Modern Abstract Analysis

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Martin Sleziak

09:09

in Calvin's Chat Room, 2 mins ago, by Bohemian relativist

@CalvinKhor I read "the real numbers are the completion of the rationals". from Wikipedia Metric Space - Types of metric spaces - Complete spaces. I actually I have read this kind of statement many times, but I keep not knowing why is so.

in Calvin's Chat Room, 1 min ago, by Martin Sleziak

@Bohemianrelativist What can be said about this depends on what you already know about metric spaces and completions and what is your definition of real numbers.

in Calvin's Chat Room, 51 secs ago, by Martin Sleziak

In fact, this can be taken as the definition of real numbers.

in Calvin's Chat Room, 22 secs ago, by Martin Sleziak

If you wish, we can briefly discuss this - but I would suggest to go to another room: https://chat.stackexchange.com/transcript/19167/2021/3/13 (To avoid having two discussions in the same room at the same time.)

If you know these two facts:
1. Rationals are dense in $\mathbb R$.
2. The metric space $(\mathbb R,d)$, where $d(x,y)=|x-y|$ is complete.

Then this already says that $\mathbb R$ is a completion of $\mathbb Q$.

The definition of the completion of a metric space $(X,d)$ is that it is a metric space $\widehat X$ which: a) is complete and b) contains $X$ as a dense subset.

A more precise formulation of b) would be that that there is an isometry between $X$ and a dense subset of $\widehat X$, but that is technicality which can actually muddle the intuition behind this notion a bit.

Bohemian relativist

09:32

@MartinSleziak I don't actually understand why "Rationals are dense in $\mathbb R$", which is actually the alternative statement of "the real numbers are the completion of the rationals", I think.

Martin Sleziak

Ok, one possible way how to say what a dense subset of $\mathbb R$ is is the following.

A set $D\subseteq\mathbb R$ is dense in $\mathbb R$ if and only if $(a,b)\cap D\ne\emptyset$ for any real numbers $a<b$.

In short: Every non-trivial interval contains a point from $D$.

So to say that $\mathbb Q$ is dense in $\mathbb R$ is the same as saying that there is a rational number between any two distinct real numbers.

Many people would accept this as obvious. (And such approach is perfectly fine in introductory courses. It is better to accept that we know what is $\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$ and what properties they have rather than to define have some way - which may be confusing for somebody just beginning to study mathematics.)

If we want to prove that $\mathbb Q$ is dense in $\mathbb R$ then the actual proof will depend on our definition of real numbers.

For example, is somebody considers real numbers as "numbers given by decimal expansion", then it is not too difficult to see that every interval contains a number where the decimal expansion ends with zeroes - which is a rational number.

For example, between 3.14159265... and 3.14200000... we definitely have the rational number 3.1416000000000000....

Bohemian relativist

@MartinSleziak excuse me, what does $(a,b)$ here mean?

Martin Sleziak

09:47

@Bohemianrelativist I use the notation $(a,b)$ for an open interval.

I.e., $(a,b)=\{x\in\mathbb R; a<x<b\}$.

Snapdragon-X

Nice room name!

Bohemian relativist

@Snapdragon-X I didn't know what this name means until I see the description.

Martin Sleziak

@Snapdragon-X Michael Greinecker is the person to blame for that, originally it was intended for functional analysis, but it§s probably good that he broadened the scopre of the room.

Nov 25 '17 at 10:07, by Michael Greinecker

room topic changed to Modern Abstract Analysis: For functional analysis, measure theory, and related areas. [functional-analysis,] [measure-theory,] [real-analysis]

Bohemian relativist

because I have never taken a pure math course.

my ex-supervisor teaches a course Function Analysis in Physics this semester.

Martin Sleziak

Ok, I'll have to go - but I will try to check this room again to see whether there are some other questions related to $\mathbb R$ as the completion of $\mathbb Q$.

Bohemian relativist

10:16

@MartinSleziak Is $a, b\in\mathbb R$, too?

Martin Sleziak

@Bohemianrelativist Yes, $a$ and $b$ are real numbers such that $a<b$.

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