Midterm 2 Flashcards

Question

How is the Bolzano-Weierstrass Theorem used?

Answer 1

It states every continuous function on a closed bounded interval is bounded.

Answer 2

Say you have a sequence of Real Numbers {a}. Then a is a Cauchy sequence if for every e > 0, there is a Natural number N so that when n, m > N we can say |an - am| <= e.

Answer 3

A cover C of X is a collection of subsets of X whose Union creates the whole space of X. Thus if Y is a subset of X, then a cover of Y is a collection of subsets whose Union is Y. Open Cover is a cover when the elements are all open sets.

Answer 4

A subcover is just a subset of the cover such that this subset still covers the entire space. Let C be a cover of a topological space X. A subcover of C is a subset of C that still covers X

Answer 5

Yes, The Function: f(x) = 1 / (1 + e ^-x) used in machine learning and neural networks.

Answer 6

A Subset D of a topological space X is Dense in A if A is in D. Thus a Set D is dense if and only if there is some point of D in each non-empty set of X. a subset A of a topological space X is called dense (in X) if every point x in X either belongs to A or is a limit point of A Every topological space is a dense subset of itself. Nowhere dense means empty or if the closure of A has an empty interior. s = Serbs, B = Bosnians. It is nowhere dense if we can make an open gate around any Bosnian and not have nay serbs within it. It is dense if we cannot make any open fences around a Bosnian that has no serbs in it. Serbs contain the Bosnians to isolated points

Answer 7

A set S is countable if there exists an injective function f from S to the natural numbers N = {0, 1, 2, 3, ...} the elements of a countable set can always be counted one at a time and, although the counting may never finish, every element of the set is associated with a natural number

Answer 8

Q is dense in R, because its limit points are all real numbers and its closure gives R. Z is nowhere dense in R because Z has no limit points and hence its closure is itself. Z does not contain any open intervals. Z is all isolated points, there is no need for limit points.

Answer 9

Set whose closure is an empty set. Any subset of nowhere dense is nowhere dense. Union of finitely many nowhere dense sets is nowhere dense. Union of countably infinite nowhere dense sets might not be nowhere dense. Complement of closed nowhere dense set is the open dense set. A nowhere dense set is always dense in itself. {Z} is nowhere dense in R Put another (equivalent) way, a set A is nowhere dense iff for every non-empty open set U there is a non-empty open set V such that V⊆U and A ∩ V=∅.

Answer 10

An interior of S is all the points of S that do not belong on the boundary of S. The interior is the compliment of the closure of the compliment of S. Another way to write it is: for s in E is interior to E if for some r > 0 we have { d(s, s0) < r} in E.

Answer 11

When it impossible to write x as the union of two disjoint open sets. When a plane has an infinite line through it then it is disconnected.

Answer 12

Between any two real numbers there is a rational. The integers are not dense in R because we can find 2 reals with no integers between them.

Answer 13

Limit can exist for partial sequences in {a}, but Lim Sup is one value for the sequence. The Lim Sup can exist and the Lim may not. Ex. {3, -7, 7, -1, 1, -1, 1..... The limit does not exist, but the Lim Sup = 7.

Answer 14

We can change it to | Lim Sup Sn Tn = s * limSup(Tn)

Answer 15

Tn + Sn means you have all elements of Tn added to all elements of Sn. Like add index 0 of Tn with index 0 of Sn and only with matching indices. Tn * Sn is the same way, but you multiply it.

Answer 16

``` >=. And the proof is that for some e > 0. t - e/2 < tk for k > N1. And s - e/2 < sk for k > N2 Then for n > Max(N1, N2). (s - e/2) + (t - e/2) < (tk + sk) s + t - e <= Lim {inf(tk + sk)} Lemma: If x - e <= w for all e > 0, then x <= w. Thus Proven True ```

Answer 17

Every Bounded Sequence in R has a convergent subsequence.

Answer 18

If a Set has a Max, Then MAx = Supremum. Max may not exist, but Supremum does. Max (a, b) doesn't exist, but Supremum (a,b) = b. Same for Infinimum.

Answer 19

Every Nonempty Subset of R that is bounded above has a least upper bound. Sup (S) exists and is a real number.

Answer 20

Metric Space (S,d) converges to s in S if the Lim n -> Infinity of d(sn, s) = 0. This can also be used to prove it is a Cauchy Sequence. e > 0 where m,n > N then d (sm, sn) < e.

Answer 21

If every Cauchy Sequence in S converges to some element in S.

Answer 22

Set of points in the Interior of E. The Interior of E is a point strictly within the boundary.

Answer 23

Then its compliment S \ E is an open set. The intersection of any closed set is closed. A set is closed if E = E^- A set is closed if it contains the Limit of every convergent sequence of points in E.

Answer 24

Then E is open in S if every point in E is interior to T. E = E^o. S is open in S. Empty Set is Open in S. Union of any collection of open sets is open Intersection of finitely many open sets is again open set.

Answer 25

The intersection of all closed sets containing E. | An element is in the Closure of E if it is the limit of some sequence of points in E.

Answer 26

It is the Set Closure(E) / Interior(E). A point is in the boundary of E if and only if it belongs to the closure of E and its compliment. Always Closed

Answer 27

A set is considered open if we can draw a ball around any point within the set and have that point be contained in the set. A set is closed if its compliment is the open set. Closed and Open are NOT Opposites. If a set is not open then it does not mean it is closed. (0,1) is open because we can draw a ball around any point in (0,1) and see that there is some interval that contains that point aka some ball. [0,1] is closed because we can see that the compliment of [0,1] - namely (-infinity, 0) U (1, infinity) is open. Neither is (0, 1] Both is Empty Set and R

Answer 28

The two element set {a, b}

Answer 29

Heine-Borel Theorem States that A subset of Rk is compact if and only if it is closed and bounded.

Answer 30

a. Interior = 1/n for n in naturals. Closure = {1/n n in naturals} U {0}. b. Interior = Empty Set. Closure = R c. Interior = Empty Set, Closure = Undefined d. Interior is r^2 < 2. Closure = [-root(2), root(2)]

Answer 31

A set strictly within the bounds of S and not on the boundary. A point q is an interior point of E if there exists a ball at q such that the ball is contained in E. The the interior set is the set of all interior points. But it really depends on what the set is a subset of. If Q is a set of Q, then the interior is Q, but if Q is a subset of R, then the Interior is empty Set since we can't take any Q and create an interval with no reals in that interval.

Answer 32

The set containing the boundaries and all the limits of the set S. Thus it will usually include the elements in the set. as well.

Answer 33

Prove that the sequence of partial sums converges to a real number. Find the limit of the sequence and see if it converges to some Real s. ∑ A converges if Lim A = 0. Breaking up into the subsequences also implies that a Series converges only if it satisfies Cauchy Criterion.

Answer 34

If ∑|A| converges then converges absolutely. If Converges absolutely then it is convergent.

Answer 35

∑ar^k = a(1 - r ^(n + 1)) / (1 - r). If you want to prove this, then make sure if starts at 0 and goes to infinity and it has an 'a' and an 'r' setup just like this. If r < 1, then ∑ar^k = a / (1 - r)

Answer 36

∑1/r^p converges if and only if p > 1.

Answer 37

Sn is a Cauchy sequence if for some e > 0 there exist an N such that n, m > N and |Sn - Sm| < e. Every converging sequence is a Cauchy Sequence. Every Cauchy Sequence Converges. Can you find N so that any pair of elements after N is within a distance of 1 from each other?

Answer 38

A Series ∑An of NON-ZERO terms converges absolutely if lim sup(|A(n + 1) / An|) < 1 diverges if Lim Inf(|A(n + 1) / An| > 1 Undetermined if = 1.

Answer 39

For Series ∑A . Let a = Lim Sup |A|^(1 / n). Converges if a < 1 Diverges if a > 1 Undetermined if a = 1.

Answer 40

∑A is a series where An >= 0 for all n. If ∑A converges and Bn <= An for all n, then ∑B converges If ∑A = infinity and Bn >= An for all n, then ∑B = infinity.

Answer 41

compare to 1 /n^2 to show it converges. p test.

Answer 42

Use Integral. 1/x from 1 to n + 1 is ln(n+ 1) which is infinity. Same proof for 1/x^2, but you don't have an ln.

Answer 43

Ratio and Root and Comparison do not apply. Terms in ∑A are nonnegative There is a decreasing function f on [1, infinity] such that f(n) = An for all n. To be decreasing, if x < y, then f(x) < f(y) If lim of the [integral from 1 to n] f(x)dx = infinity then diverges if < infinity than converges. This allows us to proves that series like ∑(-1) ^n / root(n) converges, but does not absolutely converge.

Answer 44

If a1 >= a2 >= a3 >= a4 >= a5 >= an >= 0 and lim an = 0, then the alternating series converges.

Answer 45

1. Every nonnegative number has at least one decimal expansion. 2. A Real number x has exactly one decimal or two (one ending in all 9s and one ending in all 0's). 3. A number is Rational only if its expansion is repeating.

Answer 46

0.2: 10x = 2.0 (multiply to get repeating part to the left of decimal point) Repeating already to the Right. 10x = 2.0 and thus x = 1/5 ``` 0.(2 repeating): 10x = 2.(2 repeating) (Repeating on Left) x = 0.(2repeating) repeating Right Subtract two equations: 9x = 2. x = 2/9! ``` ``` 0.1(492 repeating) 10000x = 1492.(1492 repeating) 10x = 1.(492 repeating) Thus 9990x = 1491 x = 1491/9990 ```

Answer 47

Just Do long Division or where possible try to multiply to get a power of 10 on the bottom and then you can get the decimal really easily.

Answer 48

f(x) is continuous at x0 in dom(f) if, for every sequence (xn) in dom(f) converging to x0, we have lim f(xn) = f(x0). Then f is continuous on S. S in dom(f). |f| and kf must be continuous as well.

Answer 49

Suppose Lim Xn = X0. Then we have Lim f(Xn) = Lim [2(Xn)^2 + 1] = 2[LimXn^2] + 1 = 2(X0)^2 + 1 = f(X0).

Answer 50

If f is continuous at x0 and g is continuous at f(x0), then composite g o f is continuous.

Answer 51

Then the function is bounded and f assumes its maximum and minimum values on [a,b]. Exists x0, y0 in [a,b] such that f(x0) ≤ f(x) ≤ f(y0)

Answer 52

Uniform continuity is for a set, continuity is for a point or interval. Something is Uniformly continuous for a function or a set. It doesn't make sense that a function is uniformly continuous at each point.

Answer 53

e > 0. |f(x) - f(y)| = |x^2 - y^2| = |x - y||x + y|. |x + y| ≤ 14. Then |f(x) - f(y)| ≤ 14|x - y|. ∂ = e/14. Then |x - y| < ∂ implies |f(x) - f(y)| < e.

Answer 54

it is uniformly continuous on [a,b].

Answer 55

Sn = (1/n) and then (Sn) is not a Cauchy Sequence. Thus f(Sn) cannot be uniformly continuous on (0,1)

Answer 56

it can be extended to a continuous function f´on (a,b).

Answer 57

If Lim x -> a f(x) = f(a), for all a in dom(f(x)) then it is continuous.

Answer 58

use the fact that a function is continuous on (a,b) if it is continuous on [a,b].

Answer 59

For each e > 0. there exists ∂ > 0 such that d(s, sO) < ∂ implies d*(f(s), f(s0)) < e. A function f : S -> S* is continuous if and only if f^-1(U) is an open subset of S for every open subset U of S*.

Answer 60

f(E) is a compact subset. f is uniformly continuous on E. f is bounded on E. f assumes its maximum and minimum on E.

Answer 61

Anything that can be listed as a sequence.

Answer 62

a. If Un is a sequence of dense open subsets in S, then the intersection of these sets is dense in S. b. If F is a sequence one closed subsets of S and the union of the F's contains a non empty open set, then so does at least one of the sets of F. c. Unions of sequence of nowhere dense subsets in S has a dense compliment. d. Space S is not a union of a sequence of nowhere dense subsets of S.

Answer 63

A space which is a union of two disjoint non-empty open sets is called disconnected. Take Open Sub Sets U1 and U2. n = intersection. (E n U1) n (E n U2) = ø and E = (E n U1) U (E n U2) when (E n U1) ≠ ø and (E n U2) ≠ ø. Or If E = A U B and (closure A) n B = ø and A n (closure B) = ø.

Answer 64

Sets are either First or Second Category. S is second category if it cannot be written as the countable union of subsets which are nowhere dense in S. First Category are not second category. Rationals in R are First Category Irrationals in R are Second Category Z inherited from subset of R is Second Category since every subset of Z is open in Z. Z in R is First category.

Midterm 2 Flashcards

(90 cards)