[skykooler]

"An infinite number of finite sequences" does not mean "all finite sequences". For example, the number 1.10100100010000100001... contains an infinite number of finite sequences, but does not contain, for example, any sequences with a 2, or even the sequence '10101'.

[aesthetics1]

You cannot bend the rules in this way and still equate it to the question at hand - you're introducing characters outside of base2 in your example. If you stuck within the realm of base2, '2' in its binary form certainly appears in the pattern.

Not only that, but Pi is normal (all digits distributed evenly), and your sequence clearly is not.

I understand what you are saying - indeed there are infinite finite sequences - but it just does not apply here.

[evanb]

The example isn't base two. It's base 10. It just happens to be a number that only has 0s and 1s. Like eleven. Or 1000.

[JamilD]

You can trivially modify their example to be 1.02003000400005... cycling through the digits and the argument still holds.

As another commenter pointed out, pi is not proven to be normal, but is suspected to be.

[SomeStupidPoint]

They already showed 21 (in binary, 10101) doesn't appear anywhere, even as a binary pattern.

[TOMDM]

To phrase it another way, there are infinitely many sizes of Infinity. Just because you have one infinite sequence, does not mean it is the same size as another Infinity and as such may include or exclude elements of the other Infinity.

[monochromatic]

Pi is not known to be normal.