# Tag Info

3

Let me first review what I would consider to be the standard definitions of these terms. A space $X$ is path-connected if for any $x,y\in X$ there is a continuous map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. A space $X$ is arc-connected if for any $x,y\in X$ there is an injective continuous map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. That ...

3

$\cos(nt)=T_n(\cos t)$, where $T_n$ is the $n$-th Chebyshev polynomial. So your curve can be written $x=T_m(u)$ and $y=T_n(u)$, where $u=\cos(t)$. Since $T_n(x)=T_n(T_m(u))=T_{nm}(u)=T_{mn}(u)=T_m(T_n(u))=T_m(y)$, your curve is (part of) the algebraic curve given by $T_n(x)=T_m(y)$. This is not the lowest possible degree. For instance, if $lcm(m,n)=am=bn$...

2

Use Eisenstein to deduce that $y^3-x^3(x+1)$ is irreducible. As an alternative, note that the polynomial (in the variable $y$) is primitive, of degree $3$ and has no roots. Then invoke the Gauß lemma.

2

The correct statement is that every nonempty compact metrizable space is a continuous image of the ternary Cantor set (Alexandroff-Hausdorff theorem).

1

Usual convention: \begin{align*} w &= a e^{i\theta} \\ c &= (a-b) e^{i\theta} \\ \frac{z-c}{w-c} &= e^{-i\phi} \\ a\theta &= b\phi \\ z &= (a-b)e^{i\theta}+b e^{-i\left( \frac{a-b}{b}\right) \theta} \end{align*} $w$: point of contact $c$: centre of blue circle $z$: locus of the initial point of contact, i.e. the hypocycloid ...

1

Indeed the statements are equivalent. So we have in particular that the following are equivalent for $X$: $X$ is a compact, connected, locally connected metric space. $X$ is a compact, connected, locally pathwise connected metric space. $X$ is a compact, connected, locally arcwise connected metric space. $X$ is a compact, path-connected, locally connected ...

1

That's not quite correct. For example, your curve could be empty, or a single point, or a line segment. If the Hessian matrix is positive definite on some convex region $R$, then $F$ is a convex function there. If $R$ is bounded, contains $\gamma$ and $\inf_R F < 0 \inf_{\partial R} F$, then $\gamma$ is the boundary of the convex open set $\{(x,y) \... 1 No. Actually I think you sort of didn't ask the question you meant to ask; we give a trivial counterexample to the question as asked and then a not quite as trivial counterexample to the question I think you meant to ask. First, let$J$be the unit circle, and let$K$be the circle of radius$\epsilon/2$centered at the point$1$. Every point of$K$is ... 1$C$doesn't have to be "nice".$E$is closed and bound in${\Bbb R}^2$and so there are 2 points$c,d\in E$such that their distance is maximal,$m$(distance is a bound and continuous function on the compact set$E\times E$so it has a maximum). It's pretty obvious that any such pair$c,d$with distance$m$must be on the boundary (if either or both are ... 1 I will assumme that a rotation and transformation means an isometry here, and that the curve$C$must be sufficiently nice (i.e. smooth). Suppose such a curve and a rotation and transformation exist. Let$\partial E$denote the boundary of$E$, i.e. it is the union of$C$and the line segment connecting$a$and$b$. Let$C'$denote the rotated and ... 1 No. Consider the curve $$\gamma(t) = (s \cos t, s \sin t), 0 \le t \le \frac{L}{2\pi s}$$ where$s < r$. its curvature is$\frac{1}{s}$, which is clearly unbounded. If you don't like that the path intersects itself, just make$s$a very slowly increasing function of$t$with mean$S$. Then the curvature will be approximately$\frac{1}{S}\$.

Only top voted, non community-wiki answers of a minimum length are eligible