Are you sure?

Take pi to any arbitrary length. Change one digit. A subsequence that does not occur in that expansion.

I don't think so. You have to be careful when reasoning about "infinity", since it is not something that actually exists. It's a concept that serves as a shorthand for "behavior under limit condition(s)". When we speak of "infinitely many" of something, i.e. an infinite set, we have a criterion for how the members of the set are generated, and the limiting condition is what the criterion generates when allowed to proceed without end.

In particular when you say "bigger infinity" there is only one plausible meaning for that phrase, and it's not what you think it is. The

cardinality of an infinite set is a "number" that specifies how the members of that set can be mapped onto any other set. So even though intuitively there are "more" rationals than integers—since between every two integers there are infinitely many fractions—the set of all rationals and the set of all integers have the same cardinality: there are "equally many" of them, because we can devise a method of mapping one set onto the other. In Cantor's language, they both have the same cardinality, \$ \aleph_0 \$. Any infinite set that can be mapped onto the integers is

*enumerable*: there is a way to order the set such that you can cover all of the elements by counting them.

In particular, note that the infinite set of all decimal substrings is enumerable. Proof:

I. Assemble all of the strings of length 1 in lexicographic order and count them.

0 1 2 3 4 5 6 7 8 9

II. Then append all of the strings of length 2 in lexicographic order.

00 01 02 ... 10 11 12 13 14 ... 96 97 98 99

III. Then append all of the length 3 strings, etc up to length N strings as the limit of N goes to ∞. This sequence includes all decimal substrings of any length whatsoever.

Therefore the set of all decimal substrings is enumerable and its cardinality is aleph-null.

The problem with infinity. Pi may have an infinitely long string of digits, but there will be a bigger infinity of substrings that are not in that string.

As I have shown, the set of all substrings cannot be a "bigger infinity" than the sequence of digits in pi's expansion.

∞+1 is not "bigger" than ∞. Neither is ∞+∞. To keep things clear, better to use \$ \aleph_0 \$ instead.

Take pi to any arbitrary length. Change one digit. A subsequence that does not occur in that expansion.

This is a mutilation of

Cantor's diagonal proof. Cantor proved that the cardinality of the reals, which he called the continuum or

*c*, must be greater than the cardinality of the integers, aleph-null. This amounted to a proof that the reals cannot be enumerable:

I. Consider only the reals between 0 and 1, exclusive. They have the form ".XXXXXXX"... where each X is some decimal digit. If the number is a fraction, then the expansion will be all 0 or some periodic pattern after some point, but the reals also include transcendental numbers which never repeat.

II. Suppose we have enumerated all these numbers in an (infinitely long) list. So we have a kind of grid of digits. The first row of the grid is the first number in the list, the second row is the second number, and so forth. The first column in the grid is the tenths digit of all of the numbers; the second column is the hundredths digit, and so forth.

III. So making a diagonal from the upper left, take the tenths digit of the first row, the hundredths digit of the second row, etc., to construct a number b with infinitely many digits.

IV. Now change every digit of b to produce d (for example by adding 1 modulo 10), each digit of which differs from the corresponding digit in the grid. Because d differs at at least one digit from every number in the entire grid, it cannot be in the list. But d is between 0 and 1, since it has the form ".XXXXXXX", so it should have been in the list. Therefore the assumption in (II) that we had an enumerated list of reals between 0 and 1 is false. So the reals are not enumerable and c > aleph-null.

Note that the only way this argument works is if the grid is infinitely long in both directions: both infinitely many numbers in the list, and infinitely many digits. Otherwise it can't prove anything, since there could be some position at which d does not differ from an element of the list and might indeed be in it.

Cantor also famously hypothesized that there is no

*other* cardinality between aleph-null and c, so that c represents the next set larger than the integers, or \$ \aleph_1 \$. This is called the continuum hypothesis. While most everyone believes it to be true, it has so far not been proved.

Take pi to any arbitrary length. Change one digit. A subsequence that does not occur in that expansion.

One thing you have to be careful of is arguments that prove more than intended, in other words overgeneral arguments. Since you invited me to, I take "arbitrary length" to be 1.

I. Pi to 1 digit is 3.

II. "Change one digit": okay, now I have 4.

III. "A subsequence that does not occur in that expansion": while it's true that the first digit of pi does not contain 4, it is irrelevant to the topic because 4 is contained someplace else within the expansion, which is all that I claimed (in fact, infinitely many places). And the same is true for substrings of any length, not just 1.

The property of containing every possible sequence is called

*normality*: the uniform distribution of digits. It has been known for a long time that nearly all real numbers are normal, but proving that specific numbers are normal is difficult. The known decimal digits of pi are very uniform, and those digits are known to around 10 trillion places, but uniformity of

*all* digits is unproven. We can, however, easily construct normal irrational numbers, such as

Champerowne's constant.