How does Cantor's diagonal argument actually prove that the set of real numbers is larger than that of natural numbers? [duplicate]

I have looked into Cantor's diagonal argument, but I am not entirely convinced. Instead of starting with 1 for the natural numbers and working our way up, we could instead try and pair random, infinitely long natural numbers with irrational real numbers, like follows:

• 97249871263434289... 0.12834798234890899...
• 29347192834769812... 0.23489712349789878...
• 42987412938478321... 0.23487912836784798...
• 43921649873612384... 0.58792834796781823...
• 49238749213847921... 0.58971238456497213...
• 98123489712348790... 0.58291739429587199...
• 45678294218374691... 0.09123498915832837...
• 69217346876217384... 0.23897123484839759...
• 52189347981283490... 0.34823948750038273...
• .
• .
• .

Couldn't we just as easily apply the diagonal argument to the natural numbers, and therefore generate a new natural number after completely exhausting our list? Here's another way to look at it:

• 97249871263434289... 0.0
• 29347192834769812... 0.1
• 42987412938478321... 0.2
• 43921649873612384... 0.3
• .
• .
• .
• 49238749213847921... 0.8
• 98123489712348790... 0.9
• 45678294218374691... 0.10
• 69217346876217384... 0.11
• 52189347981283490... 0.12
• .
• .
• .

If we accept Cantor's diagonal argument, doesn't the second argument now prove that there are more natural numbers than real numbers between 0 and 1? If you imagine this infinite list of real numbers, which in fact does include all numbers between 0 and 1 (or else the set [1, 2, 3, 4, 5, 6...] wouldn't contain all the natural numbers), it seems like the only distinction between the two sets are the inclusion of a preceding "0." for the real numbers. Does that magically make the set larger?

marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 4 '18 at 12:46

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• What is an “infinitely long natural number”? – José Carlos Santos Oct 4 '18 at 12:40
• List all natural numbers : $0,1,2,\ldots$. May you please, show in what way you want to "diagonalize" it in order to produce a number not in the list ? – Mauro ALLEGRANZA Oct 4 '18 at 12:42
• There is probably a slew of other duplicates that will illuminate you on the workings and failings of the diagonal argument. – Asaf Karagila Oct 4 '18 at 12:47

1 Answer

There are no "infinitely long natural numbers". Natural numbers have a finite decimal expansion.

Indeed, if you are looking at "infinite decimal expansions", i.e. possibly infinite strings of the digits 0 to 9, this set is uncountably infinite.

• ...and if you put a decimal point in front of each of them, you get all the (real) numbers between $0$ and $1$, which is the starting point for the Cantor diagonal argument. – John Hughes Oct 4 '18 at 13:52