Is $\{2^n \alpha\}$ A Dense Set?

I will show that for a specific irrational $\alpha$ , the set $S$ is not dense.

Use base 2 representation $\alpha = 0.101001000100001000001\ldots_2$ , where there are $n$ 0's between the $n$ th 1 and the $(n+1)$ th 1. Because this sequence is not an eventually repeating decimal (due to the string of 0's), it is not a rational number.

The values of $\{2^n \alpha \}$ are easily read off here. This is not dense, since it clearly doesn't contain any number in the range $[0.1011_2, 0.11_2]$ .

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Note that there are some irrational numbers where the set $S$ is dense. For example, $0.|01|00011011|000001010011100101110111| \ldots _2$ , which is formed by concatenating all binary strings of length $n$ . To see why, note that we can the number $0.a_1 a_2 a_3 \ldots _2$ within $2^{-k}$ for all integers $k$ , so the number is an accumulation point of the set. Hence, $S$ is dense.