Does the decimal representation of $\pi$ contain almost all numbers? Let $\Pi \in \Sigma^\infty$ be the decimal representation of $\pi$ where $\Sigma=\{0..9\}$. It is not known whether $\Pi$ contains each natural number. On the other hand, no number is known not to occur in $\Pi$. (Hope I've got that right)
First of all: is this property unique to $\pi$ or do all irrational numbers share it?
UPDATE: As pointed out by Brian and MJD, irrationality is completely unrelated: 1. there are irrational numbers which are known not to contain all numbers in their representation (Brian) 2. there are irrational numbers whose representation contains all numbers (MJD).
At any event, I think it should be much easier to show that almost all numbers are present, i.e. $ \exists N\in\mathbb{N}\forall k>N\exists p\in\Sigma^\star\exists s\in\Sigma^\infty: \Pi=pR(k)s$ where $R:\mathbb{N} \to \Sigma^\star$ maps each number to its decimal representation. So "allmost all" here means all except finitely many.
Has something like this been proven or disproven?
In general I don't think it's possible to construct a non-repetitive stream of symbols $\sigma \in \Sigma^\infty $ that does not contain almost all $w\in\Sigma^\star$ ("almost all" with respect to some enumeration of $\Sigma^\star$). (EDIT: This is possible as Brian points out)
 A: Every decimal that is not eventually repeating represents an irrational number, so it’s very easy to construct irrational numbers that do not have the property, e.g., $0.01001000100001\dots~$.
A: Champernowne's constant is irrational and obviously contains every integer in its digits.
Not every irrational number contains every integer in its digits.  For example, consider $0.1010010001000010000001\ldots$ or $0.101011010101011010101\ldots$ where the $n$th block is $1$ if $n$ is a square, $01$ otherwise. Obviously both of these omit nearly all sequences of digits; they even omit nearly all sequences of zeroes and ones.
A: There are many irrational numbers known not to contain some particular number. There are uncountably many irrational numbers which do not contain the decimal digit 7, for example -- and therefore don't contain 17, 73, or any numbers with a 7. Since almost all numbers contain a given decimal digit, this shows that there are uncountably many irrational numbers which fail to contain most integers as substrings.
As far as I know it is compatible with current knowledge that all digits of $\pi$ after some point are either 3 or 7 (no significance to these choices, obviously). So we can't even prove that almost all numbers appear.
