2
$\begingroup$

Let us consider an infinite binary sequence $s$ (i.e., an infinite sequence of 0s and 1s). Let us then consider set $S(s)$, which consists of all the different sequences that can be obtained by swapping the positions of the elements of $s$. For example, if $s_1 = (1, 0, 0, 0, 0, 0, \dots)$, then $S(s_1) = \{ (1, 0, 0, 0, 0, 0, \dots), (0, 1, 0, 0, 0, 0, \dots), (0, 0, 1, 0, 0, 0, \dots), \dots \}$.

Clearly, in my example case, $S(s_1)$ is a countably infinite set. However, I'm interested in whether such an infinite binary sequence $s$ exists that $S(s)$ is an uncountable set.

My intuition says that the set of all infinite binary sequences can be expressed as a countable union of sets $S(s_1), S(s_2), S(s_3), \dots$ for some sequences $s_1, s_2, s_3 \dots$, which would mean that $S(s_i)$ must indeed be an uncountable set for some $s_i$, since the set of all infinite binary sequences is uncountable. However, I cannot figure out any $s$ such that $S(s)$ would be an uncountable set.

$\endgroup$
1
  • $\begingroup$ Can you precise what you mean by swapping the elements of $s$…? $\endgroup$ Nov 11, 2021 at 20:09

1 Answer 1

1
$\begingroup$

Hint

$$s=(1,0,1,0,1,0,\dots)$$is such an element as any sequence having an infinite number of $0$ and $1$ can be obtained by reordering the elements of $s$. And the set of sequences having and infinite number of $0$ and $1$ has the power of the continuum. In particular is uncountable.

$\endgroup$
0

You must log in to answer this question.

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