Endless waiting

My love for you is older than the whales,
Stretching beyond Everest to Earth’s core,
And the very deep of the Pacific.

In the midst, a twin star is pulsating–
Boundless, limitless, beyond what is time:
The definition of pure and precious

No thing could erase it
No smudge could deface it
No fault could disgrace it

Thing is–It doesn’t look like other loves
Thing is–It cannot be like other loves
(But yet it fills every piece of my heart)

I learnt to live with it and without it–
Because time is nothing to endless stars

You taught me how to wait–I don’t like it

But…

Having waited billions of years for you
My healing heart assures me: this will do

Measure theoretic properties of Cantor set

Construction of Cantor Set

Consider the interval $E_0 = [0,1]$. Remove the interval $(1,3)$ and let $E_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$. Remove the middle third of these intervals , Continuing this way we get a sequence of compact sets $E_n$, such that

1. $E_1 \subset E_2 \subset E_3 \hdots$
2. $E_n \text{is union of } 2^n \text{ intervals, each of length }3^{-n}$

The set $P = \bigcap\limits_{n=1}^{\infty} \text{ is called the \emph{Cantor Set}.}$

Proposition:
Cantor set is totally disconnected and perfect i.e., given two distinct points $x,y \in E$, there is a point $z \not\in E\text{ that lies between }x \text{ and } y$

Proof:

Observe that in Cantor set each set $C_k$ contains $2^k \text{intervals}$

Continuity and Connectedness

We know that if $f$ is a continuous mapping of a metric space $X$ into a metric space Y, and if E is connected subset of X, then $f(E)$ is connected.Here is a short proof.

$Proof$:
On the contrary, assume that $f(E)$ is disconnected.Then there exist two disjoint sets A and B such that $A\in Y,B\in Y$.

Calculus on Manifolds,Ex:1-22

Claim: If U is open and $C \subset U$ is compact, show that there is a compact set D such that $C \subset \text{Interior}(D)$ and $D \subset U$.

Proof:Since $U$ is an open set, for every $x \in U$ there exist $\delta_{x} > 0$ such that $B(x,\delta_x)$(open ball around $x$ with radius $\delta_x$) is an open subset of U. Since $\{B(x,\delta_x/2)\}_{x \in U}$ is an open cover of U and $C \subset \bigcup\limits_{x \in U}B(x,\delta_x/2) = U$. Since C is compact, there exist finitely many $x_1,x_2,\hdots,x_r$ such that $C \subset \bigcup\limits_{i = 1}^r B(x_i,\delta_{x_i}/2)$.

Now take $D = \bigcup\limits_{i=1}^{r} \bar B(x_i,\delta_{x_i})$,since closure of open balls are closed and bounded(by the radius of the ball), their entire union is compact. Naturally it follows from the construction,$C \subset \bigcup\limits_{i=1}^r B(x_i,\delta_{x_i/2}) = \text{Interior}D$. Also $\bar B(x,\delta_{x_i}/2) \subset B(x,\delta_{x_i})$ which implies $D \subset U$.

Hence Proved.