Exercise 2.2.8

Question

For some additional practice with nested quantifiers, consider the following invented definition:

Let's call a sequence $\left(x_{n}\right)$ zero-heavy if there exists $M \in \mathbf{N}$ such that for all $N \in \mathbf{N}$ there exists $n$ satisfying $N \leq n \leq N+M$ where $x_{n}=0$
  \begin{enumerate}[label=(\alph*)] 
  \item Is the sequence $(0,1,0,1,0,1, \ldots)$ zero heavy?
  \item If a sequence is zero-heavy does it necessarily contain an infinite number of zeros? If not, provide a counterexample.
  \item If a sequence contains an infinite number of zeros, is it necessarily zeroheavy? If not, provide a counterexample.
  \item Form the logical negation of the above definition. That is, complete the sentence: A sequence is not zero-heavy if ....
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Yes. Choose $M = 1$; since the sequence has a 0 in every two spaces, for all $N$ either $x_n = 0$ or $x_{n + 1} = 0$.
  \item Yes. If there were a finite number of zeros, with the last zero at position $K$, then choosing $N > K$ would lead to a contradiction.
  \item No, consider $(0,1,0,1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0,\dots)$ where the gap between 0's grows indefinitely. For any value of $M$, for large enough $N$ the gap between zeros will be greater than $M$. Then we simply choose $N$ so that $x_N$ is the first $1$ in a streak of at least $M + 1$ 1's.
  \item A sequence is not zero-heavy if for all $M \in \mathbf{N}$, there exists  some $N \in \mathbf{N}$ such that for all $n \in \mathbf{N}, N \le n \le N + M$, $x_n 
e 0$.
   \end{enumerate}

Exercise 2.2.8

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Exercise 2.2.8

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