My proof of Dilworth's theorem

Dilworth's theorem: Given a finite poset $P$ its width $w(P)$, minimum number of chains that cover $P$, is equal to $a(P)$, the size of its maximum antichain.

My proof isn't as elegant as textbook's, but they share a key construction in the induction step.

Proof:
Clearly $a(P)\le w(P)$, so it suffices to show that $w(P)\le a(P)$.

Induct on $|P|$. Add an element $x$ to $P$ to obain $P'$. Without loss of generality we can assume $x\in \max(P')$.

If $a(P')=a(P)+1$ then we are done, because by inductive hypothesis $$w(P')\le w(P)+1=a(P)+1=a(P').$$

Suppose that $a(P')=a(P)$, and let $$C_1,\ldots,C_{w(P)}$$ be a smallest chain decomposition of $P$. If $x\gt\max(C_i)$ for some $i$ then we can add $x$ to $C_i$ and get $$w(P')=w(P)=a(P)=a(P').$$ Hence for each $i$ let $d_i$ be the minimal element in $C_i$ such that $x\ngeq d_i$.

Consider $$Q=\cup_i\{y|y\in C_i\wedge y\geq d_i\}.$$ Element $x$ is not comparable with $Q$, so $$a(Q)+1=a(Q+x)\leq a(P')=a(P).$$ Let $c_i=C_i-Q$, the remains of $C_i$ smaller than $x$. Note that $x$ is greater than $c_i$ for every $i$.

By inductive hypothesis, consider a smallest chain decomposition of Q $$C'_1,C'_2,\dots,C'_{w(Q)}.$$ We make all individual smallest element of $\{C'_i\}$ from distinct $C_i$ as follows. Whenever consecutive smallest elements of both $C'_i$ and $C'_j$ are from $C_k$, we merge them. By argument of infinite descent, the merging has to terminate without any empty $C'_i$.

Then, we glue every $c_i$ to an appropriate $C'_j$ such that exactly $w(P)-w(Q)$ $c_i$s remain. Finally we add $x$ to one of them, resulting in a chain decomposition of $P'$ of size $w(P)$.

Simple interpretation of geometric progression sum

A combinatorial proof inspired by real life event, that $1+r+r^2+\ldots+r^k=\frac{1-r^{k+1}}{1-r}$ for $r\in[0,1)$.

Proof:
A student wants to take $1$ unit of lessons, which could be divided into infinitely small amounts. To take any lesson he has to first schedule with his teacher. As a forgetful person, for every $L$ unit of lessons he takes $L(1-r)$ unit of them and misses $Lr$, which he then has to reschedule.

So, the student first schedules to have $1$ unit of lessons and misses $r$. He then (re)schedules $r$, and misses $r^2$, and so on until he schedules $r^k$ and misses $r^{k+1}$. The total amount of lessons scheduled is $1+r+r^2+\ldots+r^k$, and the total lessons taken is $1-r^{k+1}$. This proves the identity.

Q.E.D.

The isoperimetric problem in the plane

That is: among all closed curves in the plane of fixed perimeter, which has the largest area of enclosed region?

I have a sketchy proof that among smooth curves the answer is a circle. The proof probably resembles some existing ones, or may just be from my memory of a proof encountered almost 30 years ago.

My sketchy proof:

Let $C$ be an optimal curve and $A(C)$ be its enclosed area.

First of all, $C$ is non-concave, because otherwise a non-concave curve exists with the same perimeter and larger area.

Second, define diameter as a line segment that divides $A(C)$ by half. Clearly it also divides $C$ by half. Suppose a diameter $d$ intersects $C$ at points $x$ and $y$. Then the tangent lines to $C$ at $x$ and $y$ are both perpendicular to $d$, or else there is a concave optimal curve, a contradiction.

This implies that all diameters have the same length, and any two diameters intersect at their mid points, which always coincide. This implies $C$ is a circle.

Q.E.D.

The 3D version might be much more interesting!