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Gem
Charter Member
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Dec-04-00, 05:56 PM (EST) |
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"Erroneous proof (???): The real interval as a countable set."
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There must be a flaw in my thinking below. Can someone please help by pointing it out? Thanks. Theorem: The reals in the interval <0,1> are a countable set. Proof: Show that a one-to-one mapping exists between the reals in the interval and the natural numbers. Any real number
r={x: xeR, xe<0,1>} has a decimal representation given by:
r= Sum(Cn 10^-n), where n = 1 to infinity, and Cn = {0,1,...9}. Construct the corresponding natural number:
m=Sum(Cn 10^(n-1)), where n = 1 to infinity,
and Cn = {0,1,...9}. For any real number, r, there exists a corresponding natural number, m. This correspondence provides a one-to-one mapping of the reals in the interval <0,1> onto the natural numbers. Therefore, by definition, the real numbers in the interval form a countable set. Clearly, this is not true -- but what's wrong?!! A different, but related question is why can't we construct a counter- arguement for the uncountablity of the natural numbers in a way similar to the Cantor arguement for the uncountability of the reals (i.e., no list can contain a complete denumeration of the reals -- a procedure is given for constructing one not already on the list). Isn't the same thing true of the natural numbers? -- crudely thinking of them as a reflection of the reals in <0,1> "onto the other side" of the decimal point?
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alexb
Charter Member
672 posts |
Dec-04-00, 06:02 PM (EST) |
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1. "RE: Erroneous proof (???): The real interval as a countable set."
In response to message #0
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>r={x: xeR, xe<0,1>} has a decimal
>representation given by:
>r= Sum(Cn 10^-n), where n =
>1 to infinity, and Cn
>= {0,1,...9}.
>
>Construct the corresponding natural number:
>m=Sum(Cn 10^(n-1)), where n = 1
>to infinity,
>and Cn = {0,1,...9}. Do you know what Pi is? Pi = 3.14159 ... The number has an infinite decimal expansion. There are ways to find any of its digits. Divide it by 10 to get a number in the interval (0, 1): 0.314159 ... What natural number corresponds to this one according to your construction? How big is it? May you bound it by some power of 10?
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Gem (Guest)
guest
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Dec-05-00, 09:36 PM (EST) |
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2. "RE: Erroneous proof (???): The real interval as a countable set."
In response to message #1
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>>r={x: xeR, xe<0,1>} has a decimal
>>representation given by:
>>r= Sum(Cn 10^-n), where n =
>>1 to infinity, and Cn
>>= {0,1,...9}.
>>
>>Construct the corresponding natural number:
>>m=Sum(Cn 10^(n-1)), where n = 1
>>to infinity,
>>and Cn = {0,1,...9}.
>
>Do you know what Pi is?
>Pi = 3.14159 ...
>
>The number has an infinite decimal
>expansion. There are ways to
>find any of its digits.
>Divide it by 10 to
>get a number in the
>interval (0, 1): 0.314159 ...
>What natural number corresponds to
>this one according to your
>construction?
>
>How big is it? May you
>bound it by some power
>of 10? I'm inclined to say the corresponding natural number is:
m = ...951413 -- and, exactly like pi, there are ways to find any of its digits (EXCEPT the "first" one -- or "last" one in the case of pi). How big and how to bound it? Hmmm. Does it HAVE to be bounded in order to be a natural number? I suspect this is the essential idea I didn't take into account: the natural numbers are generated successively (or inductively) from zero. These numbers seem to be a "granddaddy" extension of the natural numbers -- uncountable and having the cardinality of the continuum. Interesting to contemplate. I think you've helped me understand an important distinction. Thanks for your help and for this wonderful site. |
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alexb
Charter Member
672 posts |
Dec-05-00, 09:49 PM (EST) |
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3. "RE: Erroneous proof (???): The real interval as a countable set."
In response to message #2
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> I'm inclined to say the corresponding
> natural number is: m = ...951413 -- First, according to your definition and the common convention, the number should have been 314159... which is clearly unbounded and short of being a nonsense it's surely not an integer. Every integer by definition is a finite, hence bounded, number. How can you write anything as ...951413 I can't fathom. > Does it HAVE to be bounded in
> order to be a natural number? Yes of course. By definition a cardinality is finite if it equals one of the integers. > I suspect this is the
> essential idea I didn't take
> into account: the natural
> numbers are generated
> successively (or inductively)
> from zero. That's right. > These numbers seem to be a
> "granddaddy" extension of the
> natural numbers -- uncountable
> and having the cardinality of the
> continuum. You lost me here. > Interesting to contemplate.
> I think you've helped me understand an
> important distinction. I could not wish for more. > Thanks for your help and for
> this wonderful site. You are welcome. |
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