2.

Corollary. Set m = ‘− k +1 and observe that na < m = ‘− k ≤ +1 ≤ na +1 < nb. Now if both the set of rational numbers and the set of irrational numbers were countable would you be able to get a contradiction using fact 2 and 3?

Suppose xRx0 and yRy 0. Prove that m∗(A) = 1. Let A be the set of irrational numbers in the interval [0,1]. We will show that the open interval (0,1) is uncountable. Corollary. 1) The set Q of rational numbers is countable. ♥ … Therefore, we come to the inequality na < m < nb. The natural numbers IN are not bounded. 3) The union of two countable sets is countable. Let M be the least upper bound. Prove that m* (A) = 1. (i) c is a nonzero rational number. More proving, still on Real Analysis. Question: Let A Be The Set Of Irrational Numbers In The Interval [0, 1]. Take a closed subspace $[0, 1] \cap \mathbb{Q}$ of $[0, 1]$. The proof is often referred to as “Cantor’s diagonal argument” m0 is countably additive over countable disjoint collections of sets in A. 2.6. This is my solution and if you find any errors, do let me know.

Let A be the set of irrational numbers in the interval [0, 1].

This set is closed since it just consists of all the rational numbers in between $0$ and $1$, including $0$ and $1$.

However, f is continuous for all nonzero real numbers c. We do this in two cases with sequences. By the lemma, 1/2 > 0, so there exists a natural number N so that M − 1/2 < N. But then, M < N + 1/2 < N + 1, which shows that M is not an upper bound for IN. Problems Lebesgue Measure: Let $\mu$ be set function defined for all set in $\sigma$-algebra $\mathscr{F}$ with values in $[0,\infty]$.Assume $\mu$ is countably additive over countable disjoint collections of sets in $\mathscr{F}$. Let A be the set of irrational numbers in the interval [0;1].

Let K be a compact set of real numbers and let f be a continuous function on K. Prove that there exists x 0 2Ksuch that f(x) f(x 0) for all x2K. Assume that the set of irrational numbers is countable. So if we think about the interval between 0 and 1, we know that there are irrational numbers there. This implies that we could show that every number in the set of irrational numbers has a one to one correspondance with the elements of N. Note that all irrational numbers are characterized by having an infinite number of decimal places. But we may write [0,1]=⋃x∈[0,1]{x}. By problem 14 in Section 13.1 (above), Ahas the same cardinality as N N. Therefore Ais also countably infinite. Contradiction. We already know that N Nis countably infinite. Prove that the set of all irrational numbers is uncountable.

In fact, one of them that might pop out at you is 1 over the square root of 2, which is the same thing as the square root of 2 over 2, is equal-- I shouldn't say equal, is roughly, is approximately equal to 0.70710678118. Q\[0;1] is a countable set. Prove that the set A= f(m;n) 2N N: m ngis countably infinite. A set of real numbers is said to be a G_delta set provided it is the intersection of a countable collection of open sets. Let Ibe the set of irrational numbers. Suppose (U ) 2I is an open cover of f(K). Given any ǫ > 0, there exists n ∈ IN such that 0 < 1/n < ǫ. It is complete since R is complete and [0,1] is closed. In order to show that (x+y)R(x0+y ), we must show that (x+y)−(x0 +y0) is an integer.

Look at [0,1] as a metric space (a subspace of R with the usual Euclidean distance). I read that a closed subspace of a compact space is compact, so for example, consider the unit interval $[0, 1]$ which is a compact space. Singletons are closed and have empty interior, so it follows that [0,1] must be uncountable, otherwise … And I could just keep going on and on and on and on and on and on. Proof. na < ‘− k +1. Since n is a positive integer, we devide the inequlity by n withoug changing the direction of the inequality: a = na n < m n < nb n = b. The least upper bound of a set is UNIQUE (prove it) The least upper bound of a set A is often denoted as supA or LUB(A).

2), or the interval (−1,1), or the interval (0,1). E ) = m([1 k=1 F k) = X1 k=1 m(F k) (F k’s are disjoint) X1 k=1 m(E k) (F kˆE k for each k) Problem 5 (Chapter 2, Q6). For any, x2Kwe have f(x) 2f(K). Prove that m(A) = 1. From P.31, we have m(Q\[0;1]) = 0.

bound for the natural numbers. Solution. We rst prove that f(K) is compact. Problem F01.1.

2) The set R of real numbers is uncountable. Dr Rachel Quinlan MA180/MA186/MA190 Calculus R is uncountable 143 / 222 (0,1) is uncountable This assertion and its proof date back to the 1890’s and to Georg Cantor.

4. Let R be the relation on the set of real numbers R in Example 1. Prove that m0(∪∞ k=1Ek) ≤ P∞ k=1 m 0(E k).

By the density of the irrationals, there exists a sequence of irrational numbers {x(n)} which converges to c. However, {f(x(n)} = {0} converges to 0, which is not equal to f(c) = c. Hence, f is discontinuous at x = c. Prove That M*(A) = 1. Since (x+y)−(x0 +y0) = (x−x0)+(y −y0), 3.