If the set of factors of an infinite word $\xi$ is regular, is this property stable under "shift's" of $\xi$? Let $\xi$ be an infinite binary sequence, and denote by $T(\xi)$ the set of all factors (infixes) of $\xi$. Also if $w$ is some finite prefix of $\xi$, denote by $\xi/w$ the unique $\eta$ such that $w \cdot \eta = \xi$, i.e. the tail of $\xi$ after deleting the prefix $w$. Is it possible that for $\xi$ we have two prefixes $u,v$ of $\xi$ such that
$T(\xi/u)$ is regular, but $T(\xi/v)$ is not regular?

Closely related to this question is, if $w \in X^*$ and $\xi$ is an infinite sequence, is $T(w\cdot \xi)$ regular if $T(\xi)$ is regular, and if $w$ is a prefix, does $T(\xi)$ regular imply that $T(\xi/w)$ is regular?

EDIT: I guess I have a proof that if $T(\xi)$ is regular, then $T(\xi/w)$ is regular too, but this is somewhat contrary to what Yuval Filmus wrote, so I would be grateful if someone would point out to me what might be wrong with my proof??

Okay, we introduce the notation $L / v := \{ u : vu \in L \}$ for $L \subseteq X^*$, the so called (right) quotient or derivative of $L$. First note that if $v$ is a prefix of $\xi$ then
$$
 T(\xi/v) = T(T(\xi)/v).
$$


  Proof: If $w \in T(\xi/v)$ then we have $vxw\eta = \xi$ for some $x \in X^*$. So $vxw \in T(\xi)$ or $xw \in T(\xi) / v$, which shows $w \in T(T(\xi)/v)$. For the other inclusion let $w \in T(T(\xi)/v))$, then $xwy \in T(\xi)/v$ for some $x,y \in X^*$, and by definition $vxwy \in T(\xi)$. If $vxwy$ is a prefix of $\xi$, then $xwy$ is a prefix of $\xi/v$ and we have $xwy \in T(\xi)/v$ and hence $w \in T(\xi)/v$. If $vxwy$
  is not a prefix of $\xi$, then the part $xwy$ must occur in $\xi/v$, as it begins behind the $|v|$-th position in $\xi$, hence $xwy \in T(\xi)/v$ and so $w \in T(\xi)/v$.


Okay, now the proof that $T(\xi)$ regular implies $T(\xi/w)$ is regular.


  Proof: I use that if $L$ is regular, then its set of factors $T(L)$ is also regular (I guess this is true). Then if $T(\xi)$ is regular, then $T(\xi)/w$ is regular if $w$ is a prefix of $\xi$ as regular languages are closed under quotients (this could be found in many textbooks). Taken together
  $$
 T(\xi/w) = T(T(\xi)/w)
$$
  is regular, as it is the set of factors of the regular language $T(\xi)/w$. QED
Let $A = \{0,1\}$ be a binary alphabet and let $L = A^* -\{0,1\}^*00\{0,1\}^* = (01,1)^*(\varepsilon + 0)$. I denote by $F(w)$ the set of factors of a word $w$ and by $P(w)$ the set of its prefixes.

Let $w$ be an infinite word such that $F(1w) = L$ and $P(1w)$ is nonregular. Then $F(001w) = L \cup 00P(1w)$. Since $L$ is regular and do not intersect $00P(1w)$, $F(001w)$ is not regular. You can use Yuval's approach to give an explicit construction: take an enumeration $u_0, u_1, ...$ of $L$ and consider a word $1w = 1u_01^2u_1 \dotsm 1^nu_n \dotsm$.