In his proof that Q is dense in R, he makes the claim
Hence there is an integer [;m;] (with [;-m_2 \leq m \leq m_1;]) such that [;m-1 \leq nx < m;].
Where [;nx;] is a real number and [;m_1,m_2;] are positive integers. Intuitively I can see how it makes sense but I'm not seeing how it's justified rigorously.
EDIT: This is from Principles of Mathematical Analysis a.k.a. Baby Rudin.
Posted Apr 21, 2014, 5:30 PM UTC
Take m = min{m' in Z : nx < m'} (this exists by the fact that a subset of the integers which is bounded below admits a minimum; alternatively, the fact (proved in Rudin) that a closed subset of R contains its inf). Then clearly m satisfies m ≤ m_1, and also -m_2 < nx < m. Also, if m - 1 ≤ nx was false then we would have nx < m - 1; but then m wouldn't have been minimal with that property.
Of course, immediately after posting this, I think I see the solution. If nx is an integer (call it m), then it's less than m+1. That is, [;m = nx < m+1;]. If nx is not an integer, then there is an m greater than it ([;nx < m;]) that is less than 1 greater than it, i.e. [;m-nx<1;]. So [;m-1<nx;]. That is, [;m-1<nx<m;]. Combine both cases and you get [;m-1 \leq nx < m;].
If nx is not an integer, then there is an m greater than it ([;nx < m;]) that is less than 1 greater than it, i.e. [;m-nx<1;].
This is what you have to prove; it isn't immediately obvious if you don't use the infimum or some property of the real numbers. It's like saying that one can "obviously" partition the real line into subsets of the form [n,n+1) with n an integer.
I think it follows from the least upper bound property, but you're right.
In this case, the equivalent "greatest lower bound property".
This is mostly paraphrasing u/skaldskaparmal, but...
Take the set of integers {m' | nx < m'}. This is bounded below by nx, so it has a greatest lower bound, call it m. By definition of lower bound, m-1 is not in our set (since m-1 < m). So ~(nx < m-1) implies nx >= m-1.
The "Hence" suggests there is useful information before the statement that would help.
If I remember right, you already have the fact that every set of integers bounded below has a minimum.
Using that, you form the set of integers {z in Z | nx < z} which is clearly bounded below (by nx), and so it has a minimum element m. Then m - 1 is not in the set (since m is the minimum), so it's not the case that nx < m - 1, so it must be that m - 1 <= nx by trichotomy
He had stated that [;-m_2 < nx < m_1;]. Your post was really helpful, thank you.
I don't think that had been discussed yet in the text.