Proofs in Topology and Metric Spaces (from Mathematical Analysis 2 by Zorich)

A criteria for compactness in a Metric space.

A metric space $(X,d)$ is compact if and only if for every sequence of elements one can extract a convergent subsequence.

Proof:

$\implies$ Let $\set{x_i}_{i\in{\mathbb{N}}}$ be a sequence of elements in $X$. If the sequence contains only a finite set of distinct elements than [by “the pigeonhole principle”] some element repeats infinitly many times implying the result. Thus, we assume there are infinitly many distinct elements in the sequence. Since $X$ is compact it has a finite $\epsilon$-net for any $\epsilon>04$. Thus we may pick a 1-net covering $X$. In this finite 1-net $E_1$ there exists an element $a_1$ such that there are infinitly many elements in the sequence in $\overline{B(a_1,1)}$ otherwise the sequence itself is finite.

Basic Inequalities in Analysis

Young’s Inequality

Let $a,b>0$ and $p,q\neq0,1$ be such that - $\frac{1}{q}+\frac{1}{p}=1$ than for $p>1$ we have - $$ a^{\frac{1}{p}}b^{\frac{1}{q}}\leq\frac{1}{p}a+\frac{1}{q}b $$ Implying - $$ a(\frac{b}{a})^{\frac{1}{q}} \leq (1-\frac{1}{q})a+\frac{1}{q}b $$

We note several observations:

First, we have $\frac{1}{p}=\frac{q-1}{q}$. Implying - $p=\frac{q}{q-1}=1+\frac{1}{q-1}$. This implies that $q>1$ if $p>1$.w

Second, we obtain by substituting $x-1$ to Bernouli’s Inequality: $(1+\alpha)^n\geq{n\alpha+1}$ to obtain - $$ x^n\geq{n(x-1)+1} $$

To conclude, for $x\geq0$ and natural $n$ we have $x^n\geq{n(x-1)+1}$

Basic questions in Complex Analysis (From Ahlfors)

Algebra of Complex numbers

1. Prove the isomorphism between the system of matrices (wrt matrix multiplication and addition) of the following form and the field of complex numbers -

$$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} $$

Proof: Given a complex number $x=\alpha+i\beta$ we define the correspondence $\phi(x)=\phi(\alpha+i\beta)=$

$$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} $$

Let $y=\gamma+i\delta$. Thus, $\phi(x+y)=\phi(\alpha+\gamma+i(\beta+\delta))$ =

$$ \begin{pmatrix} \alpha+\gamma & \beta+\delta \\ -(\beta+\delta) & \alpha+\gamma \end{pmatrix} = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix} + \begin{pmatrix} \gamma & \delta \\ -\delta & \gamma \end{pmatrix} = \phi(x)+\phi(y) $$

Basic properties of Metric Spaces (from Mathematical Analysis 2 by Zorich)

Proposition 1:

Let $(X,d)$ be a metric space. a. The union $\bigcup_{\alpha\in{A}}G_{\alpha}$ of the sets in any system $\set{G_{\alpha}, \alpha\in{A}}$ of sets that are open in $X$ is and open set in $X$. b. The intersection $\bigcap_{i=1}^nG_i$ of any finite number of sets that are open in $X$ is an open set in $X$. c. The intesection $\bigcap_{\alpha\in{A}}F_{\alpha}$ of the sets in any system $\set{F_{\alpha}, \alpha\in{A}}$ of sets $F_{\alpha}$ that are closed in $X$ is a closed set in $X$. d. The union $\bigcup_{i=1}^nF_i$ of any finite number of sets that are closed in $X$ is a closed set in $X$.

A proof of the Euclidean Algorithm

The Euclidean Algorithm:

Let $m,n\in\mathbb{N}$ such that $m>n$. Then we define succesively -

$$ m = q_1n+r_1 \ n = q_2r_1+r_2 \ r_1 = q_3r_2+r_3 \ .. \ r_k = q_{k+2}r_{k+1} + 0 $$

than $\gcd(m,n)=r_{k+1}$

Initial (wrong) proof: If $\gcd(m,n)=1$ than $m,n$ are foreign (or mutually prime). Otherwise, assume $\gcd(m,n)>1$. Define $r_0=n$ We prove by induction that $r_{i+1}|r_i$.

Base case - If $r_1\nmid{n}$ than we can deduce that $\gcd(m,n)=1$. Contradiction.

A question on the expansion of rational numbers in base-q

Question: A number $x\in{\mathbb{R}}$ has a periodic expansion (in any number system) $\iff$ $x$ is rational.

Proof:

$\impliedby$ Assume $x$ is rational, i.e. there are $m,n\in{\mathbb{Z}}$ such that, $x=\frac{m}{n}$ and $n\neq0$ - Assume this is the most reduced form. Than, by way of long division we obtain deteministically the decimals of its expansion in some $q$-ary number system:

We define $r_0$ to be the remainder of $m$ upon division by $n$. We define the following recursive relation obtained from the “long division” process -

A question from the 2025 UBC qualifying exam

Let $f:(0,\infty)\rightarrow\mathbb{R}$ be a twice differentiable function. Further suppose There are constants $A,B,C$ such that -

$A=\sup_{x>0}{|f(x)|}\newline B=\sup_{x>0}{|f’(x)|}\newline C=\sup_{x>0}{|f’’(x)|}$

Prove that $B^2\leq{4AC}$:

a. Use Taylor’s Theorem to expand $f(x)$ about $a$ and compute $f(a + 2h)$ for $h > 0$.

b. Rearrange the resulting expression to isolate $|f(a)|$ and bound it in terms of $A$, $C$ and $h$.

c. Now pick $h$ to minimise your bound on $|f(a)|$ and clean up.

A proof of Rolles Theorem and Lagranges finite increment Theorem

Theorem: A continous function on a closed interval $E=[\alpha,\beta]$ attains a maximum or a minimum in the closed interval.

Proof: From the definition of continuity - for any point in $[\alpha,\beta]$ there exists a neighborhood such that the function is bounded in that environment.

[Side note: For that reason the function $f(x)=\frac{1}{x}$ is not continous on any interval containing 0, since it is unbounded at any environment of the point.]

The number of binary strings on $n$ letters with no two consequitive ones

This question is from a lecture by Prof. Gil Cohen that popped on my YT feed (To my discredit, I watch YT on my free time..).

Question: “How many binary strings on $n$ letters are there with no two consequitive ones?”

Observations:

1. Naivly, For any $n$, if the number of 1s is strictly larger than $n/2$ than by the pigenhole principle we surely have two consequitive ones.
2. But ofcourse we can have consequitive 1s when the number of 1s is strictly lower than $\frac{n}{2}$. (e.g. 01100).

Some simple examples: