(4.1) Let E,F be nonempty subsets of R, E is compact and F is closed. Then there exist (e, f) ∈ E ×F such that d(E,F ) = d(e, f). proof Let α = d(E,F ) = inf(x,y)∈E×F d(x, y) be the distance between the two sets. Let > 0, and let x0 be a given element of E. Since E is compact, it is in particular bounded, and there exists r > 0 such that E ⊂ B(r, x0). Now consider the closed ball B(r + α+ , x0)...

Research indicates that people engage in rich mathematical practices in everyday activities, yet little is known about school-aged children’s mathematics learning within the family context. This paper reports results of an interview study with 20 families to understand contexts and activities that engage mathematics in the family setting. The results indicate that problem solving is frequent ac...

vasil'ev posed problem 16.26 in [the kourovka notebook: unsolved problems in group theory, 16th ed.,sobolev inst. math., novosibirsk (2006).] as follows:does there exist a positive integer $k$ such that there are no $k$ pairwise nonisomorphicnonabelian finite simple groups with the same graphs of primes? conjecture: $k = 5$.in [zvezdina, on nonabelian simple groups having the same prime gr...

where En are measurable sets. Let > 0. Since μ is outer regular, and inner regular on measurable sets of finite measure, there exist an open set On and a compact set Kn such that Kn ⊂ En ⊂ On μ(En)− 1 2N < μ(Kn) ≤ μ(En) ≤ μ(On) < μ(En) + 1 2N thus Kn ⊂ On and μ(On)− μ(Kn) < 1 N , and since X is of finite measure, we have μ(On \Kn) < 1 N consider the closed sets Kn and O c n. By Urysohn’s Lemma,...

Proof: For a contradiction, assume there are finitely many, p1, . . . , pn. Construct N = Πi=1pi. Then for any i, pi does not divide N + 1, so N + 1 is not divisible by any prime. This is a contradiction, since all numbers ≥ 2 are divisible by some prime. (This can be easily proved by induction.) Definition 1.4. We say that two numbers, a, and b are congruent mod m, or a ≡ b mod m if m | b− a. ...

1. Let z ∈ R, and consider the function fz : R→ R x 7→ x+ z We have fz is Borel-measurable since for all c ∈ R, {x ∈ R|fz(x) ≤ c} = {x ∈ R|x+ z ≤ c} = (−∞, c− z) ∈ B. We also observe that if E ⊆ R, then f−1 z (E) = {x ∈ R|x + z ∈ E} = {−z + (x + z)|x + z ∈ E} = −z + E. Thus we have: if E is a Borel set, then f−1 −y (E) = y + E is a Borel set. Conversely, if y + E is a Borel set, then −y + (y + ...

in fact we have equality. We have {is, s ∈ S} is a family of linear operators bounded on X∗, since for all `, sups∈S ix(`) = sups∈S `(s) which is finite by assumption. Therefore by the Uniform Boundedness Principle (X ∗ is a Banach space), {is, s ∈ S} are uniformly bounded, i.e. there exists M such that for all s ∈ S, ‖is‖X∗∗ ≤ M , i.e. ‖s‖X ≤M . (11.3) Let X,Y be Banach spaces. Let L ∈ B(X,Y )...

1. Exact sequences A sequence of R-module maps A φ → B π → C is exact if the kernel of π and the image of φ are the exact same submodule of B. A longer sequence is exact by definition if every subsequence of two consecutive maps is exact. a). Show that B π → C → 0 is exact if and only if π is surjective. The kernel of C → 0 is all of C, so it is exact if the image of π is all of C—that is, if π...

Recall that the “coefficient reduction” process reduces each μij to |μij | ≤ 1/2 without affecting the orthogonalized sequence and therefore without affecting the potential function. Once “coefficient reduction” has been performed, we check the second property. If there is some i with ‖bi+1‖ < 1 2‖b ∗ i ‖, we simply swap these two elements. This will change the orthogonalized vectors bi and b ∗...

(5.1) The contraction mapping principle Let 0 ≤ r < 1. Let (X, d) be a nonempty metric space and f : X → X be a function that is a strict contraction, that is, for all x, y ∈ X, d(f(x), f(y)) ≤ rd(x, y). Then f has a unique fixed point. proof Existence of a fixed point: Let x0 ∈ X, and define the sequence (xn)n by: for all n ∈ N, xn+1 = f(xn). Then (xn)n is a Cauchy sequence: first, we have by ...