General properties
For algebraic properties etc.., see algebra ref.
For Open balls, interior points, connectedness etc.. See topology ref.
Numbers
Natural numbers. Integers.
Supremum (GUB) vs infimum (GLB)
See properties of ordered sets in algebra ref.
The field of Rational numbers
Q; irrationality of \(\sqrt{2}\); \(\sqrt{2}\) as limit of rationals.
Q is an ordered field. Q is countable, as \(Z^{2}\) is countable.
Absence of least upper bound
\(\set{x \in Q, x^{2}< 2}\) does not have supremum in Q.
The field of Real numbers
Supremum property
R is a field containing Q, with the supremum property. Every \(S \subset R\) has supremum, infemum.
As Cuts from Q
Set elements of R as (Dedekind) cuts: \(a \subset Q\) which split Q into 2 parts in any way such that: if \(x \in a\), any y: \(x > y \in a\), it has some upper bound in Q, it has no max element. Eg: \(\set{x \in Q, x^{2}< 2}\). Define \(a<b\) by \(a \subset b\), \(a + b\) as \(\set{u + v| u \in a, v \in b}\) etc; map \(q \in Q\) to \((-\infty, q)\).
Archimedian property of R
If \(0<x<y\), \(\exists n \in N : y<nx\). Else, take LUB l of \(\set{nx}\), so \(\exists n: l-x<nx\), then \(l<(n+1)x\): \(\contra\).
Misc Properties
Decimal representation; its weakness: \(.\bar{9} = 1\). There is a real between any 2 rationals. So: There is a rational between any 2 reals: also by archimedian property.
There is a real number between any two rationals: scale \(\sqrt{2}\) as needed.
Extended real number system
\(R \union \set{\infty, -\infty}\): not a field.
Irrationals
Irrational numbers. Algebraic numbers: roots of polynomials with rational/ integer coefficients. Transcendental numbers: non algebraic; eg: \(e, \pi\). R dense in irrationals: Rationals are countable, but R is not. R dense in transcendentals: algebraic numbers countable.
Topological properties k cell in \htext{\(R^k\){..}}
Closed interval in R; Extend to hypercuboid in \(R^{k}\). Also see topology ref.
Nonemptiness of telescoping \htext{\(\inters\){intersection} of cells}
If \(\set{I_{n} = [a_{n}, b_{n}]}\) closed intervals in R with \(I_{n} \supset I_{n+1}\) (as if diminishing), \(\inters_{j} I_{j} \neq \nullSet\): take \(E = \set{a_{n}}\); E bounded above by every \(b_{n}\); so has sup E = x; this is in \(\inters I_{n}\).
Compactness of k-cells in \htext{\(R^k\){..}}
Take k-cell K definted by k closed intervals. K compact: For \(\contra\), take open cover \(\set{G_{i}}\) sans finite subcover. Halve these intervals to break it into \(2^{k}\) k-cells. If K is not compact: atleast one of these pieces, \(K_{1}\) is not compact. Repeat the procedure with \(K_{1}\) recursively ad infinitum. But, by theorem about diminishing intervals \(I_{n}\), \(\exists p \in \inters I_{n}\); as p interior pt of \(G_{i}\), \(\exists r, i: N_{r}(p) \supset K_{i}\): so \(K_{i}\) has finite cover: \(\contra \).
So, Bounded and closed intervals in R are compact.
Compactness in \htext{\(R^k\){..}}
S closed and bounded \(\equiv\) S Compact \(\equiv\) C: Every \(\infty\) subset E of S has a limit point of S: If S closed and bounded, it is in some k-cell, so compact. If S compact, C is true. If C true: S is closed by defunction. If C true: S is bounded: else \(\contra\): \(\set{x_{n}: |x_{n} - x_{0}| > n}\) wouldn’t have lim pt in S.
So, compact set in \(R\) has supremum in it.
Uncountability of R
\(\infty\) sequences of \(\set{0, 1}\) uncountable by Cantor’s diagonalization. So, [0, 1] uncountable. So, looking at decimal representations, R is uncountable.
If \(P \subset R^{k}, P \neq \nullSet\), P closed, every p in P is a limit point of P, then P uncountable: By a sort of diagonalization: As P has limit pts, P is \(\infty\). For \(\contra\) let sequence \(\seq{x_{i}} = P\); take \(V_{1} = N_{r}(x_{1}) \inters P \neq \nullSet\); take \(cl(V_{1})\): is compact; make \(V_{2} = N_{r}(p) \inters V_{1} \neq \nullSet\) with \(x_{1} \notin V_{2}\); do recursively; \(\inters cl(V_{i}) \neq \nullSet\), but \(\inters cl(V_{i}) \inters P = \nullSet : \contra\).
Cantor set
Take \(E_{0} = [0, 1]\); remove middle third; do this recursively; \(P = \inters E_{i}\). P is bounded and closed: so compact; also every pt is a lt pt: so uncountable; has no interval!
P has 0 measure. \why
Cardinality of \htext{\(R^{k\)}{..} is same as R}
Take \(p = (a_{1}, .. a_{k}) \in R^{k}\), represent it in binary system; map it to x such that \(x_{1:k}\) has first bits of p, \(x_{k+1:2k}\) has second bits of p, etc..
The field of Complex numbers C
Set of ordered pairs over R, with +, * redifined to form a field: more than the complex plane \(R^{2}\). An algebraic closure of R due to fundamental theorem of algebra.
Spherical form
\(z=a + ib = r (\cos \theta + i \sin \theta)\).
Use Taylor series around 0: \(\sin x = x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} \dots = \frac{e^{xi}-e^{-xi}}{2i}\); \(\cos x = 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!} \dots = \frac{e^{xi}+e^{-xi}}{2}\). So, \((\cos \theta + i \sin \theta) = e^{i\theta}\).
So, \(z = re^{i\gth}, y = r’e^{i\ga}\) related by \(y = \frac{r’}{r}e^{i(\ga-\gth)}\).
Powers of z
(De Moivre). So, (Also by induction) \([r (cos \theta + i sin \theta)]^{n} = r^{n} e^{ni\theta} = r^{n} (\cos n\theta + i \sin n\theta)\). So, \(\bar{z^{n}} = \bar{z}^{n}\).
Other Properties
Magnitude \(|z| = \sqrt{a^{2} + b^{2}}\). \(\sqrt{a+ib} = e+ih\). C has multiplicative inverse: \(\frac{1}{a+ib}=\frac{a-ib}{a^{2}+b^{2}}\) (Realify the denominator).
nth roots of 1
Divide the unit circle into \(n\) sectors: Integer powers of \(w=e^{2i\pi/n}\). \(w^{-1}=w^{n-1}=e^{-2i\pi/n}\). As \(1+w+ \dots + w^{n-1} = w(1+w+ \dots + w^{n-1})\), \((1+w+ \dots + w^{n-1}) = 0\): Also for any \(w^{x}\). \((a+ib)^{n} = re^{ni\theta}\) and \((a-ib)^{n} = re^{-ni\theta}\) are complex conjugates like \((a+ib)\) and \((a-ib)\).
Quaternions
Can’t turn \(R^{n>2}\) into a field \chk. In quarternions, you loose commutativity. In Octanions, you loose associativity.