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Cantor was Wrong: debunking the infinite set hierarchy

12 pointsby masteranzaover 4 years ago

5 comments

H8crilAover 4 years ago
This is obviously a joke by Vitalik Buterin, also it&#x27;s a good test as to whether someone actually understands the proof :)<p>If you know the proof you probably know that you don&#x27;t actually use the &quot;(n+1) mod 10&quot; mapping, you use that mapping but instead remap all 8&#x27;s into something else, like 3, in particular you don&#x27;t remap 8&#x27;s into 9&#x27;s to avoid the &quot;99999...&quot; tail problem. In fact if you like you can just use the following mapping: 0-&gt;1, {1,2,3,4,5,6,7,8,9}-&gt;0. Also, for the starting representations you simply replace all expansions with an infinite tail of &quot;9999...&quot;, as for every such representation there exists a representation that does not end in &quot;9999...&quot;. This makes the decimal expansions unique.<p>All the name-able reals are obviously countable, this has a deep connection with the Löwenheim–Skolem theorem: <a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;L%C3%B6wenheim%E2%80%93Skolem_theorem" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;L%C3%B6wenheim%E2%80%93Skolem_...</a>
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hyperpallium2over 4 years ago
Why can&#x27;t Cantor&#x27;s diagonalization argument be used to show the integers cannot be mapped to the integers?
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nabla9over 4 years ago
Funny :)<p>I hope someone comes here to argue that Vitalik is right against me.
rwjover 4 years ago
Don&#x27;t know the author well enough to know if this is a joke. However, saying something is &quot;easy to see&quot; that has not been accepted for a century must be satire...<p>In the second paragraph, there is a faulty assumption that all real numbers are computable.
comonoidover 4 years ago
Look at the date of the post :)))