Timeline for Does conjugacy preserve the set of synchronizing blocks?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 11, 2014 at 22:14 | comment | added | Dominik Kwietniak | You are right, I only commented to clarify and better understand your statement, since your first sentence may be read as: It is never the case that... | |
| Feb 11, 2014 at 22:14 | vote | accept | Dominik Kwietniak | ||
| Feb 11, 2014 at 16:22 | comment | added | Algernon | I have shown how you could find a counter-example that refutes the statement "a $k$-block $v$ ($k\ge n$) is a synchronizing block for $X$ if and only if $\psi^{(k)}(v)$ is a synchronizing block for $Y$". For that, you only need one synchronizing shift $X$ whose synchronizing blocks are all of length at least $2$. An example would be the shift space of all sequences in $\{0,1,2\}^\mathbb{Z}$ in which between every two consecutive occurrences of $11$, there is at most one occurrence of $2$. | |
| Feb 11, 2014 at 15:19 | comment | added | Dominik Kwietniak | But it may be not true that all synchronizing blocks have length at least two - there are plenty of synchronizing shifts with a synchronizing symbol (one letter word)... Moreover, identity map is always a conjugacy and it clearly preserves the set of synchronizing words. If I understand your claim correctly, you have proved that if the minimal length of a synchronizing word is at least 2, then no $n$-block map for $n\ge2$ can preserve the set of synchronizing words. | |
| Feb 10, 2014 at 12:44 | history | answered | Algernon | CC BY-SA 3.0 |