# Countability of the set of distinct sequences that all consist of the same elements

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.

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

$$s=(1,0,1,0,1,0,\dots)$$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.