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A convex set contains every line segment joining two of its points.

For a connected finite-dimensional Riemannian manifold, completeness of the Riemannian distance, geodesic completeness, and compactness of every closed bounded subset are equivalent. These conditions imply that any two points are joined by a minimizing geodesic. Pairwise existence of minimizing geodesics alone does not imply completeness: an open Euclidean ball is incomplete, yet the line segment between any two of its points stays in the ball and minimizes length.

Every line segment is a Euclidean Ramsey set.

Restrict $\phi$ to an arbitrary line segment by defining $q(t)=\phi ((1-t)x+ty)$ for $0\le t\le 1$. The chain rule and the assumed positive semidefinite matrix condition on the Hessian matrix giveThus $q^{\prime }$ is non-decreasing. For $0<t<1$, integrating $q^{\prime }$ on the two subintervals givesRearranging, and including the immediate endpoint cases, yieldsThis is precisely the definition of a convex function.

Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a regular simplex with $k+1$ vertices and side length $d$. The pigeonhole principle gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every line segment, is a Euclidean Ramsey set.

For an equilateral triangle of side length $d$, take a regular simplex with $2k+1$ vertices and side length $d$. The pigeonhole principle gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.

To prove the product theorem for Euclidean Ramsey sets, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete patternThere is a copy $Y^{\prime }\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X^{\prime }\cong X$. Every point of $X^{\prime }\times Y^{\prime }$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.

A rectangle is the Cartesian product of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a right triangle; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.

It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every ${\mathbb{R}}^m$, use the finite coloringA congruent copy has the form $a-v,a,a+v$ with $\Vert v\Vert =1$ for the Euclidean norm. The parallelogram law givesPut $n_{+}=\lfloor 2\Vert a+v{\Vert }^2\rfloor$, $n_{-}=\lfloor 2\Vert a-v{\Vert }^2\rfloor$, and $n_0=\lfloor 2\Vert a{\Vert }^2\rfloor$. The errors introduced by the three floor functions show thatIf all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the three-term unit arithmetic progression is not Euclidean Ramsey assertion.

False. Join, at one common endpoint $o$, a line segment of length $1/n$ for every positive integer $n$, and use the intrinsic path metric. This is a real tree. It is totally bounded, because outside the first finitely many arms every point lies arbitrarily close to $o$, and it is complete; hence it is compact. Removing $o$ leaves one connected component for every arm, so $o$ has countably infinite multiplicity of a point in a real tree.

Let $|f(z)|\le K$ on $\mathrm{Re}z>0$. If $\mathrm{Re}z>c$, the circle $|w-z|=c$ lies in that half-plane. The Cauchy derivative formula givesThe line segment joining $z_1$ to $z_2$ remains in $\mathrm{Re}z>c$, soThus one may take $M=K/c$, as in the lipschitz bound inside a bounded analytic half-plane.

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Frechet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

A set $C\subseteq {\mathbb{R}}^d$ is convex when every line segment between two of its points remains in the set: for every $x,y\in C$ and $0\le t\le 1$,

Choose $z_{+},z_{-}\in \mathbb{C}$ withRestrict $f$ to the line segment between them:This is continuous, with $h(0)<0<h(1)$. The intermediate value theorem gives a $t_0$ with $h(t_0)=0$. Thereforeis the required zero on a line segment.

Let $S$ be a finite Ramsey witness for $X$ under $k$ colors. Choose a finite witness $T$ for $Y$ under $k^{|S|}$ colors. Given a coloring of $S\times T$, color each $t\in T$ by the complete vectorThere is a copy $Y^{\prime }\subseteq T$ on which this vector is constant. Thus, for each $s$, the color $c(s,t)$ is independent of $t\in Y^{\prime }$. These values define a $k$-coloring of $S$, which has a monochromatic copy $X^{\prime }$. Then $X^{\prime }\times Y^{\prime }$ is a monochromatic isometric copy of $X\times Y$. This proves the product theorem for Euclidean Ramsey sets.

Every nondegenerate triangle and every line segment is a Euclidean Ramsey set. If $T$ is the given acute triangle and $I$ is a segment of length $t$ in a new orthogonal coordinate, then $T\times I$ is exactly the vertex set of the triangular prism with base $T$ and height $t$. The product theorem therefore makes it Euclidean Ramsey.