Avoid common mistakes on your manuscript.
Correction to: Aequat. Math. 96 (2022), 833–841 https://doi.org/10.1007/s00010-022-00870-w
In the original article (published by P. Pasteczka) there is a mistake (observed by Árpád Száz) in the proof of Lemma 1. In fact, the statement is false in its original form, as it is demonstrated by the following counterexample: Let \( D = [ 0, + \infty [ \times [ 0, + \infty [ \) with the standard topology,
Then \( \textbf{F} \circ T (0) = \{ 0 \} \) and \( \textbf{F} \circ T (s) = [ 0, + \infty [ \) for every \( s \in ] 0, + \infty [ \,\).
However, in Theorem 1 it is sufficient to use the lemma for the restricted case when \(D=I^p\) (with the standard topology), where \(p \in \mathbb {N}\) and \(I \subset \mathbb {R}\) is a compact interval. The proper formulation should be
Let D be a compact, metrizable topological space and \(F :D \rightarrow [0,\infty )\) be a continuous function.
If \(T :[0,\infty ) \rightarrow 2^D\) is nondecreasing, right-continuous (we consider topological limit on \(2^D\)) and such that
-
each T(x) is closed;
-
\(\textbf{F} \circ T :[0,\infty ) \rightarrow 2^{[0,\infty )}\) is left-continuous.
Then \(\textbf{F} \circ T\) is constant.
Then the following modification is required in the proof:
Instead of:
However, as D is \(\sigma \)-compact, we obtain that \(D_0\) is compact.
There should be:
However \(D_0\) is compact being a closed subset of a compact set.
As the result of this improvement, the proof of Theorem 1 should start with:
Starting section of the proof: If there exist two different \(\textbf{M}\)-invariant means \(K_1,K_2 :I^p \rightarrow I\), then there exist a compact subset \(J \subseteq I\) and a vector \(v \in J^p\) such that \(K_1(v)\ne K_2(v)\). However \(K_i|_{J^p}\) is invariant with respect to \(\textbf{M}|_{J^p} :J^p \rightarrow J^p\) for \(i \in \{1,2\}\). Thus we can assume without loss of generality that I is a compact set.
Moreover in line 26, page 836, the following sentence should be added:
For the converse implication take any element \(y \in \textbf{F} \circ T (a)\). Obviously \(0 \in F \circ T(a^-)\), thus we can assume that \(y\ne 0\). Then, there exists...
More details related to this issue can be found in [5].
Reference
Boros Z., Lovas R. L., Száz Á.: Some Relational and Sequential Results, and a True Relational Modification of a False Lemma of Paweł Pasteczka on the Constancy of the Composition of Certain Set-Valued Functions. Submitted
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The authors are grateful to Árpád Száz for finding the mistake in the original note, and his inspiration to writing this correction.
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Boros, Z., L. Lovas, R. & Pasteczka, P. Correction to: There is at most one continuous invariant mean. Aequat. Math. 97, 883–885 (2023). https://doi.org/10.1007/s00010-023-00949-y
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DOI: https://doi.org/10.1007/s00010-023-00949-y