Joy of Mathematics

The composition is a metric

December 18, 2021 Tolaso

Let $d$ be a metric , $f$ be a strictly increasing function and concave on $[0, +\infty)$ such that $f(0)=0$ . Prove that $f \circ d$ is a metric.

Continuity of infimum function

April 14, 2020 Tolaso

Let $X, Y$ be topological spaces. Let $f:X \times Y \rightarrow \mathbb{R}$ be a bounded and continuous function. Is the function $g(x) = \inf \limits_{y \in Y} f(x, y)$ continuous? Give a brief explanation.

Solution

Let $X = Y = \mathbb{R}$ with the standard topology. Define

$f(x, y) = \left\{\begin{matrix} 1 & , & x>0 \\ 0 & , & x< -e^y \\ 1+xe^{-y} & , & x \in [-e^y, 0] \end{matrix}\right.$

which is clearly continuous. But the infimum function $g$ is roughly the Heaviside function:

$g(x) = \left\{\begin{matrix} 1 & , & x \geq 0\\ 0 & , & x<0 \end{matrix}\right.$

Linear isometry

September 22, 2019 Tolaso

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ . If:

$f(\mathbf{0})=\mathbf{0}$
$\left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right|$ for all ${{\bf{u}},{\bf{v}}}$

then prove that $f$ is linear.

Solution

For convenience, identify $\mathbb{R}^2$ with $\mathbb{C}$ here. Then note that for any such function $f:\mathbb{C} \to \mathbb{C}$ , also $z_1 \cdot f(z)$ a solution for any point $z_1$ on the unit circle. Also $\overline{f(z)}$ is a solution. Note that $\vert f(1)\vert=1$ and hence we can wlog assume that $f(1)=1$ . So $f(i)$ is a point on the unit circle with distance $\sqrt{2}$ to $1$ . Hence $f(i) =\pm i$ , so w.l.o.g. assume that $f(i)=i$ . But then for any $z \in \mathbb{C}$ , both $z$ and $f(z)$ have the same distance to $0,1$ and $i$ . So supposing $z \ne f(z)$ , all $0,1,i$ lie on the perpendicular bisector between these points and in particular $0,1$ and $i$ are collinear which clearly is absurd. Hence $f(z)=z$ for all $z$ which proves the claim.

Does there exist an enumeration?

August 3, 2019 Tolaso

Does there exist an enumeration of $\{q_n \in \mathbb{Q}: n \in \mathbb{N} \}$ of $\mathbb{Q}$ such that

$\mathbb{R} \neq \bigcup_{n=1}^{\infty} \left( q_n - \frac{1}{n} , q_n + \frac{1}{n} \right)$

Solution

Yes there exists. Let $q_n$ be enumerated in the following way:

Enumerate $\mathbb{Q}\cap \big( (-\infty, 0)\cup (1,\infty)\big)$ and let them be
$\{q_{2^k}\}$ .
Enumerate $\mathbb{Q}\cap [0,1]$ and let them be $\{q_{m_k}\}$ where
$\{m_k\}$ is an increasing sequence which excludes all powers of $2$ .

Then the outer measure of

$\bigcup_{n=1}^{\infty} \left(q_n-\frac{1}{n}, q_n + \frac{1}{n} \right)$

is finite.

Intersection of closed sets

July 24, 2019 Tolaso

In the metric space $\left( \mathbb{Q} , \left| \cdot \right| \right)$ , find a decreasing sequence $\left \{ A_n \right \}_{n \in \mathbb{N}}$ of non closed subsets of $\mathbb{Q}$ such that $\mathrm{diam} \left ( A_n \right ) \xrightarrow{n \rightarrow +\infty} 0$ and $\bigcap \limits_{n=1}^{\infty} A_n =\varnothing$ .

Solution

Let us pick $A_{n}=\mathbb{Q}\cap \left[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n} \right]$ . Indeed,

$\mathrm{diam}(A_n)=\left|\sqrt{2}+\frac{1}{n}-\left(\sqrt{2}-\frac{1}{n}\right) \right|=\frac{2}{n}\xrightarrow{n \rightarrow +\infty} 0$

and

$\begin{align*} \bigcap_{n=1}^{\infty} A_n &= \bigcap_{n=1}^{\infty} \left (\mathbb{Q}\cap \left[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n} \right] \right ) \\ &=\left ( \bigcap_{n=1}^{\infty} \left [ \sqrt{2} - \frac{1}{n} , \sqrt{2} + \frac{1}{n} \right ] \right ) \cap \mathbb{Q} \\ &= \left \{ \sqrt{2} \right \} \cap \mathbb{Q}\\ &= \varnothing \end{align*}$