Why is $\{x-y: x,y\in K,\ d(x,y)\geq 1/n\}$ a compact set in a metrizable topological space with compact $K$?

Question

Assume that $K$ is a compact convex set in a Hausdorff locally convex space, and $K$ is metrizable with the induced topology. Let $d$ be a metric defining the induced topology. Show that the set $$ C_n=\{x-y: x,y\in K,\ d(x,y)\geq 1/n\} $$ is compact.

---

[Motivation:]I come up with this question when reading a proof of the following proposition

Assume that $K$ is a compact convex set in a Hausdorff locally convex space, and $K$ is metrizable with the induced topology. There exists a continuous function $u$ on $K$ which is strictly convex, i.e., $$ u\left(\frac{x+y}{2}\right)<\frac{u(x)+u(y)}{2} $$ for any $x,y\in K$ with $x\not=y$.

The following is the proof:

First observe that any function $u$ of the form $u(x)=|\varphi(x)|^2$, with $\varphi\in X^*$, is convex (show that $u$ is continuous and mid-point convex). Moreover, for such a function we have $$ u\left(\frac{x+y}{2}\right)<\frac{u(x)+u(y)}{2} $$ provided that $\varphi(x)\not=\varphi(y)$. The desired function will be constructed as $$ u(x)=\sum_{n=1}^\infty|\varphi_n(x)|^2,\quad x\in K $$ where the functionals $\varphi_n$ are chosen so as to

• separate the points of $K$ and
• $|\varphi_n(x)|\leq 1/n$ for $x\in K$.

It will suffice to show that for every $n\geq 1$ there exists a finite set $F_n\subset X^*$ such that, for $x,y\in K$ with $d(x,y)\geq 1/n$, there exists $\psi\in F_n$ with $\psi(x)\not=\psi(y)$.

$\color{red}{\textrm{Indeed, observe that the compact set}}$ $$\color{red}{ C_n=\{x-y: x,y\in K,\ d(x,y)\geq 1/n\}} $$ does not contain the origin, hence $V_n\cap C_n=\emptyset$ for some absolutely convex neighborhood $V_n$ of zero.

Choose a finite set $A_n\subset C_n$ so that $A_n+V_n\supset C_n$, and for every $x\in A_n$ choose a functional $\psi_x$ such that $\Re\psi_x(x)>\Re_x(v)$ for all $v\in V_n$. The set $F_n=\{\psi_x: x\in A_n\}$ satisfies our requirements.

Answer 1

Thanks to the comments, it suffices to note that the set $$ S=\{(x,y)\in K\times K\mid d(x,y)\geq 1/n\} $$ is closed (and thus compact in $K\times K$), and $$ C_n=f(S) $$ with $f(x,y)=x-y$ which is a continuous function.

Answer 2

Let $(x_m,y_m)\in K\times K$ be such that $d(x_m,y_m)\geq 1/n$ and $x_m-y_m\to z$ in metric. By compactness of $K\times K$, there exists convergent subsequence $$ (x_{m_k},y_{m_k})\to (x,y) $$ for some $(x,y)\in K$. By continuity of the function $f(x,y)=x-y$, $$ x_{m_k}-y_{m_k}\to x-y. $$ Thus $x-y=z$. Note also that $$ d(x,y)=\lim_{k\to \infty} d(x_{m_k},y_{m_k})\geq 1/n. $$ It follows that $z=x-y\in C_n$. Now we can conclude that $C_n$ is closed in $K$.