Prove there are uncountably many syndetic sets Here is a sketch of my thought of proving it but I encounter some problems. Following the thought of Cantor's diagonal argument so firstly using indicator functions $1_s$, (S is a syndetic set) translate S into 0-1 sequence. S1 responds to a 0-1sequence, S2 responds to a 0-1 sequence and goes on. Then assume there is a one-to-one relation between these syndetic set(0-1 sequences) and $\Bbb{N}$. Finally construct a new syndetic set(maybe choosing one element from each 0-1 sequence,but how?) which does not occur in the enumeration. But I just cannot figure out how to construct such 0-1 sequences. Can anyone help or have any other solution? Remark: < or in other word, syndetic set is a set which just has bounded gaps.

For each $\sigma=\langle b_n:n\in\Bbb N\rangle\in\\{0,1\\}^{\Bbb N}$ let

$$S(\sigma)=\\{2n:n\in\Bbb N\\}\cup\\{2n+1:b_n=1\\}\;;$$

clearly $S(\sigma)$ is syndetic, and the map $\sigma\mapsto S(\sigma)$ is injective. Thus, there are $2^\omega=\mathfrak{c}$ syndetic subsets of $\Bbb N$.