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 (y-\varepsilon,y+\varepsilon), y\in[0,1]. 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 -\infty and +\infty are adjoined to R . . . The result is the so-called set of extended real numbers, [-\infty,\infty]." 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 x\in R, one can construct an open ball that is in R. This shows that R is open.