1
$\begingroup$

We consider the measurable space $(M,\mathcal{B}(M))$ where $M=\{0,1\}^{\mathbb{N}}=\{\omega_1,\omega_2,\dots\},\omega_i=0$ or $ \omega_i=1$ i.e the space of all binary sequences indexed by the naturals and $\mathcal{B}(M)$ is the product topology (the coarsest topology which which makes all the projections continuous).

Let $\pi_n: \Omega \to \{0,1\}^n$ given by $\omega \to (\omega_1,\omega_2, \dots)$ be a projection and $\mathcal{F_n}=\pi_n^{-1}(\mathcal{B}(\{0,1\}^n))$.

Why is $\mathcal{F}_n$ finite? the way I reason, I am quite convinced that it should be uncountable. Clearly $\mathcal{B}(\{0,1\}^n)$ is the power set on all n sized binary sequences. I mean even $\pi_5^{-1}((0,1,0,1,0))$ is the set of all $ \omega \in \{0,1\}^{\mathbb{N}}$ whose first 5 entries are $0,1,0,1,0$.

I am very confused as to why is $\mathcal{F}_n$ is finite.

$\endgroup$
6
  • $\begingroup$ If I understand correctly, the elements of $\cal F_n$ are not finite. What is the issue? $\endgroup$
    – Not Mike
    Mar 9, 2018 at 11:38
  • $\begingroup$ The issue is that I think they aren't finite but if the book I am reading is to be believed they are finite. And this means I am some really wrong ideas and understanding of this setting $\endgroup$
    – user3503589
    Mar 9, 2018 at 11:40
  • 1
    $\begingroup$ There are only finitely many elements of each $\cal F_n$. However, each member of $\cal F_n$ is not finite. Is it possible you misinterpreted which item was asserted finite? $\endgroup$
    – Not Mike
    Mar 9, 2018 at 11:49
  • $\begingroup$ @NotMike So if I understand you correctly for example corresponding to $\pi^{-1}{0}$ is a set which contains an uncountable number of sets of binary sequences with the first element as $0$? $\endgroup$
    – user3503589
    Mar 9, 2018 at 11:54
  • $\begingroup$ That is the correct interpretation of $\pi^{-1}_1( 0 )$. However, I feel it's worth pointing out that $\pi^{-1}_1 \{0,1\}= \{ \pi^{-1}_1(0), \pi^{-1}_1(1)\}$ $\endgroup$
    – Not Mike
    Mar 9, 2018 at 11:59

1 Answer 1

Reset to default
1
$\begingroup$

If $f:X\to Y$ where $Y$ is a finite set then every subcollection $\mathcal V\subseteq\wp(Y)$ is a finite set so that also $f^{-1}(\mathcal V):=\{f^{-1}(V)\mid V\in\mathcal V\}$ is a finite set.

This can be applied on $\pi_n: \Omega \to \{0,1\}^n$ where $\{0,1\}^n$ is a finite set, and $\mathcal{B}(\{0,1\}^n)\subseteq\wp(\{0,1\}^n)$.

$\endgroup$
6
  • $\begingroup$ the inverse $f^{-1}(V)$ for $V \in \mathcal{V}$ is a set in X and if $f$ sends uncountably many elements in X to the same $V$, then why will the inverse image be finite? $\endgroup$
    – user3503589
    Mar 9, 2018 at 11:44
  • 1
    $\begingroup$ The inverse image will not be finite then but the collection of inverse images will, and that is what $\mathcal F_n$ is defined to be. We must not confuse $f^{-1}(V)$ and $f^{-1}(\mathcal V)$ $\endgroup$
    – drhab
    Mar 9, 2018 at 12:01
  • $\begingroup$ So if I understand you correctly for example corresponding to $\pi^{-1}$ is a set which contains an uncountable number of sets of binary sequences with the first element as 0? and so on for all finite $\pi_n^{-1}\mathcal{B}\{0,1\}^{n}$ will have a maximum of $2^n$ elements $\endgroup$
    – user3503589
    Mar 9, 2018 at 12:06
  • 2
    $\begingroup$ The only thing relevant here is the question: is $\mathcal B(\{0,1\}^n)$ a finite set? As a subcollection of $\wp(\{0,1\}^n$ it is. Then consequently for every function $f:X\to\{0,1\}^n$ the collection of preimages of elements of $\mathcal B(\{0,1\}^n)$ under $f$ is also finite. For every element of $\mathcal B(\{0,1\}^n)$ there is only one preimage. $\endgroup$
    – drhab
    Mar 9, 2018 at 12:18
  • $\begingroup$ Thank you for taking out the time to help me out. I really appreciate it. $\endgroup$
    – user3503589
    Mar 9, 2018 at 12:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .