Prove about sequences using König's Tree Lemma I am having problems proving the following:

A sequence $S$ only consisting of $1$'s and $0$'s is called 'non-stuttering' if there exists no segment in $S$ that repeats three times (e.g. $00101101$ is non-stuttering, but $110010010011$ is because '$100$' is repeating three times) Suppose that for any $n \in \mathbb{N}$ there exists a sequence that is non-stuttering. Prove that there exists an infinite sequence $\hat{S}$ that is non-stuttering.  

I want to prove this using König's Lemma but I am not sure how to apply it. Let $T$ be the rooted tree of all non-stuttering sequences where every node has at maximum two children ($1$ and $0$). My idea is that on every node $t_n$ there has to be a child of $t_n$ with infinitely many descendants because for every $n \in \mathbb{N}$ there excists such a sequence. But this doesn't quite add up. Because isn't there a possibility that i.e. I am looking at $t_3$ and I know there has to exist a $t_4$ somewhere. But how can I tell that it is a child of $t_3$? Couldn't it be that I chose a wrong path at $t_2$ so that I could get a non-stuttering sequence of length $3$ but not of lenght $4$?
I am quite confused at the moment and an explanation would be much appreciated!
 A: The useful thing about Konig's lemma is exactly that it gets you out of the situation you describe. You're essentially trying to build a path (= infinite non-stuttering sequence) directly; with Konig, you can avoid this entirely.
We do this as follows. The crucial fact is:

If $\sigma$ is an initial segment of $\tau$, and $\tau$ is non-stuttering, then $\sigma$ is also non-stuttering.

This means that the set of finite non-stuttering sequences of $0$s and $1$s forms a tree! Now, a non-stuttering sequence of length $n$ corresponds to a node of height $n$ on this tree - given that there is such a sequence for each $n$, what does this tell you about the cardinality of $T$? (Konig's lemma then gets you the existence of a path.)

In fact, a general principle is true: if $P$ is a property of finite or infinite binary strings which is closed under initial segments - that is, if $\sigma\prec\tau$ and $\tau$ has $P$, then $\sigma$ has $P$ - then the following are equivalent:


*

*There is a sequence of length $n$ with property $P$, for each $n\in\mathbb{N}$.

*There is an infinite sequence such that each finite initial segment has property $P$.
The direction bottom-to-top is proved by "cutting off" an infinite sequence at a given length, and the top-to-bottom direction uses Konig's lemma exactly as above.
Usually we're interested in "continuous" properties - ones which hold of an infinite string iff they hold of each of its finite initial segments. Non-stuttering is a continuous property; "finite", however, isn't. Continuity in this sense can be generalized beyond the natural numbers, by consider more complicated infinite ordinals; a property of ordinal-indexed sequences is continuous if for all limit ordinals $\lambda$, a length-$\lambda$ sequence has the property iff all its strict initial segments have that property.
