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j0k3R_
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Homework Statement
Let A = {m + nx | m, n in Z} and x irrational. Show cl(A) = R.
j0k3R_ said:yes
also one more thing, iam feeling little bit uneasy when you give "subtract then divide everything in interval"; it is valid to say we have x in (changed interval), then x(operation) in (interval * operation)? because maybe i am thinking "there exists some pair (m, n)"
cellotim said:My proof is correct. I was not changing the number m + nx, m and n integers into p + qx, p,q rationals, but subtracting then dividing to get x. The proof of denseness only requires that every open set around a member of the reals contains a member of A, one member, which I constructed in my proof.
Furthermore, the reals are only a closed set if they contain -infinity and +infinity, which, in my graduate courses was referred to as the extended reals. A set is closed if and only if it contains all its limit points. Therefore, whether the reals are a closed set depends on how you define them.
Dick said:I define the 'reals' as the 'reals'. The usual one. Standard topology. The real line contains all of it's limit points without being extended. I reread the proof. As near as I can tell it looks like you are taking the set A to be {m+nx : m and n integers, x ANY irrational}. That would make A the set of ALL irrationals. We already know that is dense. The set A they are referring to is for x a fixed irrational. Like all numbers of the form m+n*sqrt(2).
j0k3R_ said:"Can you see where to go from there? (Hint: think pigeonhole principle and try to show the numbers are dense on [0,1])."
can you illustrate for me how ot use pigeonhole principle here
cellotim said:Yes, I see my mistake. In which case we are looking for an m such that mx - floor(mx) is in the open set [tex](y-\varepsilon,y+\varepsilon), y\in[0,1][/tex]. On that point you are correct.
On the second, I think not. This is largely semantics, but perhaps important. Let me give you a quote from Dudley's Real Analysis. "[T]he two points [tex]-\infty[/tex] and [tex]+\infty[/tex] are adjoined to R . . . The result is the so-called set of extended real numbers, [tex][-\infty,\infty][/tex]." There is a wikipedia entry on it: http://en.wikipedia.org/wiki/Extended_real_number_line" .
cellotim said:Take an integer n and the numbers 0*x, 1*x, ..., n*x of which there are n+1. Now partition [0,1] into [0, 1/n), [1/n,2/n), ... ,[n-1/n,1] intervals of which there are n. By the pigeonhole principle, two of the fractional parts of the n+1 numbers must fall into one of the n partitions.
cellotim said:For any [tex]x\in R[/tex], one can construct an open ball that is in R. This shows that R is open.
In mathematics, cl(A) refers to the closure of a set A, which is the smallest closed set that contains all elements of A. R represents the set of real numbers. In this context, we are trying to prove that the closure of set A is equal to the set of real numbers.
Proving that the closure of set A is equal to the set of real numbers is important because it helps us understand the properties and characteristics of both sets. It also allows us to make connections between different mathematical concepts and to solve more complex problems.
One strategy is to show that every element in R is also in the closure of set A, and vice versa. Another strategy is to use the definition of closure, which states that the closure of A is the intersection of all closed sets that contain A.
No, the closure of set A can never be a proper subset of R because by definition, the closure must contain all elements of A and R is the set of all real numbers. This means that cl(A) must be equal to or a superset of R.
Yes, this concept is commonly used in topology, which is a branch of mathematics that studies the properties of spaces and the relationships between them. It is also used in engineering and computer science, particularly in the analysis of algorithms and data structures.