Is it true that the set of all 0-1 sequence with no two consecutive 1 is uncountable

Question

My answer is yes. It is true that the set of all 0-1 sequence with no two consecutive 1 is uncountable. Because think about it in the situation of coin tossing that either head or tail appearing with the probability of 1/2. And use indicator function $1_s$, S={the tossing is head}, then it is a normal 0-1 sequence because any 0-1 word of length n appearing with a limiting frequency $1/2^{n}$ and now translate the set of normal 0-1 sequence into normal numbers in binary system, then the set of normal numbers will form a full measure in [0,1], then it is uncountable. (I don't know if this step is valid or not, the set of full measure is uncountable?)

Answer 1

You can map $\{0,1\}^{\mathbb N}$ one-to-one onto your set: for any sequence $s$, $f(s)$ is obtained from $s$ by replacing each $1$ by $1,0$. Thus e.g. $f([0,1,1,0,0,1\ldots]) = [0,1,0,1,0,0,0,1,0\ldots]$. Thus your set has the same cardinality as $\{0,1\}^{\mathbb N}$, namely $\frak c$.