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 April 25th, 2018, 01:09 PM #11 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 346 Thanks: 26 Math Focus: Number theory f(x) maps uncountable x to uncountable f(x); x is equal to or greater than f(x). g(y) maps countable y to countable g(y); y is equal to or greater than g(y). h(z) maps finite z to finite h(z); z is equal to or greater than h(z). ?
April 25th, 2018, 01:15 PM   #12
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 Originally Posted by Loren f(x) maps uncountable x to uncountable f(x); x is equal to or greater than f(x). g(y) maps countable y to countable g(y); y is equal to or greater than g(y). h(z) maps finite z to finite h(z); z is equal to or greater than h(z). ?
As a professor of mine once joked, we can only have 26 maps!

I really don't see what you're doing. Can you give an example? What is y for example? What does it mean that y is greater than g(y)?

When you say y is greater than g(y) do you mean arithmetically greater? Or the cardinality is greater? Your exposition is very murky. It's certainly true that the cardinality of a set must be greater than or equal to the cardinality of any function's image of the set, since each element of the domain gets mapped to exactly one element of the range, and more than one element of the domain may be mapped to the same element of the range.

Or do you mean these are decreasing maps? If f(x) = x - 1, then f(5) = 4 and in general y > f(y).

Which meaning are you intending?

Last edited by Maschke; April 25th, 2018 at 01:21 PM.

April 25th, 2018, 01:55 PM   #13
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 Originally Posted by romsek what it sounds like he's after is a function that maps a discrete set into a continuous one. I don't see how that could be done.
The continuum is an odd beast indeed. Even nature isn't able to produce a true continuum. Not by a long shot.

Take two numbers as close as you like. They can be so close that expressing them takes up every bit of matter in the universe and they differ only in the last bit.

There are still infinite numbers between the two.

Do this with a countable number of universes.

Still infinite numbers between the two.

Countable sets can never have elements so close together that there isn't a further continuum between the closest members.

April 25th, 2018, 02:06 PM   #14
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Quote:
 Originally Posted by romsek what it sounds like he's after is a function that maps a discrete set into a continuous one. I don't see how that could be done.
What's a continuous set? In any event the identity map from the reals with the discrete topology to the reals with the usual topology maps a discrete set into a continuous one, if by continuous set you mean something like the reals with the usual topology. A complete linearly ordered set for example.

Quote:
 Originally Posted by romsek Countable sets can never have elements so close together that there isn't a further continuum between the closest members.
I got lost trying to parse that. There are infinitely many rationals between any two rationals. But there aren't any integers strictly between 3 and 4. So countable sets may be discrete or they may be densely ordered.

April 25th, 2018, 02:52 PM   #15
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 Originally Posted by Maschke What's a continuous set?
An interval, or the countable union thereof. A set with non-zero measure.

Quote:
 Countable sets can never have elements so close together that there isn't a further continuum between the closest members.

Quote:
 I got lost trying to parse that. There are infinitely many rationals between any two rationals. But there aren't any integers strictly between 3 and 4. So countable sets may be discrete or they may be densely ordered.
I should have probably included some verbiage about the underlying set the countable set is embedded in.

My point was that given any countable set of real numbers, generally the rationals, there are infinitely many numbers between any two of them no matter how small the distance between them is.

This shouldn't be any sort of point of contention, it's simply an expression of the density of the reals on the rationals. (and vice versa)

You're arguing just to argue.

April 25th, 2018, 03:35 PM   #16
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 Originally Posted by romsek An interval, or the countable union thereof. A set with non-zero measure.
The set $\{1, 2, 3, 4, 5\}$ has measure $5$ under the counting measure but could in no way be considered a continuous set. Can you clarify? Do you mean Lebesgue measure on the reals? That's a very particular special case that the OP said nothing about.

The OP specifically asked about countable and uncountable sets in general, and said nothing about Lebesgue measure or the real numbers.

Quote:
 Originally Posted by romsek I should have probably included some verbiage about the underlying set the countable set is embedded in. My point was that given any countable set of real numbers, generally the rationals, there are infinitely many numbers between any two of them no matter how small the distance between them is.
But the OP said nothing about the real numbers. Perhaps that's what they had in mind but I wish they'd confirm that point.

Now by distance, you are assuming the sets in question have a metric. And by using the word between, you are assuming there's an order. There are many countable ordinals whose order structure as ordinals is quite wild. IMO you are assuming way more than the original question asked.

What if I biject the rationals to the natural numbers and use that order? Then there is no rational between any two others. So again, you are assuming the usual order on the real numbers, things that the OP said nothing about.

I do agree that the OP may be thinking about the real numbers. I'm not much of a mind reader. I feel that responses to the question should help the OP clarify their own ideas, not just slather on more unspoken assumptions. My opinion.

Quote:
 Originally Posted by romsek This shouldn't be any sort of point of contention, it's simply an expression of the density of the reals on the rationals. (and vice versa)
Yes but those extra assumptions (in my opinion of course) blur and obfuscate the issues raised by the OP's question, which said nothing about the real numbers or order properties.

Quote:
 Originally Posted by romsek You're arguing just to argue.
I can see how you might think that. On the contrary, I'm arguing because people are assuming things the OP never mentioned, such as distance, order, and density. The OP asked if you can somehow map a countablye set on an uncountable set using a countable sequence of maps, and the answer to that is no, as I explained earlier.

I don't think it helps the OP's thought process to assume all their unspoken assumptions, without pointing out that these are special cases. If they use the word map, that has a very specific meaning in math and it doesn't hep the OP to pretend you can somehow map a countable set on an uncountable one by taking limits. That is very far off the point of what the OP asked, even if that's (perhaps, perhaps not) what they meant.

Perhaps when you say I'm arguing just to argue, what you really mean is that I can't read people's minds, I can only read what they write.

Last edited by Maschke; April 25th, 2018 at 03:38 PM.

 April 25th, 2018, 07:22 PM #17 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 346 Thanks: 26 Math Focus: Number theory Maschke, I respect your authority that there is no way of expressing the uncountable reals (no topological "wormholes") by operating on the countable reals, but that the reverse exists -- both relations perhaps by definition. Is continuity between the two what needs to be (dis)proved ? May one say that sets of any two cardinalities exhibit continuity? I dare say that real numbers between the two sets might be fundamentally different, unlike not only in sets but also in numerical character -- not just singularities. Can one "approach" the continuum (lower limit given by 2^Aleph-null?) more closely by approaching 1+^Aleph-null from above?
April 25th, 2018, 08:15 PM   #18
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 Originally Posted by Loren I respect your authority
I plead guilty to excessive pickiness, which might have been the case here. I found your question ambiguous and confusing, but others apparently understood you better than I did. I should probably quit while I'm behind here.

Quote:
 Originally Posted by Loren that there is no way of expressing the uncountable reals (no topological "wormholes") by operating on the countable reals, but that the reverse exists -- both relations perhaps by definition.
It's certainly true that we can express every real as the limit of a sequence of rationals. In fact that's essentially how we construct the reals in the first place: as the rational sequences themselves. But this is in no way a "mapping" or a countable sequence of iterated mappings. There's just no way to make your idea work with mappings. So if the idea of mapping is important to you, I'm right to raise these issues. But if the sequence limit is appealing to you, then I'm reading too much into your use of the word mapping.

Quote:
 Originally Posted by Loren Is continuity between the two what needs to be (dis)proved ? May one say that sets of any two cardinalities exhibit continuity?
Sets don't have continuity. I made this point earlier and was accused of arguing for the sake of arguing. Yet my point is correct and calling sets "continuous" greatly confuses the issue. I must push back on the notion that a set may be regarded as continuous. There's no such definition in math. If I'm being too picky, so be it. I often am.

Quote:
 Originally Posted by Loren I dare say that real numbers between the two sets might be fundamentally different, unlike not only in sets but also in numerical character -- not just singularities.
Cardinal equivalence is in fact a very weak notion of set equivalence. For example the open unit interval (0,1) has the same cardinality and the same topological structure as the entire real line; even though these are obviously very different sets. For example (0,1) is bounded and the reals aren't. And as I mentioned earlier, the countable ordinals can get very strange. There are countable sets that are very hard if not impossible to visualize, despite being cardinally equivalent to the natural numbers.

Quote:
 Originally Posted by Loren Can one "approach" the continuum (lower limit given by 2^Aleph-null?) more closely by approaching 1+^Aleph-null from above?
I don't understand that notation. If x is any transfinite cardinal, 1 + x = x so addition can't help.

In fact the only simple operation that jumps cardinalities is exponentiation (or taking powersets, same thing). In fact Paul Cohen, who proved the independence of the Continuum Hypothesis, said that he believed that $2^{\aleph_0}$ must be far greater than $\aleph_0$ because exponentiation is a very powerful operation. For example between $2^4 = 16$ and $2^5 = 32$ we have many cardinalities. Why shouldn't it be the same for the transfinite cardinals? Of course this is only a heuristic argument and in no way a proof; but it does show how a great set theorist thinks about the size of the continuum. Cohen pointed out that taking successors is a weak operation and exponentiation is strong; so we can not naively expect the cardinal successor of $\aleph_0$ to be $\aleph_1$.

Last edited by Maschke; April 25th, 2018 at 08:18 PM.

 April 25th, 2018, 10:09 PM #19 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 346 Thanks: 26 Math Focus: Number theory Thank you for your tolerance; I am learning gradually. Sorry for the Katzenjammer. My continuum extends the typical lower limit given by 2^Aleph-null. Generalize it more closely by approaching 1^Aleph-null (not a sum, but a limit of Aleph-null's infinitude as an exponent of its base 1+ [from above] where the former diverges more rapidly that the latter converges).
April 26th, 2018, 09:28 AM   #20
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Quote:
 Originally Posted by Loren Thank you for your tolerance; I am learning gradually. Sorry for the Katzenjammer. My continuum extends the typical lower limit given by 2^Aleph-null. Generalize it more closely by approaching 1^Aleph-null (not a sum, but a limit of Aleph-null's infinitude as an exponent of its base 1+ [from above] where the former diverges more rapidly that the latter converges).
I'm afraid none of that makes any sense. $1^{\aleph_0} = 1$ no matter how you slice it. $1^n$ for any natural number $n$ is $1$, and the limit of the sequence $1, 1, 1, 1, \dots$ is $1$.

On the other hand, the set-theoretic definition of $1^{\aleph_0}$ is that it's the set of functions from $\aleph_0$ to $1$. The set $1$ is defined as $\{0\}$.

There's only one such function, namely the constant function that maps each element of $\aleph_0$ to $0$.

So that's two proofs that $1^{\aleph_0} = 1$.

Last edited by Maschke; April 26th, 2018 at 09:53 AM.

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