(To others who may care) Anyway, the above is a minor digress as currently I am reading about the nested interval theorem to remind myself on how to prove the compactness of closed intervals in the reals
Hmm... so the outline of the proof combining these sources is as follows:
1. The reals are linearly ordered. Pick a closed interval $[a,b]$.
2. Generate a sequence of intervals by picking some $c \in [a,b]$ and $d \in [a,b]$ such that $c \leq d$
3. All bounded sequences will converge to its infimum or supremum. Therefore these are contained in any interval constructed
4. In the case of creating nested intervals by halving, the infimum of the sequence with 1st term a and supremum of the sequence with 1st term b coincides, to some $y$. Since [y,y] is be definition a singleton, we have the required result
Bolzano-Weierstrass Theorem
2. Generate nested intervals by halving a and b
3. By nested interval, the arbitrary intersection exists and is some number y
4. Let U be a neighbourhood of y defined by $(y-r,s-y)$. Pick one of the closed intervals I form the nested intervals such that $I\subset U$. Regardless of $U$ there are uncountably many points in these intervals, hence y is an accumulation point. Since [a,b] is arbitrary, it followed every real number is an accumulation point
... and I don't want to use contradiction...
1. Let $C$ be a cover of $[a,b]$
2. Any such $C$ must obey the property that $\{a,b\} \subset C \cap [a,b]$
3. Since the arbitrary union of any open set is open, it follows there exists no sequence of unions of open intervals with infimum a or supremum b that can contain a or b.
4. Now consider some open interval $(c,d)$ such that $c < a, d > b$. $(c,d)$ can be covered by some infinite cover $C$.
5. By Bolzano-Weierstrass Theorem and that arbitrary union of open intervals are open, all points on the real line are accumulation points, therefore $C$ cannot consists of a subsequence of open intervals with some supremum or infimum $x \in [a,b]$ that includes $x$. (In more intuitive terms, the open cover cannot contains an infinite "tail" converging to some point x in [a,b] without leaving x uncovered)
6. Therefore there are no countable sequence of open intervals within $[a,b]$, and thus any infinite sequence in the infinite cover $C$ must be located outside $[a,b]$
7. Therefore the union of open intervals within $[a,b]$ is finite. Meanwhile for the region $(c,e)$ and $(f,d)$ where $e < a,b > f$, the infinite bounded sequence of intervals can form a union to give finite open intervals as a sub cover. for said region.
8. Therefore $[a,b]$ always have a finite sub cover and is compact
5
Let $A$ be a subset of $R$ which consist of $0$ and the numbers $\frac{1}{n}$, for $n=1,2,3,\dots$. I want to prove that $K$ is compact directly from the definition of compact.
So, given any open cover of $A$, I should be able to find a finite subcover. Proving a set is compact is much difficult...
A more elegant proof here
typo: more elegant proof strategy
Well, we know that $e^{ni} = \cos(n) + i\sin(n)$.
We also know that $\sin(-n) = -\sin(n)$ and $\cos(-n) = \cos(n)$, so we can figure that $e^{-ni} = \cos(n) - i\sin(n)$, and thus $e^{ni} - e^{-ni} = 2i\sin(n)$.
This is a classic extended definition of $sin$, where $sin(x) = \frac{ie^{-ix} - ie^...
We define $\epsilon\text{-LCM}(a,b)$ for $a,b \in \mathbb R$ as the least non-negative real number such that there exists non-zero $n,m \in \mathbb Z$ with $|\epsilon\text{-LCM}(a,b) - na| \le \epsilon$ and $|\epsilon\text{-LCM}(a,b) - mb| \le \epsilon$. This is always defined if $\epsilon \gt 0$...
Im looking for examples of $A,B,C,D$ such that :
$A$ halts If $B$ does not halt.
$C$ does not halt If $D$ halts.
