We say that a real number ? is an isolated point of a set ? if ? is an element of ? and there exists ? > 0
such that ? is the only element of ? that is in the interval (? − ?, ? + ?)
(a) Prove that every element of the set ? = {1,2,3,… } is an isolated point of ?.
(b) Prove that if ? is an isolated point of dom(?), then ? is continuous at a
We write R+ for the set of positive real numbers. For any positive real number e, we write (-6, 6) = {x a real number : -e < x <e}. Prove that the intersection of all such intervals is the set containing zero, n (-e, e) = {0} EER+
We write R+ for the set of positive real numbers. For any positive real number e, we write (-6, 6) = {x a real number : -e < x <e}. Prove that the intersection of all such intervals is the set containing zero, n (-e, e) = {0} EER+
real analysis 1,3,8,11,12 please 4.4.3 4.4.11a Limits and Continuity 4 Chapter Remark: In the statement of Theorem 4.4.12 we assumed that f was tone and continuous on the interval I. The fact that f is either stric tric. strictly decreasing on / implies that f is one-to-one on t one-to-one and continuous on an interval 1, then as a consequence of the value theorem the function f is strictly monotone on I (Exercise 15). This false if either f is...
Suppose we tried to apply our real analysis definitions/methods to the set of rational numbers Q. In other words, in the definitions, we only consider rational numbers. E.g., [0, 1] now means [0, 1] ∩ Q, etc. In this setting: (a) Find an open cover of [0, 1] that contains no finite subcover. Hint: Fix an irrational number α ∈ [0, 1] (as a subset of the reals now!) and for each (rational) q ∈ [0, 1] look for an...
Suppose we tried to apply our real analysis definitions/methods to the set of rational numbers Q. In other words, in the definitions, we only consider rational numbers. E.g., [0, 1] now means [0, 1] n Q, etc. In this setting: (a) Find an open cover of [0, 1] that contains no finite subcover. Hint: Fix an irrational number a € [0, 1] (as a subset of the reals now!) and for each (rational) qe [0, 1] look for an open...
Suppose we are given a line segment of length 1. We say that a real number x is constructible if we can construct, with unmarked straightedge and compass, a line segment of length (So for example, -3 is constructible since we can construct a line segment of length 3 by joining three segments of length 1 along the same line.) Prove that the set G of constructible numbers is a field. (Constructible numbers are real numbers, so you don't need...
Number 6 please S. Let ) be a sequence of continuous real-valued functions that converges uniformly to a function fon a set ECR. Prove that lim S.(z) =S(x) for every sequence (x.) C Esuch that ,E E 6. Let ECRand let D be a dense subset of E. If .) is a sequence of continuous real-valued functions on E. and if () converges unifomly on D. prove that (.) converges uniformly on E. (Recall that D is dense in E...
Given that H is a set and that x is an interior point of the set R∖H, prove that x is not close to H. So, I have an example where they are opposite of the question, I just don't know how to spin it around. Suppose that S is a set of real numbers and that x is a given number. Then the following two conditions are equivalent to one another: 1.The number x is close to the set...
Recall that (a,b)⊆R means an open interval on the real number line: (a,b)={x∈R|a<x<b}. Let ≤ be the usual “less than or equal to” total order on the set A=(−2,0)∪(0,2). Consider the subset B={−1/n | n∈N,n≥1}⊆A. Determine an upper bound for B in A.. Then formally prove that B has no least upper bound in A by arguing that every element of A fails the criteria in the definition of least upper bound. Note: make sure you are addressing the technical...
Show that is nowhere dense in if and only if is not an isolated point of . Note: you may need these definitions. The set is said to be nowhere dense in if . Also, is an isolated point of if there exists small enough so that for all . We were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this imageWe were unable to transcribe this...