Let $\mathbb{N}$ denote the set of non-negative integers. We say that a sequence $f:\mathbb{N}\to \{0,1\}$ is *normal* if every finite $\{0,1\}$-sequence appears in $f$.


Let the *swapping operation* $\sigma:\mathbb{N}\to \mathbb{N}$ be defined by swapping each even integer with its successor - that is, $\sigma(2n) = 2n+1$ and $\sigma(2n+1) = 2n$ for all $n\in \mathbb{N}$.


**Question.** Is there a normal binary sequence $f:\mathbb{N} \to \{0,1\}$ such that $f \circ \sigma$ is not normal?



Is normalcy preserved under the swapping operation?
If every string appears in $f$ consecutively, then every string appears consecutively in $f \sigma$.


(So *yes* in answer to the title question, *no* in answer to the question as phrased in the question body :)


(Thanks to Alessandro Della Corte for corrections in the comments below!)


To see this, suppose that every string appears consecutively in $f$. Let $S$ be a string of even length. Since $S 0 S$ appears consecutively in $f$, it follows that $S$ appears consecutively in $f$ starting from an even position. (We could equally have used $S1S$.)


Since every string $S$ of even length appears consecutively in $f$ starting from even position, it's also the case that $S \circ \sigma$ appears consecutively in $f$ starting from an even position. So every string $S$ of even length appears consecutively in $f \circ \sigma$ (starting from even position). As every string is a consecutive substring of a string of even length, it follows that every string appears consecutively in $f \circ \sigma$.


