increases from $0$ on $(0,\pi/2)$, so $f(x) >0$ on $(0,\pi/2)$. Next $$f(k\pi+x)+f(k\pi-x) = \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) + \left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x$$ and each of the two halves is positive on $(0,\pi/2)$. Recall $$\sum_{k=1}^\infty \left(\frac{1}{(k\pi+x)^2}+\frac{1}{(k\pi-x)^2}\right) = \csc^2 x - \frac{1}{x^2}$$</p> <p>$$\sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)=\frac{1}{x}-\cot x$$</p> <p>$$\sum_{k=1}^\infty\left(\frac{-1}{k\pi+x}+\frac{1}{k\pi-x}\right)\cot x=\frac{\cot x}{x}-\cot^2 x$$</p> <p>Our answer is the sum of three integrals:</p> <p>$$\int_0^{\pi/2} \left[\left(\frac{1}{x^2}-\frac{\cot x}{x}\right)+\left(\csc^2 x-\frac{1}{x^2}\right)+\left(\frac{\cot x}{x}-\cot^2 x\right)\right]dx = \int_0^{\pi/2} 1\;dx = \frac{\pi}{2}$$</p> http://mathoverflow.net/questions/130595/the-pth-power-of-a-distance-function-is-twice-continuously-differentiable-for-p/130602#130602 Answer by Gerald Edgar for The pth power of a distance function is twice continuously differentiable, for $p>2$? Gerald Edgar 2013-05-14T15:55:34Z 2013-05-14T15:55:34Z <p>How about in $\mathbb R$ the open set is $(-2,-1) \cup (1,2)$. Study $\beta(x)$ at $x=0$.</p> http://mathoverflow.net/questions/129413/what-fields-can-be-used-for-an-inner-product-space/129431#129431 Answer by Gerald Edgar for What fields can be used for an inner product space? Gerald Edgar 2013-05-02T15:21:15Z 2013-05-02T18:11:24Z <p>Of course if you insist on condition $\langle \mathbf{x},\mathbf{x}\rangle > 0$, and not merely $\langle \mathbf{x},\mathbf{x}\rangle \ne 0$, then you must have an order. </p> <p>Let $F$ be a formally real field. Then $$\langle \mathbf{x}, \mathbf{y}\rangle = \sum_{j=1}^n x_j y_j$$ can be a reasonable inner product on $F^n$. According to an ordering for $F$ (indeed, any ordering, since there may be more than one) we have $\langle \mathbf{x}, \mathbf{x}\rangle > 0$ if $\mathbf x \ne \mathbf 0$. </p> <p>Another part that you quote is what would be required for metric completeness. Do you want that? If $F$ is a proper subfield of $\mathbb R$, then even the one-dimensional space is not complete. </p> <p>Something weaker than completeness will be enough to carry out the Gram-Schmidt process. It requires only that square-roots of $\langle \mathbf{x}, \mathbf{x}\rangle$ exist.</p> http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial/129112#129112 Answer by Gerald Edgar for Identity of binomial series with factorial. Gerald Edgar 2013-04-29T16:13:36Z 2013-05-02T02:54:43Z <p>Darij's comment... </p> <p>Tthe truncated exponential series: $$e_p(z) = \sum_{n=0}^p\frac{z^n}{n!}$$ Your sum is $$p!x^pe_p(1/x)$$</p> http://mathoverflow.net/questions/128601/continuous-automorphisms-of-q/128620#128620 Answer by Gerald Edgar for Continuous automorphisms of Q* Gerald Edgar 2013-04-24T14:18:37Z 2013-04-24T14:18:37Z <p>I assume by "continuous" you mean in the topology inherited from the usual topology on $\mathbb R$. A continuous homomorphism will be uniformly continuous (in the natural uniformm structure for a group). And therefore extend continuously to the complettion, which will be $\mathbb R^\ast$, the nonzero reals under multiplication. Same for the inverse map, since it is assumed continuous as well. So we get an extension that is a continuous automorphism of $\mathbb R^\ast$. Such an automorphism is of the form $x \mapsto x^c$ on $x>0$, where $c$ is a nonzero constant. (And appropriately for $x&lt;0$.) But the only maps of this form that map rationals bijectively to rationals are $c=1$ and $c=-1$, I guess.</p> http://mathoverflow.net/questions/128259/is-function-from-topological-group-to-metric-space-borel/128277#128277 Answer by Gerald Edgar for Is function from topological group to metric space Borel? Gerald Edgar 2013-04-21T19:05:58Z 2013-04-21T19:05:58Z <p>Let's try this. I'm not using the hypothesis in your second paragraph, so maybe I am missing something. </p> <p>Suppose $G$ is pseudometrizable but not metrizable. The the closure of ${e}$, (the identity), is a closed subgroup $N$ of $G$. And every open set in $G$ either contains $N$ or is disjoint from $N$. Then this same thing is true for every Borel set. On the other hand, for any two points of $X$, there is an open set that contains one but not the other. So, whatever bijection $f$ we choose, it is not Borel.</p> http://mathoverflow.net/questions/127719/textbooks-on-asymptotic-expansions/127722#127722 Answer by Gerald Edgar for textbooks on asymptotic expansions Gerald Edgar 2013-04-16T17:20:25Z 2013-04-16T17:20:25Z <p><a href="http://www.amazon.com/Asymptotics-Summability-Monographs-Surveys-Mathematics/dp/1420070312/" rel="nofollow">http://www.amazon.com/Asymptotics-Summability-Monographs-Surveys-Mathematics/dp/1420070312/</a></p> <p>Asymptotics and Borel Summability, O. Costin</p> http://mathoverflow.net/questions/123744/formal-writing-numbers-under-10/123756#123756 Answer by Gerald Edgar for Formal writing: numbers under 10 Gerald Edgar 2013-03-06T13:10:20Z 2013-03-06T13:10:20Z <p>Also in engineering watch for units... "The two rods had diameter 5 mm." "A 6-volt battery was used."</p> http://mathoverflow.net/questions/123647/defining-definite-integral-using-indefinite-integral/123653#123653 Answer by Gerald Edgar for Defining definite integral using indefinite integral. Gerald Edgar 2013-03-05T18:59:48Z 2013-03-05T18:59:48Z <p>In case the exceptional set $C$ is empty, I have seen this called <em>the Newton integral</em>. </p> <p>It can exist in cases where the Riemann integral does not exist, also in cases where the Lebesgue integral does not exist. There is an integral (the <em>generalized Riemann integral</em> or <em>gauge integral</em> or <em>Kurzweil integral</em> or <em>Henstock integral</em>) that does have that theorem: if $f = F'$ everywhere, then the integral exists and satisfies $\int_a^b f(x)dx = F(b)-F(a)$. An elementary reference: <em>Generalized Riemann Integral</em> by R. M. McLeod.</p> <p>If $f$ is Riemann integrable, then it is true that there is a null set $C$ (but not, in general countable) such that $F'(x)=f(x)$ for all $x \in [a,b]\setminus C$ where $$F(x) = \int_a^x f$$<br> and this $F$ is of course continuous. The same for Lebesgue integral.</p> <p>In measure spaces $(X, \mathcal F,\mu)$, the usual thing is not to connect function $f$ to another function $F$ where $F'=f$, but to connect a function $f$ and a signed measure $\nu$ where $$\int_A f d\mu = \nu(A)\qquad\text{for all A \in \mathcal F}$$</p> http://mathoverflow.net/questions/123482/is-there-a-constructive-proof-of-cantorbernsteinschroeder-theorem/123484#123484 Answer by Gerald Edgar for Is there a constructive proof of Cantor–Bernstein–Schroeder theorem ? Gerald Edgar 2013-03-03T17:48:00Z 2013-03-03T17:48:00Z <p>Failure of the law of the excluded middle is hard for me to imagine. But here is a try... </p> <p>That König proof has this: </p> <blockquote> <p>For any particular $a$, this sequence may terminate to the left or not</p> </blockquote> <p>But perhaps we also have to consider the case of both: it terminates to the left and also does not terminate to the left... ???</p> http://mathoverflow.net/questions/123252/extension-of-measures-from-the-ball-sigma-algebra-to-the-borel-sigma-algebra/123260#123260 Answer by Gerald Edgar for Extension of measures from the ball sigma-algebra to the borel sigma-algebra Gerald Edgar 2013-02-28T19:52:30Z 2013-02-28T19:52:30Z <p>Take a set $X$ of power $\aleph_1$, with the discrete metric where two distinct points have distance $1$. The balls are singletons and the whole space. The ball sigma-algebra is the countable and co-countable sets. Let countable sets have measure zero, co-countable sets have measure 1. </p> <p>Now all subsets are open, so the Borel sigma-algebra is the power set. There is no extension of this measure!</p> http://mathoverflow.net/questions/122237/maximal-ideals-of-the-algebra-of-measurable-functions/122303#122303 Answer by Gerald Edgar for Maximal ideals of the algebra of measurable functions Gerald Edgar 2013-02-19T13:41:30Z 2013-02-19T13:41:30Z <p>If, instead of <em>all</em> measurable functions, you use the <em>bounded</em> measurable functions, then you probably want to consult the literature known as "lifting" ... for example:</p> <p>Topics in the Theory of Lifting (Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge)</p> http://mathoverflow.net/questions/122115/problems-from-the-scottish-book/122122#122122 Answer by Gerald Edgar for problems from the scottish book Gerald Edgar 2013-02-18T01:20:05Z 2013-02-18T01:20:05Z <p>The <a href="http://www.amazon.com/Scottish-Book-Mathematics-Cafe/dp/3764330457/" rel="nofollow">book version</a> edited by Daniel Mauldin (from 1982) has commentaries on the problems as of that date.</p> http://mathoverflow.net/questions/122006/second-order-difference-implies-differentiability/122016#122016 Answer by Gerald Edgar for Second order difference implies differentiability Gerald Edgar 2013-02-16T20:12:11Z 2013-02-16T20:12:11Z <p>I am assuming you mean $|f(x+2h)-2f(x+h)+f(x)|\le |h|^{3/2}$. </p> <p>Well, what if $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$? Certainly your inequality holds then. But (according to the Axiom of Choice) there are badly discontinuous functions like this.</p> http://mathoverflow.net/questions/120238/good-examples-of-random-variables-whose-image-is-not-a-measurable-set/120249#120249 Answer by Gerald Edgar for Good examples of random variables whose image is not a measurable set? Gerald Edgar 2013-01-29T19:25:13Z 2013-01-29T19:25:13Z <p>An analytic set that is not a Borel set...see <a href="http://www.math.niu.edu/~rusin/known-math/97/measure" rel="nofollow">this post</a> from long ago.</p> <p>Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.</p> http://mathoverflow.net/questions/120121/mean-value-theorems-for-the-haar-integral/120132#120132 Answer by Gerald Edgar for Mean value theorems for the Haar integral? Gerald Edgar 2013-01-28T18:06:30Z 2013-01-28T18:06:30Z <p>Say $\mu$ is a Borel probability measure on a connected set $A$ in a topological space. Let $f : A \to \mathbb R$ be continuous. Then the mean value $\int_A f\;d\mu$ is equal to $f(a)$ for some $a \in A$. Proof: the mean value is between the sup of all values and the inf of all values, so (by connectedness) it is a value of the function. </p> <p>Of course the desired result fails for non-connected sets. Even in the two-point group we get a counterexample.</p> http://mathoverflow.net/questions/119696/translation-of-a-non-standard-analysis-formulation/119700#119700 Answer by Gerald Edgar for Translation of a non-standard analysis formulation Gerald Edgar 2013-01-23T22:05:12Z 2013-01-23T22:05:12Z <p>How about this ...</p> <p>Let's say a finite set $V \subseteq \mathbb N$ is a <strong>Feldmann set</strong> iff there are infinitely many $n$ such that for all $u \in V$, $n+u$ is prime. </p> <p>For example, $\{0,2\}$ is a Feldmann set iff there are infinitely many twin primes.</p> <p>Your equivalent statment: there are arbitrarily large Feldmann sets.</p> http://mathoverflow.net/questions/119127/apollonian-gasket-and-the-degree-of-convergence/119129#119129 Answer by Gerald Edgar for Apollonian gasket and the degree of convergence Gerald Edgar 2013-01-17T02:09:03Z 2013-01-17T15:13:35Z <p>This critical value is $\alpha_0 \approx 1.3056867$ ... For $\alpha > \alpha_0$ the series converges, for $\alpha &lt; \alpha_0$ it diverges. </p> <p>Before the inexpensive computer, it was difficult to tell whether the critical value is ${}> 1$ or not.</p> <blockquote> <p>Boyd, David W. The sequence of radii of the Apollonian packing. Math. Comp. 39 (1982), no. 159, 249–254. </p> </blockquote> <p><a href="http://www.ams.org/mathscinet-getitem?mr=658230" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=658230</a></p> <p><em><strong>added</em></strong> </p> <p>mentioned in the comments... arbitrarily packed disks, not necessarily touching as in Apollonian packings. The critical value (= dimension of the residual set) is shown to be ${}> 1.02$.</p> <blockquote> <p>Larman, D. G. On the Besicovitch dimension of the residual set of arbitrarily packed disks in the plane. J. London Math. Soc. 42 1967 292–302. </p> </blockquote> <p><a href="http://www.ams.org/mathscinet-getitem?mr=209982" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=209982</a></p> http://mathoverflow.net/questions/117613/measurable-sets-and-valuation-theory/117623#117623 Answer by Gerald Edgar for Measurable sets and Valuation Theory Gerald Edgar 2012-12-30T13:25:15Z 2013-01-13T20:26:36Z <p><strong>edit, January 13</strong> </p> <p>Write $|\cdot|$ for the extended $2$-adic absolute value. I am assuming you mean $A = \{ (x,y) : |x|&lt;1, |y|&lt;1\}$, but since you say valuation'' maybe that is not right. Anyway, your set $A$ is simply related to my set $A$, perhaps by taking complements or multiplying by a constant.</p> <p>Note that $A$ is a group under addition, since $|-x| = |x|$ and $|x+y| \le \max\{|x|,|y|\}$.</p> <p>Assume (for purposes of contradiction) that $A$ is measurable.</p> <p>If $A$ has positive measure we get a contradiction: indeed, the set $A - A$ contains a neighborhood of zero (for the usual topology). But $A-A = A$ and $A+A=A$, so $A$ is the whole plane, which is false.</p> <p>Now consider sets $A_n = 2^{-n}A = \{(2^{-n}x,2^{-n}y): (x,y) \in A\}$. These sets are also groups under addition. The map $(x,y) \mapsto 2^{-n}(x,y)$ is an affine bijection, so all sets $A_n$ are Lebesgue measurable. But also note that multiplication is continuous with respect to $|\cdot|$, and $|2^n| \to 0$, and $A$ is a neighborhood of zero for the $|\cdot|$ topology. So for any $(x,y) \in \mathbb R^2$ there is $n$ so that $2^n(x,y) \in A$, and that means $(x,y) \in A_n$. Thus $$\bigcup_{n=1}^\infty A_n = \mathbb R^2 .$$ A union of measurable sets. So some $A_n$ has positive measure. Get a contradiction as before.</p> http://mathoverflow.net/questions/118687/rfc-for-definite-integral-connection-to-second-derivative/118726#118726 Answer by Gerald Edgar for RFC for definite integral connection to second derivative Gerald Edgar 2013-01-12T12:43:57Z 2013-01-12T16:32:20Z <p>Hint ... write $u(x) = f''(x)$, so that the condition is $$\int_0^T u(t) g(x,y)dt = \int_0^x\left[\int_0^y u(s) ds\right]dy$$</p> http://mathoverflow.net/questions/118301/are-all-irrational-elementary-numbers-conjectured-to-be-normal/118315#118315 Answer by Gerald Edgar for Are All Irrational Elementary Numbers Conjectured to Be Normal? Gerald Edgar 2013-01-07T22:21:27Z 2013-01-07T22:21:27Z <p>Perhaps connected to general conjectures published <a href="http://projecteuclid.org/getRecord?id=euclid.em/999188630" rel="nofollow">here</a> ... "On the Random Character of Fundamental Constant Expansions", David H. Bailey and Richard E. Crandall<br> <em>Experiment. Math.</em> Volume 10, Issue 2 (2001), 175-190.</p> http://mathoverflow.net/questions/19356/how-has-what-every-mathematician-should-know-changed/118298#118298 Answer by Gerald Edgar for How has "what every mathematician should know" changed? Gerald Edgar 2013-01-07T18:11:38Z 2013-01-07T18:11:38Z <p>In Littlewood's <em>Miscellany</em> there is an essay "A Mathematical Education" where he describes the situation before 1907.</p> http://mathoverflow.net/questions/118008/which-metric-spaces-have-this-superposition-property/118015#118015 Answer by Gerald Edgar for Which metric spaces have this superposition property? Gerald Edgar 2013-01-04T02:04:49Z 2013-01-04T02:04:49Z <p>Like Euclidean geometry, also hyperbolic geometry has this extension property: an isometry defined on any subset extends to an isometry of the whole space. As I recall from long ago, in the book<br> Busemann &amp; Kelly <em>Projective Geometry and Projective Metrics</em><br> it is shown (among that class of geometries) there are very few of these spaces. </p> http://mathoverflow.net/questions/117759/streamlined-probability-measure-for-tossing-infinitely-many-coins/117761#117761 Answer by Gerald Edgar for Streamlined probability measure for tossing infinitely many coins Gerald Edgar 2012-12-31T22:25:04Z 2012-12-31T22:25:04Z <p>This version is due to Emile Borel ... </p> <p>Sequence of heads &amp; tails encoded as 0s and 1s, then sequence is taken to represent a number in $[0,1]$ in its binary expansion. The measure is Lebesgue measure. </p> <p>So you still need to know that Lebesgue measure exists. </p> http://mathoverflow.net/questions/117735/approximating-erf-by-tanh/117745#117745 Answer by Gerald Edgar for Approximating erf by tanh Gerald Edgar 2012-12-31T17:37:42Z 2012-12-31T17:57:10Z <p>First, \begin{align} 1-\mathrm{erf}(x) &amp;= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &amp;= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align} Subtract: $$\mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt$$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.</p> http://mathoverflow.net/questions/117609/injectiveintregrable-mapping-from-mathbb-r3-to-mathbb-r/117625#117625 Answer by Gerald Edgar for Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ Gerald Edgar 2012-12-30T13:42:03Z 2012-12-30T13:42:03Z <p><strong>beginning</strong> </p> <p>Start with the unit cube $E$ in $\mathbb R^3$ and the unit interval $[0,1]$ in $\mathbb R$. Choose an injective map $\phi : E \to [0,1]$. This is the remaining question: do this so that $\phi$ is Riemann integrable. That is, the set of discontinuities has measure zero.</p> <p>Then cover $\mathbb R^3$ by disjoint unit cubes, make them disjoint by including boundary points in only one of the possible cubes. Call these $F_n, n \in \mathbb N$. Choose disjoint intervals $I_n$ going to zero fast enough. Define $f$ using $\phi$ with translation and dilation to map $F_n$ injectively into $I_n$. If the original $\phi$ has set of discontinuities of measure zero, then this pieced-together function $f$ does too, since its set of discontinuities can be at most the discontinuities of a copies of $\phi$ in the interiors of the $F_n$, together with the boundary planes of the $F_n$. So $f$ is (improperly) Riemann integrable.</p> http://mathoverflow.net/questions/117494/math-for-a-cake/117559#117559 Answer by Gerald Edgar for Math for a cake Gerald Edgar 2012-12-29T20:22:09Z 2012-12-29T20:22:09Z <p>Maybe just make the cake in the shape of a golden rectangle, and use two colors of icing to show the decomposition into a square and a smaller golden rectangle.</p> http://mathoverflow.net/questions/117332/a-monotone-relation-for-s-numbers/117350#117350 Answer by Gerald Edgar for a monotone relation for s-numbers Gerald Edgar 2012-12-27T19:55:52Z 2012-12-27T19:55:52Z <p>I just tried a few random matrices... \begin{align} &amp;A=\begin{bmatrix} -0.1 &amp; -0.4\cr -0.4&amp;0\end{bmatrix}, \qquad B=\begin{bmatrix} 1.5 &amp; -0.5 + i\cr -0.5 - i&amp; 3.5\end{bmatrix}, \cr &amp;\|A+iB\| = \frac{7}{2}+\frac{\sqrt{61}}{10}\approx 4.28 \cr &amp;\|2A+iB\| = \frac{7}{2}+\frac{\sqrt{29}}{10}\approx 4.04 \end{align}</p> http://mathoverflow.net/questions/117233/lebesgue-differentiation-theorem-beyond-euclidean-spaces/117258#117258 Answer by Gerald Edgar for Lebesgue differentiation theorem beyond Euclidean spaces Gerald Edgar 2012-12-26T14:10:49Z 2012-12-26T15:52:31Z <p>A standard reference on derivation theory:<br> Hayes &amp; Pauc, <em><a href="http://dx.doi.org/10.1007/978-3-642-86180-2" rel="nofollow">Derivation and Martingales</a></em> (Springer 1970)</p> <p>(plug) there is a chapter on derivation in:<br> Edgar &amp; Sucheston, <em><a href="http://dx.doi.org/10.1017/CBO9780511574740" rel="nofollow">Stopping Times and Directed Processes</a></em> (Cambridge Univ Pr 1992) </p> http://mathoverflow.net/questions/130735/approximate-closed-form-solution-for-a-recurrence Comment by Gerald Edgar Gerald Edgar 2013-05-15T15:51:55Z 2013-05-15T15:51:55Z Needs some motivation or reason anyone should care about it. Perhaps some interpretation involving a random walk or ??. http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures/130488#130488 Comment by Gerald Edgar Gerald Edgar 2013-05-14T21:48:19Z 2013-05-14T21:48:19Z So the set $\{\delta_x : x \in [0,1]\}$ is not compact; it is discrete. http://mathoverflow.net/questions/130549/principal-value-of-integral Comment by Gerald Edgar Gerald Edgar 2013-05-14T14:22:06Z 2013-05-14T14:22:06Z @Carlo: that is perhaps a sensible definition. Can you cite a textbook that uses that definition? http://mathoverflow.net/questions/130581/can-an-accumulation-point-be-an-eigenvalue Comment by Gerald Edgar Gerald Edgar 2013-05-14T14:18:58Z 2013-05-14T14:18:58Z $0$ is always in the spectrum (for a compact operator on an infinite-dimensional Hilbert space). For some such operators $0$ is an eigenvalue, but for others it is not. http://mathoverflow.net/questions/130549/principal-value-of-integral Comment by Gerald Edgar Gerald Edgar 2013-05-14T12:18:46Z 2013-05-14T12:18:46Z Since there are multiple poles, I do not know what &quot;principal value&quot; means. Perhaps you can provide a definition? http://mathoverflow.net/questions/130492/higher-dimensional-convex-hull Comment by Gerald Edgar Gerald Edgar 2013-05-13T16:59:48Z 2013-05-13T16:59:48Z For your first paragraph, specify that $S$ is a finite set. Then <i>vertex</i> and <i>edge</i> make sense. http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures/130488#130488 Comment by Gerald Edgar Gerald Edgar 2013-05-13T16:56:45Z 2013-05-13T16:56:45Z @Lutz... you are right. Because andy has the wrong definition for the weak* topology. The usual way to define it is to use only continuous $Z$. If we use all measurable $Z$, as andy does, then you get a much stronger topology. Moreover, andy didn't even say that $Z$ should be bounded. So, let's give him a chance to say whether he wants to correct the definition. http://mathoverflow.net/questions/130470/existence-of-dominating-measure-for-weak-compact-set-of-measures Comment by Gerald Edgar Gerald Edgar 2013-05-13T13:28:35Z 2013-05-13T13:28:35Z Indeed, in Davide's example, the map $x \mapsto \delta_x$ is a homeomorphism from $[0,1]$ onto $\mathcal P$. http://mathoverflow.net/questions/130418/polynomial-zero-within-a-square Comment by Gerald Edgar Gerald Edgar 2013-05-12T22:37:29Z 2013-05-12T22:37:29Z $p(z)=z^8+k$, for $k$ large, has no zeros inside a large square. But still $|p(0)|$ smaller than the specified points on the unit square. http://mathoverflow.net/questions/130418/polynomial-zero-within-a-square Comment by Gerald Edgar Gerald Edgar 2013-05-12T16:48:17Z 2013-05-12T16:48:17Z Try polynomial $z^8+2$. http://mathoverflow.net/questions/130177/prove-that-the-sum-of-a-certain-infinite-series-is-1 Comment by Gerald Edgar Gerald Edgar 2013-05-12T13:10:48Z 2013-05-12T13:10:48Z Denominator 16 ... maybe it is somehow related to these: <a href="http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> http://mathoverflow.net/questions/130368/continuty-of-volume-of-a-convex-set-in-rn/130373#130373 Comment by Gerald Edgar Gerald Edgar 2013-05-12T12:39:09Z 2013-05-12T12:39:09Z If convexity is not required, then there are simpler counterexamples. The unit cube is the limit of a sequence of finite sets. http://mathoverflow.net/questions/130368/continuty-of-volume-of-a-convex-set-in-rn Comment by Gerald Edgar Gerald Edgar 2013-05-12T12:37:32Z 2013-05-12T12:37:32Z If you mean for $L$ to be convex, add that to the text of the question. Then we can use the fact that the boundary of a convex set has measure zero. http://mathoverflow.net/questions/130247/closed-form-for-derivatives-zetan1-2 Comment by Gerald Edgar Gerald Edgar 2013-05-10T13:50:21Z 2013-05-10T13:50:21Z Let's hope we get an answer here. Things on Mathworld (and Wikipedia, and so on) stated without citation are not entirely reliable... http://mathoverflow.net/questions/58193/leibnizian-calculus-textbook Comment by Gerald Edgar Gerald Edgar 2013-05-10T13:27:03Z 2013-05-10T13:27:03Z @Aerik: It is my experience that in elementary courses (like calculus) it is a Bad Idea to deviate from the textbook in any way.