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I will construct a bijection from the open interval [math](0,1)\subseteq{R}[/math] to [math]\mathbb{N}[/math]. It works as follows:
Let [math]x\in (0,1)\subseteq\mathbb{R}[/math]You give me the number of decimal digits [math]d[/math] needed to represent [math]x[/math], and I will give you the interval in [math]\mathbb{N}[/math] it appears on. This interval will have the exact number of natural numbers as the number of possible reals that require [math]d[/math] digits to represent (thus bijection).
[math]d=1[/math]: [math]x\in [0, 10^1)\subseteq\mathbb{N}[/math]
[math]d=2[/math]: [math]x\in [10^1, 10^3)\subseteq\mathbb{N}[/math]
[math]d=3[/math]: [math]x\in [10^3, 10^6)\subseteq\mathbb{N}[/math]
[math]d=4[/math]: [math]x\in [10^6, 10^10)\subseteq\mathbb{N}[/math]
...
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What about infinite decimals
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>>8923897
kek
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>>8923897
What about infinite naturals. Infinity aint no shit nigga
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>>8923991
They are different infinities
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>>8924010
t. Cantor
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bump for btfo cantor
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>>8923850
Deeply retarded post. You haven't even defined what any of the numbers maps to. But an attempt to define a bijection for any of those intervals will fail because there are only NINE one-digit numbers on (0,1): .1, .2, .3, .4,... .9.
While there are TEN natural numbers in the interval [0,10): 0, 1, 2, 3, 4, 5... 9.
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>>8924561
blah blah off by one error fug off
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>>8923850
Firstly you didn't define any function, secondly at no point you demonstrated that anything is bijective, thirdly, you appear to be defining a function between R and something like P(N)???? which appears to be unrelated to what yo wanted to show, fourthly your Latex is awful, fifthly you are trying to show something which is wrong.
Thread posts: 11
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