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 August 26th, 2017, 09:31 PM #1 Senior Member   Joined: Jun 2014 From: USA Posts: 364 Thanks: 26 A Philosophical Rant Stemming From R/Q If: 1) $x, y \in \mathbb{R}$, 2) $x, y \notin \mathbb{Q}$, 3) $x - y \notin \mathbb{Q}$, 4) $X = \{ x + q : q \in \mathbb{Q} \} \cap ( \,0,1) \,$, and 5) $Y = \{ y + q : q \in \mathbb{Q} \} \cap ( \,0,1) \,$, then there does not exist a $k \in \mathbb{R}$ such that $\{ k + r : r \in X \} = Y$ because: 1) $k = 0 \rightarrow X = Y \rightarrow x - y \in \mathbb{Q}$, 2) $k > 0 \rightarrow \exists a \in X \text{ such that } a > 1 - k \rightarrow k + a > 1 \rightarrow k + a \notin Y$, and 3) $k < 0 \rightarrow \exists a \in X \text{ such that } a < -k \rightarrow k + a < 0 \rightarrow k + a \notin Y$. If both $X$ and $Y$ had a least element, $x’$ and $y’$, respectively, then $k$ would exist as defined above because $k$ would equal $( \,y’ + q) \, - ( \,x’ + q) \,$ for any $q \in \mathbb{Q}$ where both $x’ + q$ and $y’ + q$ are in $( \,0,1) \,$. Alternatively, $k$ might be defined as follows: If 1) $x’’ \in X$, 2) $y’’ \in Y$, and 3) $\{ x’’ - r : r \in X \} = \{ y’’ - r : r \in Y \}$, then $k = y’’ - x”$. There is of course a rational number between any two given irrational numbers. Where the rational numbers are countable and the irrational numbers are not, it seems as though, from a cardinality standpoint, there must be ‘many’ more irrational numbers than rational numbers. The standard $<$ ordering of the reals is not a well ordering, but assuming for a moment that it were, we might picture $[ \,0,\infty) \,$ as $0$ followed by an initial uncountable segment of irrational numbers, followed by a second rational number, followed by a second uncountable segment of irrational numbers, followed by a third rational number, followed by a third uncountable segment of irrational numbers, and so on. In pairing this non-standard notion of a well ordering with the idea that a $k$ might exist as defined above for each possible $X$ and $Y$, then we would see that for any two elements $x’$ and $y’$ of the initial uncountable segment of irrational numbers there would be a $k$ such that $y’ - x’ = k$. For completeness, we might also assume there is a $k$ for each $x’$ in the initial uncountable segment of irrational numbers such that $x’ - 0 = k$. We could then let $A = \{ \text{ the set of all such non-standard elements } k \}$ and let $\mathbb{R}^{A} = \{ r + k : r \in \mathbb{R} \land k \in A \}$. The set $\mathbb{R}^{A}$ would lack the archimedian property and would not be closed under multiplication or division, but would be well ordered under the standard $<$ ordering and, by definition, more 'complete' than $\mathbb{R}$ itself. In separating from the standard view, the above crosses into the philosophical. It is at best an attempt to describe the method by which a collection of points can achieve two dimensional space (at least in my humble point of view). I realize I may have crossed into the twilight zone, but these concepts have been bugging me ever since I found this forum and learned the basics of real analysis. I guess I'm not sure why. Thoughts? Last edited by AplanisTophet; August 26th, 2017 at 09:37 PM.
 August 27th, 2017, 05:11 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra It's difficult to read some of your post on my phone because of the way the mathematics is rendered, but I find it useful to consider cardinality as a property of sets that relates something other than a measure of the number of elements in a set. Thus the reals are uncountable, but that doesn't necessarily mean that there are "more" of them than the rationals. It just means that we can't list them. More precisely, they can't be put into 1-1 correspondence and this describes a qualitative property of the two sets. This also helps with understanding the dichotomy whereby we intuitively feel that there are many more rationals than naturals despite the fact that both sets have the same cardinality.
August 27th, 2017, 05:58 PM   #3
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Quote:
 Originally Posted by v8archie Thus the reals are uncountable, but that doesn't necessarily mean that there are "more" of them than the rationals. It just means that we can't list them. More precisely, they can't be put into 1-1 correspondence and this describes a qualitative property of the two sets. This also helps with understanding the dichotomy whereby we intuitively feel that there are many more rationals than naturals despite the fact that both sets have the same cardinality.
I agree with your sentiments towards cardinality where the term 'more' isn't really defined in this sense while cardinality is given a precise mathematical definition. It is merely an intuitive interpretation that there are more irrationals than rationals based on a cardinality argument. That intuition can run somewhat wild, though in considering how 'large' a countably infinite set can get (think Hilbert's Hotel or an ordinal such as $\epsilon_0$, etc.), it seems there must be an awful lot of irrational numbers so as to ensure the dichotomy is not just an illusion (compare the countable ordinal $\epsilon_0$ to the first uncountable ordinal $\omega_1$).

There is more than just cardinality too, but it doesn't seem helpful. For example, the Lebesgue measure of any countable set of real numbers is 0, but there are also uncountable sets of real numbers that have a measure of 0 (e.g., the Cantor set). To pick on my OP's non-standard assumptions, the following would constitute a Vitali Set with an undefined measure:

Quote:
 Originally Posted by AplanisTophet ...0 followed by an initial uncountable segment of irrational numbers, followed by a second rational number, ...
Next, we obviously have sets of reals that have positive measures. In trying to toy with the notion of what is meant by 'more' irrationals than rationals, we really don't have much to go on.

The difference between a countable and an uncountable set cannot come down to a single element (due Hilbert's Hotel type arguments) but, at the same time, what's the difference between $\omega_1$ and the ordinal immediately preceding it in the hierarchy of ordinals? That, to me, makes no sense. We don't seem to be able to come to a conclusion regarding the continuum hypothesis either, but I'm leading towards there being no cardinal between $\aleph_0$ and $2^{\aleph_0}$.

I've strayed quite a bit from the concepts contained in my OP now though, which is not really my intent, so will stop this 'side rant'.

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