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I'm looking at the sub $\sigma$-algebra example on Wikipedia, and I don't understand the notation that is used.

The example defines the $\sigma$-algebra $G_n = \{ A \times \{H,T\}^{\infty}\ : A \subset \{H,T\}^n \}$.

My understanding is that, for example, for the case $n=1$, we may consider the set $\{(H)\} \times \{H,T\}^{\infty}$. Using the definition of Cartesian product, an element in the set could be $((H), (H, T, H, \ldots))$. However, this ordered pair is not an element of $\Omega$. I presume some kind of implicit concatenation, so that the pair really represents the sequence $(H, H, T, H, \ldots)$, which is an element of $\Omega$. Hence, $\{(H)\} \times \{H,T\}^{\infty}$ really represents the set of all sequences that begin with $H$.

But the idea of concatenation seems to against the definition of the Cartesian product. Am I missing something here?

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  • $\begingroup$ Note that $\{(H)\}$ should read $\{H\}$ since $\{(H)\}$ is not a subset of $\{H,T\}$. This explains your problem. $\endgroup$
    – Did
    Commented Feb 4, 2015 at 7:16
  • $\begingroup$ I don't think that's true. At least in the way it's defined, the subset $\{H,T\}^1$ is really $\{(H),(T)\}$. $\endgroup$
    – pyrrhic
    Commented Feb 4, 2015 at 7:19
  • $\begingroup$ No, for example, $(H)\notin\{H,T\}$. $\endgroup$
    – Did
    Commented Feb 4, 2015 at 7:20
  • $\begingroup$ I believe your comment is valid if we take out the "parentheses" surrounding the sequences. However, it confuses me as to what happens when parentheses are included around given sequences, as defined on the Wikipedia page. It would seem that they define separate sequences which can be denoted in a binary ordered pair. $\endgroup$
    – pyrrhic
    Commented Feb 4, 2015 at 7:20
  • $\begingroup$ @pyrrhic Note that $A \subset \{H,T\}^{n}$. In your example, you took $n=1$ and considered $A=\{(H)\}$. What Did pointed out is that $\{(H)\} \notin \{H,T\}$. This is why you picked an invalid $A$. $\endgroup$
    – madprob
    Commented Feb 4, 2015 at 9:50

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