Set with a metric \(d: S\times S \to R^{\geq 0}\). Metric obeys non negativity, positive definiteness, symmetry, \(\triangle\) inequality. Eg: Euclidian k space: \(R^{k}\): every point is a vector.
Absolute value: \(|x| = d(x,0)\) for some 0.
Open ball around p of radius r
Aka r- neighborhood (nbd) of p: \(N_{r}(p) = \set{x \in S: d(x,p) < r}\). Similarly define r-nbd of set of points S. A uniform nbd of S contains some r-nbd of S.
Set of open balls defines a topological space: topology from nbds. Similar topologies for vector spaces, manifolds.
Open ball in \(R^{k}\) is convex.
Interior point p of S
If \(\exists r: N_{r}(p) \subset S\). \((0,0) \in N_{2}(1,1) \subset R^{\geq 0} \times R^{\geq 0}\) is interior pt. All others are boundary points. Thence defined interior of S: int(S), and boundary of S: bd(S) = cl(S) - int(S). If S has a non-empty interior, it is \textbf{solid!}
[0, 1] has an interior wrt R, but not wrt \(R^{2}\): then every pt is in boundary.
Limit point p of set S
\(\forall r: N_{r}(p)\) contains a pt in S other than itself.
\(\forall r : |N_{r}(p)| = \infty\): Else, can find small \(r’\) with \(N_{r’}(p) = \set{p}\). So, a finite set has no limit points.
p is the limit of some Cauchy sequence: Keep reducing r and pick \(q \neq p \in N_{r}(p)\) in each step.
Every interior pt is a limit pt, but not vice versa. For \(E\subset R\), sup(E) is a limit pt.
If p is a lt pt of E, \(\exists\) convergent seq \((s_{i})\) in S with \(s_{i} \to p\). Set with 1 limit pt: A convergent sequence in R.
Closure of E
cl(E): E with all its limit pts. Also: cl(E) = S - int(S - E).
Diameter of E
\(diam(E) = \sup_{p, q} d(p, q)\). diam(E) = diam(cl(E)): by \(\contra\), using triangle inequality.
Sets in S: Topology
Nature of the boundary
Open set S
Aka Open space. For every \(p \in S\) is an interior point. Diagramatic representation: [] and () in R, dotted an undotted lines in \(R^{2}\). Eg: dotted dumbbell in \(R^{2}\).
Open sets \(S_{i}\): \(\union S_{i}\) is open. \(S = \inters_{i=1}^{k} S_{i}\) is open: for any \(p\in S\), pick r small enough to ensure \(\forall i: N_{r}(p) \in S_{i}\).
If \(S \subset Y \subset X\): S open wrt \(Y\) iff \(\exists G \subset X\), G open wrt \(X\) and \(S = G \inters Y\): \pf if G open wrt Y, \(G\inters Y\) open wrt Y; If S open wrt Y, take \(\union_{p \in S} N_{r}^{X}(p)\) where r is radius which shows interiorness of p in S.
Closed set S
Set with all its limit points. So, finite sets closed. \([n, \infty )\) closed. Same as S with all its boundary points.
\(S\subset X\) closed iff \(S’\) open (good trick to show closedness). \(\inters\) of closed sets \(S_{i}\) is closed: \(\union S_{i}’\) is open. Similarly, \(\inters_{i=1}^{k} S_{i}\) is closed.
cl(E) is closed: as \((cl(E))’\) is open.
Non-oppositeness of Openness and Closedness
Eg: \(\phi\) and R are both open and closed. (0, 1) open wrt \(R\) but nor wrt \(R^{2}\). Half open intervals in R are neither open nor closed.
Boundedness of set S
A is bounded if \(\exists r, p: A \subset N_{r}(p)\).
Compactness
Open cover of S
Bounded Open sets \(\set{G_{i}}\) with \(\union G_{i} \supset S\). Subcovers: Subsets of open cover which also cover S.
Definition
Every open cover of S has a finite sub cover. In \(R^{d}\), compactness \(\equiv\) closed and bounded.
Properties
Finite S is compact. R is not compact: Take \(G_{n} = (n-\frac{2}{3}, n+\frac{2}{3})\), \(\set{G_{n \in Z}}\) is an open cover, but no finite or even proper subcover. Similarly, \([n, \infty]\) closed but not compact.
Any compact set S is closed: Any \(p \in S’\) is interior pt in \(S’\): \(\union_{q \in S} N_{r}(q): r = \frac{d(p-q)}{2}\) is an open cover of S, within it is some finite subcover; so \(\exists N_{r’}(p) \subset S’\).
Closed subset E of compact set S is compact: Take any open cover of E; add open set E’ to it to get open cover of S; some finite subset of this without E’ is also open cover of E.
Finite union of compact sets is compact.
If F closed and K compact, \(F \inters K \subset K\) compact: \(F \inters K\) is closed.
If \(\set{K_{i}}\) is (possibly \(\infty\)) set of compact sets and if \(\inters\) of every finite subclass \(\neq \nullSet\), \(\inters K_{i} \neq \nullSet\). Assume \(\inters K_{i}=\nullSet\); Take \(K_{1}\); every \(p \in K_{1}\) is \(\notin \inters_{i \neq 1} K_{i}\); so \(p \in \union_{i \neq 1} K_{i}’\); so finite subset of \(\set{K_{i}’}\) is an open cover of \(K_{1}\); so some finite \(\inters\) of \(\set{K_{i}}\) is \(\nullSet\): contradiciton.
So, if \(\set{K_{i}}\) compact, \(K_{n} \supset K_{n+1}\): \(\inters_{i} K_{i} \neq \nullSet\). Does not hold for open sets: Take \(G_{n} = (0, n^{-1})\).
If E is an \(\infty\) subset of compact set K, E has a limit pt in K: Else every \(p \in K\) would have some \(N_{r}(p) = \set{p}\); \(\union N_{r}(p)\) is an open cover of E without a finite subcover. Also, if every \(E \subset K, |E| = \infty \) has a lt pt in K, K is compact. \why
If \(\set{K_{i}}\) compact, \(K_{n} \supset K_{n+1} \neq \nullSet\), \(\lim_{n \to \infty} diam(K_{n}) = 0\), then \(\inters K_{n}\) is 1 pt: else \(\contra\).
Connectedness and completeness
Connectedness
A, B separated if \(A \inters cl(B) = cl(B) \inters A = \nullSet\). Eg: (0, 1) and (1, 2) but not (0,1] and (1, 2). S is connected if it is not \(\union\) of separated sets.
\(E \subset R\) connected iff it is an interval.
Dense set
Contains points in the neighborhood of every point.
Completeness of S
Limit of every Cauchy sequence \((s_{n})\) wrt metric = some point \(s \in S\).
Any closed set in complete metric space S is complete. Also, any compact space is complete.
Sigma algebra of open sets
Aka Borel Sigma algebra. This is the sigma algebra \((X, \bS)\) formed by the closure wrt \(\union, \inters, \bar{X}\) of all open sets in \(X\). All sets in \(\bS\) are called Borel sets.
Covering and packing Number
Let the space have norm \(\norm{}\), and let \(C\) be a set in it.
Covering number \htext{\(N(\eps, C, \norm{)\)}{..}}
\(\eps\) covering \(F_\eps\): Set of \(\eps\) balls which contains \(C\). Covering number \(N(\eps, C, \norm{}) = \min |F_\eps|\).
Covering entropy
Aka metric entropy. \(\log (N(\eps, C, \norm{}))\).
Total boundedness
If \(N(\eps, C, \norm{})\) is finite for all \(\eps\), \(C\) is totally bounded. Else, \(C\) is non totally bounded: for every \(n\), there is some \(\eps: N(\eps, C, \norm{}) > n\).
For D dim sphere
\(\frac{Vol(sphere(r_1))}{Vol(sphere(r_2))} = (\frac{r_1}{r_2})^D\). Let \(vol(B(f’, \eps)) = k \eps^{D}\). Then, \ \(k(R+ \eps)^{N} \geq N(\eps, C, \norm{})k \eps^{D} \geq kR^{D}\). Thence, \(\log (N(\eps, C, \norm{})) \approx D \log(\frac{R}{\eps})\).
Packing number \htext{\(M(\eps, C, \norm{)\)}{..}}
\(\eps\) packing is a set of points \(\set{g_i}\) with \(g_i\in C; \norm{g_i - g_j} \geq \eps\). The maximal \(\eps\) packing: packing number.
Relationship with N
\(M(2\eps, C, \norm{}) \leq N(\eps, C, \norm{}) \leq M(\eps, C, \norm{})\). 2nd ineq: For maximal packing \(\set{g_i}\), \(\forall h \in \)C\(: \norm{g_i-h} \leq \eps\). 1st ineq: For maximal \(2 \eps\) packing: Any \(\eps\) ball has \(\leq 1\) \(g_i\).
Use
Often easier to find than covering number; thence can bound covering number.
\(R^{k\)}: Topological properties }
See complex analysis ref.
Sequence \((s_{n)\) in S}
For properties of sequences in fields and vector spaces, see complex analysis and linear algebra ref.
Cauchy sequence
After some point, elements get closer as sequence progresses: contraction or Cauchy criterion: \(\forall m, n> N: d(p_{m}, p_{n}) < \eps\) or diameter of tail of seq tends to 0. Limit of sequence may not exist in S. Like convergence without needing a limit.
Any cauchy seq S in compact set \(X\) converges: As \(X\) compact, S has limit pt in X, also limit of S is unique.
Bounded sequences
Range is bounded.
Convergent sequence
Convergence to limit c: \(\forall i>N: d(x_{i}, c) < \eps: x_{n} \to c\). Divergence. Limit is unique. If \(x_{n} \to c\), every \(N_{r}(c)\) has all but finitely many \(x_{i}\).
Any convergent sequence is bounded. \(1^{n}\) convergent but has finite range. If range not 1, it is \(\infty\).
All convergent sequences are cauchy sequences.
Every subsequence of a convergent sequence converges to the same limit. If every subsequence of a sequence converges to the same limit, the sequence is convergent.
Sequence \((s_{n})\) in compact S has convergent subsequence: If S compact, every \(\infty\) subset has limit pt p; make seq out of \(s_{i}\) in decreasing \(N_{r}(p)\).
Subsequential limits
Take seq \(s_{n}\), subsequential limits form closed set E: Take any limit pt p of E, can find subseq limit e close to it, so can find \(s_{n}\) close to it; so p is in E.
Function across metric spaces: f:X to Y
See algebra ref for general properties of functions. Also ref on analysis of functions over R and C.
Limit of f
\(\lim_{x\to p}f(x) = q: \forall \eps, \exists \del: 0< d(x,p) < \del \implies d(f(x), q) < \eps\): f has a limit at p. q is unique. Visualize as balls in X, f(X).
\(\forall (p_n), p_n \to p, f(p_n) \to q \equiv lt_{x \to p} f(x) = q\): show \(\implies\) by \(\contra\). So, can use properties of sequences. So, get \(\lim f+g, f(x)g(x), f/g\).
Continuity of f:X to Y
f continuous at \(p \in E\) if \(\forall \eps \exists \del: d(x,p)<\del \implies d(f(x), f(p))< \eps\). If f has limit at p, continuity iff \(\lim_{x \to p} f(x) = f(p)\): f defined only over p has no limit at p but is continuous. Continuity over \(E \subseteq X\).
If f continuous at p, g continuous at f(p), then f(g(x)) continuous at p.
f continuous over \(X\) iff \(\forall\) open \(V \subseteq Y\), \(f^{-1}(V)\) open in X: Visualize interior pts, match \(\del\) balls in \(X\) with \(\eps\) balls in Y.
If f continuous, \(X\) compact, then f(X) compact: Take open cover \(\set{V_{i}}\) of f(X); \(\set{f^{-1}(V_{i})}\) is open, covers X; so take finite subcover; get \ \(f(X) \subseteq \union_{i=1}^{k} f(f^{-1}(V_{i})) \subseteq \union_{i=1}^{k} V_{i} \).
If f continuous, bijection, then \(f^{-1}\) is cont: f(V) open iff V is open.
If f continuous, \(E \subseteq X\) connected, then f(E) connected: else if f(E) separated into A, B but \(f^{-1}(A) \union f^{-1}(B)\) not separated, \(cl(f^{-1}(A)) \inters f^{-1}(B) \neq \nullSet\) or \(cl(f^{-1}(B)) \inters f^{-1}(A) \neq \nullSet\); then continuity of f violated, so \(\contra\).
Uniform continuity over X
$$\forall p, q \in X \forall \eps>0, \exists \del: \ d_{x}(p,q) < \del \implies d_{y}(f(p), f(q)) < \eps\(. \)1/x\( continuous, but not uniformly cont over \)R\(: consider points near \)0\(; neither is \)x^{2}$$. A measure of whether gradient gets very big.
If f continuous, \(X\) compact, then f uniformly cont: As \(Y\) compact: Given \(\eps\), take \(\forall p \in X: g(p)\), radius which guarantees \(\eps/2\) closedness to \(f(p)\); make open cover \(\set{N_{g(p)}}\); get finite subcover; take max \(g(p)\); use \(\triangle\) ineq to guarantee \(\eps\) closedness anywhere.
Also see the more powerful notion of absolute continuity in the complex analysis survey.
Bounding steepness
Aka Lipschitz continuity/ smoothness. Lipschitz condition: \(d(f(x), f(y)) \leq L d(x, y)\). \(L\) is lipshcitz constant. Note that it implies the usual notion of continuity.
But, it does not imply differentiability! When differentiable, there is a relationship with the derivative, see complex analysis ref.
A generalization
Holder continuity: Holder condition of order a: \(d(f(x), f(y)) \leq L d(x, y)^{a}\).
Sequence of functions \((f_{n: \)X\( \to Y)\)
Consider the properties of sequence of functions from any set to a metric space, which is described in the survey on basic mathematical structures.
If \(x\) is a limit pt of \(E \subseteq X\), \(lt_{t \to x}f_{n}(t) = A_{n}\), then \(A_{n}\) converges, \(lt_{t \to x} f(t) = li_{n \to \infty} A_{n}\). Pf: \(d(f(t), A) \leq d(f(t), f_{n}(t)) + d(f_{n}(t), A_{n}) + d(A_{n}, A)\): make 1st and 3rd terms small by picking large \(N\), make 2nd term small by picking large t.
So, if \((f_{n})\) continuous, f continuous: see \(lt_{t \to x}f_{n}(t) = f_{n}(x)\), get \(lt_{t \to x} f(t) = lt_{t \to x} f_{n}(x) = f(x)\).